TIGHT BINDING
BOOK DAMAGE BOOK
03
>
DO
164443
OSMANIA UNIVERSITY LIBRARY '
Call
No^-///7";>//
Accrasi(m
This book should be returned on or before the ckte
No
last
J 2,
m marls*t-bei o W
.
OXFORD UNIVERSITY PRESS AMEN HOUSE,
E.C,
4
London Edinburgh Glasgow New York Toronto Melbourne Capetown Bombay Calcutta Madras
HUMPHREY MILFORD PUBLISHER TO THE UNIVERSITY
PIRST EDITION IQI2 REPRINTED PHOTOGRAPHICALLY IN GREAT BRITAIN IN 1937 BY LOWE AND BRYDONE, PRINTERS, LONDON, FROM CORRECTED SHEETS OF THE FIRST EDITION
PREFACE THE
aim has been
book suitable to the beginner who wishes to acquire a sound knowledge of the more elementary parts of the subject, and also sufficient for the The syllabus for candidate for a mathematical scholarship. author's
Honour Moderations
at
to produce a
Oxford has been taken as a maximum
limit.
The main
principles observed in its construction are (1) to make of to the the student (2) knowledge :
utilize the previous
;
subject self-dependent; (3) to arrange the bookwork in such (4) to logical order that any portion can be readily found illustrate difficulties by worked-out problems, each selected with ;
a definite object (5) to graduate the exercises and to select only those which can be done by the preceding bookwork. ;
The
solutions of illustrative examples are not always the most elegant possible the probable capacity of the student at each ;
stage has been carefully considered.
During twelve
years' daily experience of teaching this subject the author has noted the difficulties common to students. For example, the average student has no idea of the form of '
an equation
'
thus, asked to find the equation of a line through a given point perpendicular to a he begins, Let gpven line, = mx + c be the line, then its "m"', &c., whereas, if he had y a clear understanding of form, he would readily write such an :
c
equation down. In order to give the -reader confidence in analytical methods, familiar properties are used as illustrations and well-known
wherever they arise naturally out of the analysis. Thus the circle is fully dealt with; methods and ideas are A large number of thereby illustrated earlier than usual. facts are noted
exercises are given in this part of the work so that the pupil can make the foundations sure the reader with special mathe;
A2
PREFACE
4
matical ability can omit many of these. Some of the work on the circle, especially that dealing with the circles of the triangle,
the author believes, new. tried to avoid obtaining analytical results by quoting geometrical results with which the reader may be This process makes some acquainted in Geometrical Conies. is,
The author has
pupils lose confidence in Analytical Geometry; others welcome it as a dodge enabling them to avoid a real understanding of in either case the result is bad. the principles of the subject All properties of the conic are developed by analytical processes Thus, for instance, the equation of following on definitions. ;
the axes of the coordinates,
is
general conic, either in Cartesian or Areal obtained from the simple definition of an axis
as a straight line about which the conic is symmetrical the This equation foci are subsequently shown to lie on the axes. :
deduced from those giving the foci by making the statement that the foci lie on the axes, a statemejit which, most probably, the reader would fail to justify except by an appeal to Geometrical Conies. Briefly, this book attempts to answer the question, What do the general equations of the first and second degree represent ?^ of the axes
is
not
'
1
What
equations represent certain known curves ? The chapter on the circle, however, comes before the general discussion of the equation of the second degree; the purpose rather than,
(
'
being to familiarize the student with the work before the more serious attack, and to cater for those examinations their syllabus to the line and circle.
A few details may be noted
:
abridged notation
is
which limit
insisted
upon
as probably the best introduction to quite general coordinates.
author's treatment of the parabola Vax + \/6//= 1 is original and will, he hopes, commend itself to teachers who have realized the difficulties boys find with the usual work. Parametric co-
The
ordinates are given their rightful prominence. In the first draft of this book point and line coordinates were treated concurrently :
convinced, however, of the relative importance of the former, the author changed his scheme it is hoped that the introductory :
chapter on line coordinates will prove useful. Special care has been given to the introduction of imaginary points, points at at infinity. The last chapter, is devoted infinity, and the line
PREFACE
5
and here tangential equations are freely of the many proofs given are new. author's first and unlimited thanks are to Mr. A. E. Jollifte,
to Areal coordinates,
used
;
The
M.A., Fellow and Tutor of Corpus Christi College, Oxford whenever a difficulty, either of arrangement or of method, has arisen he has given most helpful advice, and it is largely due to his aid :
and encouragement that
this work has been completed. read through and thoroughly criticized both the manuscript and the proofs. The author is entirely responsible for the form and accuracy of the work, but it is right to state that Mr. Jolliffe most generously placed a quantity of his own
Mr.
Jolliffe also
work
at the author's disposal thus the practical methods of drawing conies and some of the best paragraphs in the later ;
chapters are adapted from his manuscript. Chapter IV was submitted this part inserted at his suggestion, and he kindly of the work to other mathematical authorities for their criticism.
H. Wykes, M.A., spent much time and care reading the manuscript, and his suggestions were often adopted. Miss Isabella Thwaites, scholar of Girton College, Cambridge, and Mr. W. E. Paterson, M.A., have kindly read the proofs. The author is glad also to recognize the unfailing courtesy and kindness extended to him by the Clarendon Press. The author hopes he has produced a book that will not only Mr.
P.
make
the subject interesting to schoolboys, but will be a valuable companion to which later on the undergraduate will often refer
and from which he
will not readily part.
A. C.
BRADFORD, 1912.
J.
CONTENTS PAGE
CHAPTER
I
THE POINT
9
CHAPTER THE EQUATION OF THE FIRST DEGREE
CHAPTER EQUATIONS OF HIGHER DEGREES.
II
.....
28
III
CHANGE OF AXES
.
84
.
CHAPTER IV ANALYTICAL NOTATION, A REVISION AND EXTENSION
.116
.
CHAPTER V THE CIRCLE
125
CHAPTER THE GENERAL EQUATION
VI
OF THE SECOND DEGREE
CHAPTER
.
.
.
VII
THE PARABOLA
257
CHAPTER CENTRAL CONICS.
222
VIII
THE ELLIPSE AND THE HYPERBOLA
.
.
303
CHAPTER IX 378
POLAR COORDINATES
CHAPTER X LINE COORDINATES AND TANGENTIAL EQUATIONS
.
.
.
390
CHAPTER XI MISCELLANEOUS THEOREMS
.......
CHAPTER
430
XII
TRILINEAR AND AREAL COORDINATES
452
ANSWERS
527
INDEX
..........
545
CHAPTER
1
THE POINT THE method
1.
processes (a)
(I)
of algebraical analysis involves three distinct
:
The conditions
problem are represented by and algebraical expressions equations. The processes of algebra are applied to these expressions and of a geometrical
equations to obtain (c)
new
results.
These new results are translated back into geometrical guage. object of the
The
perform readily the
bookwork given
first
is
lan-
to enable the student to
and third operations.
The tendency
in this
lose sight of the geometrical significance: the student subject should take the greatest pains to acquire the habit of connecting is to
every algebraical detail with its geometrical interpretation. One of the most elementary relations between Geometry and
The number the expression for the area of a rectangle. of square units in the area of a rectangle, whose sides are a and b units of length respectively, is the product ab. The algebraical ex-
Algebra
is
pression ab may thus be said to represent the geometrical quantity, the area of a rectangle.
Algebraical proofs of the propositions in Euclid, Bk. II, are based idea. The logic of the process is here illustrated.
on this
j
// a
whole ivith
straight line be divided into
line is
equal
to the
sum
any two parts
of the squares
the
square on the
on the two parts, together
twice the rectangle contained by the tivo parts.
THE POINT
10
AB
is the straight line divided at C, let AC, CB be a and b units of length respectively. is represented algebraically by (a) The area of the square on the product (a + b) x (a + b).
If
AB
(6)
Now
(c)
But
(
=
a + 6)x(a + 6)
(a
+ 6) 2
this result represents the square
the rectangle contained by the square on is equal to
AB
Hence The algebraical *
tion
thus
;
on
A C and
CB + twice
if
',
AC+ the
square on
CB.
&c.
idea of sign is represented geometrically is the length of CB measured from C to
by
B
B
length measured from
to
A
C is
A
&_--C
b
b (")
(i)
Thus in Fig. (i) AB is of length (a + 6), in The absence of this idea in Euclid accounts
Fig.
of length (a b). of pairs
(ii)
number
for the
of propositions which are algebraically equivalent II. 12 and II. 13. II. 5 and II. 6 II. 7 ;
the
b.
B
C
direc-
+ 6,
:
e. g.
II.
4 and
;
For instance, in the example given above, if Using Fig. (ii) proposition II. 7 is derived. (a) The square on AB is represented by (a~b) (a = a 2 + b*-2ab. (b) (a~b)~ (c) This represents the sum of the squares on
b
were negative *
7>).
AC
and
BC
less
AC
twice tho rectangle contained by and BC. The drawing of graphs is a further useful step towards connecting the subjects of Geometry and Algebra. Graph drawing is now taught in all schools, of the process. 2.
and we assume that the reader has some knowledge
Cartesian Coordinates.
Rectangular and Oblique Axes.
P
The
in a plane is indicated by its distances position of a point measured in fixed directions from two chosen intersecting straight lines Ox, Oy (the coordinate axes or axes of reference) ; the axes are called rectangular
oblique.
In both
the lengths
when
Oy are perpendicular, otherwise PN be drawn parallel to Oy (the axis of y), called the x and y coordinates. Thus, if Ox,
cases, if
ON,NP
ON = h and NP =
ft,
are
P is
the point
(h, k).
The same convention with regard nometry: lines
made as in trigomeasured upwards (NP) are positive, downwards (N'P') to sign is
THE POINT
11
and again
negative,
to the right (ON) positive, to the left (ON') a and the point IJ/ is ( a, 1) where ON'
negative:
e.
N'P'~
The angle xOy
I.
g.
usually called
is
to.
P(h,k)
N
1
N
The
its ordinate. a,
b
'
is called
^-coordinate of a point
as
'
We
shall refer to
the point P(a,
l>)
'
its abscissa,
the point
P
the ^/-coordinate
whose coordinates
are
'.
Note. In the majority of pioblems it is more convenient to use rectangular axes, especially when the lengths of lines or the magnitude of for angles (i. e. metrical properties) are involved, because the expressions these quantities are much simpler when w however, that the solution of a problem
is
a right angle
:
it
may happen,
much
simplified in other due to the, more inconvenience that the axes use of the oblique by
ways clumsy formulae
is
so
is outweighed consequently the student should make himself familiar with the formulae in the more general case. It may be well to note here that as a rule the student has free choice of axes of ;
the first step is to decide what lines in the reference in any problem axes of reference, the only restriction convenient the most will make figure a variable line must not be chosen for an that they must be fixed ;
being axis nor a variable point for origin. ;
Polar Coordinates.
3.
Ofpok)
P
The (i) (ii)
is indicated by position of a point a fixed point 0, called the pole, from its distance OP (r)
the angle
(0)
axis OZ.
which
OP (the
radius vector) makes with a fixed
THE POINT
12
When
the position of P, the radius vector
is positive, to find
in a direction from the position OZ and revolve about the of a clock the to that of the hands opposite through angle if distance r, if positive, is measured along this radius from It should be noted that with negative, in the opposite direction.
must
start
;
this convention of sign the points indicated (
TT
r,
4.
+ 0),
by
(r,
0),
(
?rH-0),
r,
27r) are identical.
(ry
The polar coordinates of a point referred
to
the
line
OZ
are
connected by simple relations with the Cartesian coordinates referred to rectangular axes through 0, along and perpendicular to OZ.
(xy
Let
P be
then
the point x
Conversely
(r,
0)
= ON = r = v/a^-f
or
(,r,
r cos
T/'
;
y) ;
:
y
= ^VP =
=
and
r sin -
tan" 1
3x
Thus with these #
=
lines of reference the graphs of
2?/
1
and
r cos
=
2r sin
1
are identical.
Examples 1.
I
a.
With rectangular axes mark the positions of the points (2, 1*5), (0, 3), -3) (-2, 4-5). Note graphically that they are collinear. What
(4, 0), (8,
equation do they all satisfy ? 2. With axes inclined at 12(T, note the positions of the points (6, 1),
(-2, -3),
(5, 4), (2,
-I) (-8,
Which of these points are collinear ? 3. Mark the positions of the following (.
and
M
(0, 0),
4).
(a, fir), (a,
points
:
-Jir), (-a,
- jrr),
find their Cartesian coordinates referred to rectangular axes
through
the pole, one of which coincides with the polar axis. 4. Find the polar coordinates, referred to Ox, of the following points, whose Cartesian coordinates referred to rectangular axes are (3, 4), (5, 5), (2
A
:
2)(-2v/3,
2),
(-2-A
~2), (2V3, -2).
THE POINT The make
18
sides of a square are 4 units long choose coordinate axes so as the coordinates of the corners as symmetrical as possible, and state their coordinates. 5.
to
6.
The
;
sides of a parallelogram are 4 and 6 units of length its corners referred to suitable axes.
;
find the
;
find the
coordinates of
7. The diagonals of a rhombus are 2 and 4 units of length coordinates of its corners referred to suitable rectangular axes.
8.
Draw
9.
(vi)
Draw graphs
= JTT,
(iv)
(v)
=
x+y of
x+a
=
0,
=
0,
x
(i)
=
cosec B
are
6.
To
=
a2
(ii)
;
tan
(a, CX), its
the pole,
t/,
(iv)
Cartesian coordinates
the #-axis
making an
prove
:
y
y*
=
0.
to rectangular axes through angle 30 with the polar axis, are (x y)
x1 4
x
=
referred
(i)
3, (iii)
=
The polar coordinates of a point
10.
=
y
(ii)
0,
(i)
2r+ 5
= 4,
referred to (a) rectangular, (b) oblique axes. r cos & +2 0, (iii) 3r-sec 1, (ii) r sin
the graphs of
4#,
y
(v)
a =
(y 1/8
+ x)/(x*/$ - y}.
find the distance between two points whose coordinates are
given. I.
Cartesian Coordinates, Oblique axes.
M
Let the points be P (;r t to 0>/, and QL parallel to
LPLQ ==
then
y^ and Q
,
TT-
PQ 2 =
/.
or
J'Q
Note
i.
=
When
(^
- ^8
2 )
,
y.,).
Draw 7W,
0,r,
^/,
;
=
iP = NP-MQ = y therefore in the triangle
(
l
0.V- OTJf =
-y>>
.r
t
-a?2
.
;
PLQ, PQ* = iP + QL*-2LP. QL cos PZ
i
2
+ (yt-y 2 2 + 2 )
(^1-^2) tVi-^a) cos
the axes are rectangular, since
=
\TT,
)
THE POINT
14 ii. If the point P (#, then the coordinates of
Note (a,
fc),
--~y) is
always at a distance c from a fixed point satisfy the equation
P always
f-b) cos
co a>
=
2 2 a (oblique axes) and (a?-a) -f (t/-6) =c (rectangular axes). The equations therefore represent a circle whose centre is the point (a, ft), and whose
radius II.
is c.
Polar Coordinates.
Let the points be P(rx and from the triangle P0$ ,
X)
and Q
(r 2
,
2 ),
then the Z.POQ
PQ 2 - r^ + r^-gr^ cos (,-5 2 PQ = v/ r 2 + ra - 2 ^ r2 cos (^ -
l
2>
c
from
),
or
{
Note. If the point (a, a), then
the point
(r, 6)
its
r2
t
moves
so as to be
fl
2 )}
.
always a distance
coordinates satisfy the equation
+ a 2 -2ra
cos (0-a)
=
o
2 ,
which equation therefore represents in polar coordinates the whose centre is (a, a) and whose radius is c. 6.
To ywd Me coordinates of a point which
bettveen tivo given points in
a given
circle
divides the distance
ratio.
y
N
The method and axes.
result are the
same
M
for rectangular
and oblique
THE POINT Let the given points be P(x y } and Q(^/ 2 ), and m. (x, ?/) divide PQ internally in the ratio 1 l
l
E
15 let
the point
:
UN parallel PR _ LN
Now
draw PL, QM,
then
.
l
EQ~~NM' X
By drawing that
=
?/
Hence
Note Note
E is the
poin t (v
The mid-point
.
l
-
~~~:
through P,
parallels
l_ _ x-x^ '*'m~~x~-x'
= mxmx
Ix
te2
.*.
to Oy,
??/ 2
E
Q, -f
we
to Ox,
mi/t
:
^
.
/~hm
-
l -'/
/-f
*
w
can show similarly ,
A
*-
-
l-\
m
f
)
'
[Jfo + tfj), J(yi + y 2 )]. outside PQ, i.e. when it divides PQ externally in the ratio //m, this ratio must be considered negative, and m written negative in the result.^ For in the ratio PR/RQ the length RQ is i.
ii.
\When
of
PQ
the point
is
J? lies
measured in the opposite direction
Note
to the length PI?.
D divide
AB
a straight line internally and externally in the same ratio, the points CD are said to be Harmonic conjualso the four points A, B, C, are said to form a Harmonic gates of A, Jj it Note follows that the From (ii) Range. points If the points C,
iii.
D
:
harmonic conjugates of the points (x l9 form a harmonic range.
are
Note Q
(#2
iv.
#2)
^
l
:
w
'
we nave shown
1 is
(# 2
,
t/ 2 ),
and these four points
If (x, y] is a point dividing the distance
n * ne ra ^
wi
This
y^),
-
r
,r
-a?
the line PQ. Therefore, provided (x, y) its coordinates satisfy the equation
x (y
1
-y^-y
(x l
P
(x ly
t/j),
~^ ^y"^ a
true for all values of the ratio
or
between
that
y a -y" /
is
:
ni
provided the point
(x, y) is
on
P
Q,
on the .straight line joining
- x2 ) + x^ - x^
=
;
:
then the condition that the point (*, y) should lie on the equation 'and is called the equation of the straight line PQ. line ;P
this
is
system of coordinates
may
be in any special case.
THE POINT
16
Illustrative
Ex.
TJie base
1.
a fixed straight
line
:
of a triangle
Examples.
is
find the locus
of
fixed its
ami
vertex moves alomi
its
centroid.
Take the
fixed base produced as axis of x and the given fixed line as axis Let the coordinates of the fixed points J9, C be (/, 0), (wi, 0), and sup-
of y.
pose the variable vertex is (0, X). The coordinates of E, the mid point of BC, are {| (Z-f m), 0}. The coordinates of G which divides in the ratio 2 1 are {-J
AE
Hence
:
for all positions of A, the ^-coordinate of
=
to J (J-f w), i.e. Gr lies on the line* x is, i.e. to the given fixed straight line.
Ex.
Show
2.
on a straight conjugate of If
Q
that
line,
Q
the
o^ the line PR,
lies
its
7
the
(i)
P
to
ivith respect
ratio
/
Also i. ft.
Q
+
lies
~-
i.e.
*
:
QIt<
.
is
parallel to the
(ii)
7? (7, the
1)
lie
harmonic
coordinates must be of the form 1
4 2
l
+ tn
on PR, and Z\>
-3
its
;
-
-2, then 3/ :
The harmonic conjugate ratio 2
PQ
4 w), J X]
m
I '
4m} m )'
/ 4-
in
-
if
+ wi), which
:
(1
constant and equal
and R.
(
}
is
P(2, -4), Q(4, -2),
points
and find
(
G
QR = of
()
2w/
2
:
;
o.
with respect to /'and
It
divides
PR
in the
coordinates are there fuio
(-8, -14}. The abbreviation *the
equation
is
.
.
.'.
line
.
.
.'
is
used throughout for 'the lino whose
THE POINT
17
Examples Ib. (The axes are rectangular unless otherwise stated.)
Find the coordinates of the centroid of the triangle whose vertices
1.
are (1,5), (-3, 7), (5, -9). 2. Show that the points
(2, 4),
(2+
y% 5),
(2, 6)
are the vertices of an
equilateral triangle.
that the point {2 a cos 2 J0, 2#cos 2 (jTr-f i#)} distance from the point (a, a) for all values of 6. 4. The distance of a point (x, y) from (12, 8) is double
Show
3.
is
at a constant
its distance from must ? the coordinates x, y satisfy equation (3, 2). 5. Find an equation satisfied by the coordiuates of all points distant 5 units from (4, 3). 6. Find the ratio in which the point (2, 3) divides the distance between the points .4 ( 7, 1J), B 14, 5), and find its harmonic conjugate with respect to A and B. 7. Find the coordinates of the mid-point of the line joining (4, 3) and Where is its harmonic conjugate with respect to these points 2, 1).
What
(
(
situated ?
Straight lines are drawn from a fixed point to meet a fixed straight Show that their points of trisection lie on one of two fixed straight
8. line.
lines. 9.
(1, 53) lies on the line joining divide this distance ?
Shfcw that the point
In what ratio does
What
is
it
(5, 3)
and (-2,
the harmonic conjugate of this point with respect to
7).
the
other two ? 10.
and
Find the
sides of the triangle
whose
vertices are (1, 3), (3,
1), (6, 4),
its
greatest angle. 11. Find the coordinates of the centre of gravity of five equal particles situated at the points (2, 4), (-1,7), (8, 11), (-8, -5), (4, 8). 12. Find the distance of the mid-point of the line joining (a, ft), (b, a) from the origin. 13. Prove that the points (1, 1), (4, 4), (4, 8), (1, 5) are the corners of
and find the lengths of its diagonals. Prove algebraically that the joins of the mid-points of the sides of a quadrilateral form a parallelogram. (Take the diagonals for axes.) 15. Find the condition that the points (o lf bj, (# 2 & 2 ) should lie on a
a parallelogram, 14.
,
straight line through the origin. (Oblique coordinates.) 16. Find the polar coordinates of the six vertices of a regular (side a) taking one vertex and side as pole and initial line. 17. Points X, Fare taken in the side duced respectively so that BX: XC =
XAYiB 1267
a right angle.
R
hexagon
BC of a triangle and in BC BY: YC ~ AB AC. Show :
prothat
THE POINT
18
To find
7.
the area
of any rectilineal figure, given the coordinates
its vertices.
of
I.
If A,
B
be the points (X L
L
Let LAOjc =
1?
BOx = 0,2
area
y L ), (x 2J y 2 ),
M
1
;
Then
perpendicular to Ox.
,
OAB.
find the area
M
1
EL parallel to LAOB = 2 f
draw AL, since
to
Oy
;
A M, BM'
A
OAB = \ OA OB sin (e^-O^ = V 05 sin 2 04 cos 0, - OA = J {BM'.OM-AM. OJf' .
.
{
COS w
sin
X
.
~
OB cos COS
a}
W
or in determinant notation
area
OAB =
1 sin
<
Note. So many results in analytical geometry are simplified by the determinant notation that the student is recommended to acquaint himsufficient knowledge of self with it before proceeding with the subject two or three hours. in determinants for this purpose can be acquired XXXI11. Vide Hall and Knight's Higher Algebra, Chap. Cor. If the points 0, A, B are collinear the triangle OAB has zero area :
:
hence the condition that the points is
xy l -x l y^
(x lt y x )
(0, Oj, (x, y),
should be collinear
0.
(x, y) of any point and conversely any point ft) (0, 0) Hence this equation lies on this straight line.
This condition must be satisfied by the coordinates
on the straight line joining
whose coordinates
xy\- x\y =0 and (xv ft).
is
satisfy
and (x v
:
called the equation of the straight line joining the origin
In the above method of finding the area of the triangle OAB no question of sign arises, because the angle a is taken greater than Ol in the figure. But when the coordinates of two points A, B are given in a general form
THE POINT
19
we do not know which of the angles A Ox9 BOx is (a? 2 ya ), the greater, and the result for the area will differ in sign according to which angle is chosen the greater in the figure.
such as (xlt yj
,
We adopt the following When the
first
points
convention
:
the expression for the area of the triangle OAB is written down, term contains the ^-coordinate of A. Thus A and B being the
(*,,
t/i),
(#2 y a ), ,
OAB = OB A =
w (tf^-tf^i) o> (x^ l a;^) and consequently area <X4# = -area
area
urea
T
Let as
Me
2o
II.
area
ABC be the
drawn
in
^ sin
-J
5
sin
;
o/ Me
whose vertices are
triangle
points join OA, OB, OC, then remembering that the figure the area OAG is positive and therefore ;
OCA negative, A ABC =. A OAB+ AOJSC-f A OCA = i sin co (x^x^y^) + (x y^
area
2
{
j
of in determinant notation y\
sin
co
Note. This expression gives the area positively if, as we go round the triangle in the order A B, C, the rotation is positive in accordance with the convention used in trigonometry, and vice versa. y
Cor.
i.
When
the axes are rectangular the area of the triangle whose
vertices are (x l9 y^, (#a
,
y,), (ar s
,
t/ 3 )
is
B2
THE POINT
20 Cor.
If
ii.
(a;, //)
any point P on the straight line then the area of the triangle PQR is
are the coordinates of
joining the points Q(x l9
i/J,
R(xz
,
f/ 2 ),
Hence
zero.
x y
1
= x (y l - y 2 ) - y (#1 - a-2 )
or
This
is
4-
0,
x^ - ar^y = ,
0.
therefore the condition that the point (#, y) should lie on the in other words, it is the equation of this
straight line joining (xv yj (x2 y a ) ,
straight line.
(Of.
7,
;
Cor.)
II L The method applies similarly to the area number of sides. Thus a pentagon, for example
of a figure of
any
:
Area
=
$
ABODE = &OAB+ AOBC+ A 0<7D+ AODE+ &OEA.
sincoj^!^
^y^ + ^a
a?
3
y 2) + (
3 ^4
a;
4
yj + (o5^
IV. The case when the polar coordinates (r^ 6J, (r,2 ^ 2 ), (r3 3 ) of theT the vertices are given is left as an exercise for the student consequent formula is of no value particular cases should be worked ,
,
:
:
out from
first principles.
We
have seen that the coordinates of a point lying on the straight line joining two given points satisfy a certain equation, and 8.
conversely that all points whose coordinates satisfy the equation lie on the straight line. This straight line is called the locus of points
which satisfy the equation. .Any equation containing the variables x and y represents a locus. To every value of x definite corresponding values of y can be found, one value if y occurs only in the first power, two values if y occur in the second power. Thus the coordinates of any number of points can be found which can be indicated
THE POINT
21
on a graph. If r is made to increase continuously, the corresponding values of y will as a rule change continuously a line can be drawn ;
which passes through
these points, and the curve so drawn is the locus of the called point whose coordinates satisfy the given all
In the present work all forms of equations of the first and second degree are examined, and the properties of the corre-
equation.
sponding loci found. At present, when the student is asked in examples to find the equation of the locus' of a point under given the student is conditions, the equation is all the answer required '
;
not of course at this stage expected to recognize the locus which the its form can be roughly found by drawing its equation represents also note that when the coordinates of a point are may graph. :
We
given in some special /orw, this is equivalent to giving an equation Thus the point (ucostf, a sin 0), 1>y the coordinates (.*', y\ 1 2 2 for different values of 0, is on the locus -f y' = a
satisfied
->
.
Again, points whose coordinates are
=
the locus ry c2 by the word form ,
'
of the form (cA, c A) lie on whatever value A may have. The idea conveyed
'
is
of the highest importance.
Illustrative
Ex. lateral
1. The joins of the mid-points of the opposite sides of a quadriand the join of the wid-jioints of its diagonals Used each other.
Choice of axes.
It
an advantage in four vertices by
is
this case to indicate the
quite general coordinates: the quadrilateral and /', Q, points of the sides, and (*'?>
!As)i
Examples.
C r4
tor. if It,
AliCD
are the midt/J,
(.r,,
the vertices
/At)
S
(> 2
A BCD,
,
then,
having found the coordinates of the in idpoint of PR, those of QS follow by changing the suffixes
1, 2, 3,
4 in cyclic order.
The coordinates of Pand I
\
0*i + * 2 )>
R
are
I (in
+ 2/2)}
,
{
y
Therefore those of the mid-point of Pit are
U fo
-f
#2 + :r3 -f x<\
The same result is therefore true for the mid-point The cooidinates of L and M are similarly Therefore those of the mid-point of {
} (xl
hence the mid-points of
+ arf + a?3 -f a?4 ), P/2, ^.V,
of QS.
LM MG -}
f
/A
-I
y2
-f
y/ s -f //
LM are identical.
THE POINT
22
Ex.
Two
2.
the area
the mid-point
Take OP, (0,
J>).
sides
OP,
OR
of the quadrilateral
of a quadrilateral
is constant,
OPQR
find the equation
if
:
ofOQ.
as axes of .r and y and let P be 'the point coordinates of the mid-point of OQ be (#,
OR
If the
arc fixed
of the locus of
(a, 0), 7? ?/),
sin o>. Area OPQ = \ sin w 2y // = = fc# 2 bx sin o>. o> Area 0
the point Q are
those of
.
.
Ex.
P is
is
A BCD w
a parallelogram whose diagonals meet at 2 2 2 2 + J5C2 + 4 any point, prove that PA 1 P7? + PC + P7) '= 3.
AW
:
if
PO 2
.
We
of the use give two solutions of this question to illustrate the effect of different coordinate axes Take the diagonals for axes of reference since the diagonals of :
;
(i)
bisect
a parallelogram (a, 0), (0,
(-a,
fc),
0), (0,
each other, the coordinates of the vertices are let P be the point (x, y\ then -fc) ;
PC = (x + 2
P^
=
4
2
-f
PJ5 2 -f PC*2 2
(,r
2
-f ;/
+
-f
PD*
2a^/ cos
=
) -f
2
r/)
-f ?/
2
-f
2/
(ir -f a]
I
cos w, 2
2
4(x*+if + 2
) 4 2 (a 4- fc ). 2 2 (a -f ^ 4 2afc cos ca) -f (a* + V-2al cos
.THE POINT Take
(ii)
lines
>erpendicular to fives for
23
through the intersection of the diagonals parallel and a side respectively as axes. The symmetry of the figure
the vertices^/*, k\ (-/,
k), (-fc,
-fc),
(Z,
-fc)
C(-h r k)
Examples
I
c.
(The axes are rectangular unless otherwise
stated.)
Find the coordinates of the points of trisection of the straight line joining the point P(l, 2) to
The
2, 4), (3, 1) is divided into five straight line joining the points ( Find of division. coordinates of the the equal parts. points 3. The coordinates of three points P, Q, R are (1, 1), (3, 5), and (G, 11). 2.
that Q is a point lying between P and R on the straight line PR. Find also the coordinates of a point S on the straight line between P and E such that PQ = 3 SR. 4. Find the coordinates of the six vertices of a regular hexagon referred to rectangular axes through its centre, one of the axes lying along a diagonal. 5. Squares are described on the sides AB, AC, BC of a triangle rightangled at A. Find the coordinates of the corners of these squares and the
Show
points of intersection of their diagonals
AB 6.
c,
when AB,
AC are axes
of reference.
b.
Find the polar coordinates of the
when 7.
AC =
six corners of a regular hexagon, the pole and a diagonal the initial line. the area of a triangle whose vertices are the points (5, 7),
the centre
Find
is
(-2, -1),(0,8). 8.
Find the area of the quadrilateral whose vertices are
(-14,
6),
(-5, -7).
(4, 5), (2,
-6),
THE POINT
24
Find the cosine of the angle which the straight line joining the points b\ (a', I') subtends at the origin.
9. (a,
10.
The coordinate axes being inclined
at 60, prove that (a, 0), (0, 2r<), an equilateral triangle. 11. The centre of a circle is (3, 5), one end of a diameter is (7, 3) what are the coordinates of the other end of this diameter ? Find the radius. 12. Prove analytically that the three straight lines joining the vertices of (2 a, a] are the corners of
;
a triangle to the mid-points of the opposite sides have one point of trisection in
common.
Find the coordinates of a point which is equidistant from (a, b) and (2, Z>), and whose distance from the origin is -, 14. Find the coordinates of a point which is equidistant from (0, 0), (0, a), and (3 a, 4 a). 13.
15.
Draw
the graphs of the equations
(i)
distances of the point (8, 8) from them. 1G. Prove that the point (acos#, asinO)
x is
y
5, (ii)
at the
=
-7, and find the
same distance from the
origin whatever value 8 may have. Show that the point (a-\ ?'cos$, b + rsinO) lies on a fixed circle whose centre is (a, b) whatever value may have.
What
equation
is
by coordinates of
satisfied
this form, a,
7>,
and
r
being
fixed?
Show that the
17. is
the same for
Draw
18.
all
distance between the points {-f5,& + 9], {a 4 and b. Find the distance.
the graphs of
xy
^xy + if + rcoa(0 + JTT) = x^
(i)
(ii)
19.
2,
ft-f
5]
values of a
;
2.
The coordinates of three points AJ^ Care
(8, 6), (7, 7), (0,6);
show that
they are equidistant from the point (4, B). Find an equation satisfied by the coordinates of any point on the circle ABC. 2 20. Find the distance of the point (a tan 0, 20tan0) from (, 0), and from = x 0. the locus represented by 4 a What equation will the coordinates in this form satisfy, a being a constant ?
Find the ratio of the distances between the pairs of points (ii) (r/costf, fcsintf), {(ct -Iff )<*>*$, (i) (rtsintf, 6cos0),(0, 0) 22. Express algebraically that the point (a-y) is equidistant from 21.
;
and
l-
(a, b)
a, b).
(
What
23.
angle does the line joining (\/3, a) and
b
(?>,
the origin ?
Write down an equation
24.
moves lines
so that the difference of its
is
by the coordinates of a point which distances from two intersecting straight
satisfied
constant.
(Take the lines as axes the locus.)
:
if
they are inclined at 30 trace the graph of
25. The polar coordinates of two points are (rlf 0^), through the pole bisecting the angle which they subtend their join in P. Find the polar coordinates of P.
26.
The distances
of the collinear points P,
/?,
(r2
,
tt
2)
:
the line
at the pole
Q from
meets
the origin form
THE POINT a harmonic
If the coordinates of
series.
Prove that
/?.
P,
Q
/*,
Q
25 are
(
0), (b> 0),
find those of
OR
internally and externally in the same ratio. 2 2 points whose coordinates are (aw 2am), (am~
divide
P
and Q are two and S is the point (a, 0). Prove (i) that PSQ arc collinear, is constant for all values of MI. (ii) 1/XP-f \/$Q 28. Choose the most convenient axes to represent the veitices of an equilateral triangle AB(' and find the coordinates of the centre (P) of its 27.
20w?~
,
,
l
),
t
circumcircle.
If
Q
is
any point
in the plane, find the ratio of
29. Find the area of the triangle whose vertices are (a, 70), (2 a, 40) and the origin. Also of the triangle whose vortices are (a, 10"), (3, 40), (5a, 100). 30. Show that the points (a, a), (fc, Av/) subtend a right angle at the Prove also that the triangle whose vertices are (3, origin. 2), ( 5, + 4), (9, G) is
right-angled.
+ atan 2 0, 2 a tan 0) is equally distant from the ( and axis the of point (2fl, 0), y for all values of 0. 32. Show that the middle points of the non-parallel sides of a trapezium and the middle points of its diagonals lie on a straight line parallel to the Prove that the point
31.
parallel sides.
Find the equation of tho locus of a point P which moves so that the tin* rectangle formed by the axes and the perpendiculars from P
33.
area of to
them
is
of constant area.
Prove that the distance of the point 34.^ 2 b2 = V2 (tic, 0) is a 4 ac cos #, if a
(a cos #, I sin 6}
from the point
.
Find the condition that the coordinates of the middle point of the
35.
3c. straight line joining (a, />), (26, 2 a) should satisfy the equation 2x + 30. Find the equation of the locus of the middle point of straight lines drawn from a given point to any point in a given straight line.
2y=
In any triangle 2 2
37.
that
A B + A G* =
ABC,
if
AD + 2 BD 2
D
is
the mid-point of BC, prove analytically
2 .
Find the coordinates of the centroid of the triangle t^), (.r 2 y 2 ), (.r,, y.<), and hence prove
38.
vertices are (#,,
ABC
whose
,
AB* + ]*
3
(AG* +
<7
2
-f
CG
2
).
39. The sum of the squares on the diagonals of any quadrilateral is double the sum of the squares on the lines joining the middle points of the
opposite sides.
The diagonals of a trapezium are drawn and the mid-points of the show analytically that there is a point common parallel sides are joined to these straight lines which divides each of them in the same ratio. (Take the third line and one parallel side for axes.) in the points 41. A variable line cuts any two lines intersecting at 40.
:
P
y
(J
so that
OP+ OQ =
21.
Find the equation of the locus of
its
middle
point. 42. If P,
Q
are two points (*
y,), (.r^
y 9 ), and PL,
QM
be drawn parallel
THE POINT
26 to the axis of (ii)
t/,
find the area of
PQLM when
(i)
the axes are rectangular,
oblique.
R are the points fo, t/j), (#2 t/ 2 ), (#8 ys ), and PL, QM, 1LY are Find the areas PLMQ, RNMQ, drawn perpendicular to the axis of x. of the triangle PQR. for area an the and deduce PLNR, expression 44. Show that the three points (-1, -2), (4, 1), (9,4) lie on a straight line, and find the ratio of the distances of the second from the other two. 43. P, Q,
,
Find the condition that the point
45.
joining (8,4) (-5,
,
should
(x, y)
lie
on the straight
line
2).
ABCD is
a rectangle, P any point in the plane. PA* Prove + PC* = PB2 + P/)a 47. The coordinates of two points Z>,
.
Find the coordinates of a point
R
such that
RQ = %QP
PR/RQ. 48. The points C, D divide the line AB harmonically. Show that (i) the lengths AC, AB, AD are in //.P; is the mid-point of AB, OA* = 00 (ii) if (iii) 2/C/)= l/BD+1/AD. 49. The coordinates of four points aro (1, 3), (3, 5), (4, 6), .
(i)
(ii) (iii)
and the
OD
ratio
;
(7, 9).
Prove they are collinear. Prove they form a harmonic range. In what ratio
is
the lino joining the
first
and third divided
l>y
the
second ?
Find the equations of the straight
(iv)
linos joining the origin* to the
points. (v) Find the coordinates of the points on oach of these straight lines whose abscissae are 10, and prove that these points form another harmonic
range. 50. (
r
i
Find the condition that the three points whose polar coordinates are (*a ^a)> ( rs ^s) should be collinear.
^i)
51.
Find the ratio of the areas of
t,wo triangles
whoso angular points are
2
(i)
(ii)
(a\ ,2a\), {aX/u, a
52. Express algebraically that the distance of the point (x, y) is
What
is
(3,
4)
the graph of the equation ?
A, B are fixed points on the axes find the equation of the locus of point P such that 2 OP = AP* 4 BP given that OA = a, OB = b. 54. Two sides of a quadrilateral OP, OR are fixed, and the diagonal OQ is 53.
a
from
5 units.
:
2
2
,
of constant length c: find the equation of the locus of the middle point of the join of the mid-points of the diagonals, given that OP a, OR = b. 55. Two vertices of a triangle are (x l j/i), (# 2 y 2 ) and the centroid is (x, y)
=
,
find the coordinates of 56.
Find two points
A(xlt yj,
B(#t, y8 )
:
if
,
:
the other vertex. C, D which are harmonic conjugates of the points perpendiculars are drawn from the four points to the
THE POINT
27
feet of the perpendiculars form a harmonic a?, prove the range. general propeity can you deduce from this result ? show that D, a point of trisection of AC 57. ABCD is a parallelogram
axis of
What
:
and the mid-point of AB, are
The
58.
vertices
&sin0-f $7r),
of
(acostf-f
$TT,
a
collinear.
triangle frsin
=
0+
are TT).
(acos#, (i)
Prove
5sin0), its
centroid
is
the origin.
show that the triangle is equilateral. Find its area, (iii) If a (ii) 21 has its ends on the axes of reference: 59. A straight line of length find the equation of the locus of its middle point.
Draw
the graph when Z= 3. point P has coordinates which satisfy the equation distance from the origin is r find the angle POx.
A
60.
and
b
its
x+ v/ 3y-~2r=0,
:
Find the distance between two points whose polar coordinates are
61.
(6,iir),(-3, ITT). 62. P is a point inside a parallelogram ABCD such that the area double the area PBAD: find the equation of the locus of P.
PBCD is
A
point moves so that the ratio of its distances from the points 2 3 find the equation of its locus. Where docs the locus meet the axis of x ? 63.
(-,
0), (a, 0) is
64.
(#3
,
:
:
A, B, C ...K are n points whose coordinates are (x^ y,K ...(*, y n ). is bisected at
(ar 2
,
t/ 2 ),
t/ 3 )
AB
divided at so that
P3
(n-
1)
so
P,Cis divided at P2 so that 2PjP2 = P2 C, P2 Z) is that 3PoP3 = Pn D, and so on until Pn ^K is divided at Pn _, P,,
Pn _^Vi = pn -i K
-
Find the coordinates of
PM-1
.
CHAPTER
II
THE EQUATION OF THE FIRST DEGREE 1.
WHEN
two points on a straight
that the coordinates
the points (x l9
(x,
y^), (x 2
,
y)
y2 )
line are
known, the straight
We
have shown in Chapter I, 6, of any point on the straight line which joins
line is completely determined.
satisfy the equation
or in determinant notation
x
y
1
=0. 2/2
This equation
we
proceed to
is
of the first degree in the variable coordinates (x, y} that every equation of the first :
show conversely
degree in x and y represents a straight line, or, in other words, that every pair of values of x and y which satisfy a given equation of the first degree represents a point on a definite straight line.
The most general form where the
of the equation of the first degree
coefficients or constants of the equation
I,
/?,
C
is
can have
any values, positive or negative. 2.
The equation Ax-}- By + C
Method
Let
y^
=
represents
# 2 ), (#.$, y coordinates satisfy the equation then I.
(x l9
(r 2
,
t}
)
a
straight line.
be any three points whose
:
Eliminate the constants A,
7?,
and
G',
= Hence the
triangle
represented by
then
0,
formed by joining any three points on the locus
Ax + Sy + c =
(j
THE EQUATION OF THE FIRST DEGREE has zero area, line
:
Method
any three points on the locus therefore a straight line.
is
B
If
II.
is
not zero the point (0,
Ax + By + C =
equation
on a straight
lie
i.e.
the locus
29
^)
satisfies
the
0,
on the axis of y it is therefore the point where the locus and Let this point be Q, represented by this equation cuts the axis of y. lies
and
:
for convenience let k
Q Clearly
^
=
--.
can be positive or negative
;
if
due regard
the signs the following argument holds whether tity positive or negative. (b)
we
is
paid to
take this quan-
Rectangular axes.
y
Suppose
P (Xj
Join
and draw QA,
PQ
y) is
any other point whose coordinates
PA
satisfy
parallel to the coordinate axes,
then
The given equation can be written
C
A or
"B
9
AP_ _4
hence
QA~~ consequently for
all
positions of
B
P the
AP ratio
-
1
,
is
constant, hence
(JA.
,P lies
on a straight
If JB
= 0,
line. [Euclid VI. ] the equation reduces to
40 + 0=0, which evidently represents a
straight line parallel to the axis of y.
THE EQUATION OF THE FIRST DEGREE
80 Note.
k were negative and
If
then the length of
OQ
is
+ ^
P in
some other quadrant, say the
and since the ^-coordinate of
,
P
is
fourth,
negative,
y
NP
the numerical value of the length
and
QA =
-y, hence the length
is
AP
is
x.
A
C But, as above,
A -
AP -=-
- A and .
-
i.e.
B
B
QA
The student should go through the proof with the point P in the other quadrants, remembering that the general coordinates (#, y) contain the signs thus (x, y) might represent, for example, (-3, 4) or (-5, -6). :
Let the straight line PQ make an angle 6 with the axis of x LPQA = 6 then LQPA = o>-0. In the triangle PQA
Cor. i.
c.
;
QA _
sin
AP~
(<0
hence
sin B
Therefore
(A cos
co
tan 6
(6)
(o-0)
_ -
If
^
sin0
=
J5)
J. sin 1 sin
= -f
or
;
o>
.
cos
J
0.
6>
^tcos o>-Z?'
in rectangular coordinates
^ P - _^
tan 6 _ :
~&i~
^*
should be carefully noted that the value of tantf does not contain the constant C hence the direction of the straight line It
:
independent of C, and depends only on the ratio of the coefficients of x and y. Hence if A and B are kept constant while different values are given is
to C, the resulting equations will represent a series of parallel straight lines all
1 making an angle tan" (-
any particular position of Q.
line is fixed
--
)
with the axis of x: the position of
by the value of
C, for this
determines the
THE EQUATION OF THE FIRST DEGREE Examples II
31
a.
(The axes are rectangular except in questions marked with an asterisk.) *Find the equation of the straight line joining the points
1.
(a
-f 1,
6
+
3),
(a, b).
*Find the equations of the straight lines joining the following pairs (a) (1,1), (5,5); (b) (-2, 3), (-7, -9)
2,
ofpoints:
;
(c)
(0, 1), (1, 0)
(0, 0), (6,
(d)
;
3)
;
(e) (O.OUO.a); (f) (0, 0), (b, 0). *What are the equations of the coordinate axes? Draw the graphs of: 3#4 4y = 12,
3.
4,
= =
0,
=
12;
-6,
and show that the distance between each pair of consecutive
lines
is
the same.
What
5.
make with
angles do the following lines
the axe$ or x r'l v
respectively ?
( C)
= 7;
(d)
x+y
(a) y-f
A/3*
'
(b)
,
Draw
-8.
these lines.
*If the axes of coordinates are inclined at 60,
6.
following lines
What
7.
passes 8.
(i)
make with
(a)
y-*0;
(ft)
y+x^
-2;
(c)
2* + y =
(rf)
(v/3~l)t/-2.r-7.
1;
the equation of a straight line parallel to 2x + through the point (0, 4) (ii) through the origin ? is
(a)
(b)
Some
3^/
=
6
which
;
Find the angle between the pairs of
3.
what angles do the
the o:-axis ?
y-4/S~x
=
lines 0,
fSy-x = 0; y+y3# = 5, >v/3y-;r = 1.
special forms of the equation of a straight line.
Special attention should be paid to the results found for rectangular axes as before stated, oblique axes are not often necessary in the :
more elementary parts of the subject. The results are, however, worked out for future reference.
for oblique axes
THE EQUATION OF THE FIRST DEGREE
32,
Form which
it
Tlie equation
I.
intercepts
on
Let the straight line that
OA =
a,
straight line
;
OB = its
of the straight
line in
terms of the lengths
the coordinate axes.
1.
PQ cut the coordinate axes Now the points (a, 0), (0,
equation therefore bx + ay
in
A
and
B
so
b) are on the
is
= ab,
or
(i)
thus regard to sign must be observed 3 on 4 and I'/?' lii the making intercepts the equation of straight 1the axes is i# + # This result is true for rectangular and oblique axes. .B.
The usual convention with
;
=
Form
II.
'Flic
equation of the straight line in
and the angle pendicular on it from the origin x-axis. the with
this
terms of the per-
perpendicular
nyiJccs
(Rectangular Coordinates.)
10
Let the length of the perpendicular be p and the angle made by it with the x-axis be ex. This angle must be measured from Ox as
THE EQUATION OF THE FIRST DEGREE initial line as in polar coordinates
33
thus the polar coordinates of the
:
foot of the perpendicular on the straight line are (p, OK). and J2. Letjiue^traight line meet the axes in
A
OA p sec a, OB = JM cpsec
Hence
==
ex.
Therefoie by
the equation
(i)
x cos
to
# sin
y. -f
In this form of the equation
vary from
'
p cosec y.
jjsecon
or
is
p
is
ex
p.
(ii)
supposed
to be positive
and
ex
to
2?r.
If the line does not lie in the positive quadrant, as in Fig.
(ii), it
will
be found that the signs of the intercepts are given correctly by the thus in the third quadrant sin y signs of the sine and cosine of y :
and cos
OK
are both negative, and the intercepts
OA,
OB
are both
negative.
Cor. i. In the case of oblique axes, if the perpendicular makes angles with Ox and Oy, it can be shown in exactly the same way that the
0( 9 ft
equation of the line
Cor.
ii.
is
x
cos
y
-f
y cos
/3
= p.
The general equation Ax +
A= B= >- =
Let
then consequently
cos
On
=
sin
On
=
x+
Hence
r-^
*
By+CQ
rsin
A+ 1
on,
J5
=
2 ;
,
.
-- y = --- -
L
^
It follows that
127
-----
2 /-I + B*
in the required form, if the sign of the radical the right-hand side positive.
is
Note.
this form.
rcoson,
-f
to the straight line
can be put in
is
chosen so as to
make
the length of the perpendicular from the origin
Ax -f By -f C =
is
THE EQUATION OF THE FIRST DEGREE Form which
The equation of
III.
makes with
it
the straight line in terms
and
the x-axis<
the length
of
of the angle on the
the intercept
axis of y.
and the intercept
Let the angle be (a)
c.
Oblique Coordinates. P(x,y)
P (x,
Let
y)
be any point on the straight
cuts the "?/-axis
;
therefore
OQ
Q
line,
the point where
it
c.
AP = y-c, QA = r,
Now
hence in the triangle APQ,
yc
sin
x
sin
or the coordinates of any point
equation
c
sin 6
(b)
on the straight line
s{ n r
+
-
sn
satisfies the
co
y=mx
+ c represents a straight line cutting oft' from the axis of y and making an angle 6 with the axis of .r given
The equation
Cor, a length
by
x
0)
(oj
(x, y}
w
sin (w
6).
Rectangular Coordinates. Since
=
%-n
in this case, the
equation becomes
and
=W
y
= x tan +
c,
represents a straight line inclined at an angle tan l ?/
-f c
m
to the axis of
x and cutting
length c from the axis of
y.
1
oft
a
The
that formed by a straight angle line starting from the initial posiis
the given x tan 377 -f c.
is parallel to
line
y =
line.
Ox and
revolving in the until it positive direction about Thus P'Q in the figure is the straight tion
THE EQUATION OF THE FIRST DEGREE Cor.
i.
make an angle
If the straight line
a length
oft'
c
from the axis of
x
its
.r,
y tan Q -f
equation
6 with the axis of
\j
35 and out
is
c,
and conversely the equation
= my c tan" m with x
makes an angle cuts off a length
-f
1
c
the axis of
from the axis of
y,
and
This angle a straight
x.
must be measured in a similar way from the position Oy and revolving :
line starting
in the negative direction until parallel to the given line moves through the angle 0,
Cor.
li.
The
straight lines
y
=
i/
.r
tan B
jc
tau (6
c
\
}
} Trj-f <1\
are evidently perpendicular to each other.
These can be written
= x tan
//
=
y or,
-a- cot
with tho notation of this paragraph, y
=
tu# 4 C | '
^ j,=
-+,?) w
Thus, for example, the straight lines
ax + by + c to -f (? =
)
(nj
)
are perpendicular.
This can be expressed otherwise thus
the straight linea
:
= mx + c y = nix +d
y
mm'
are perpendicular if
)
,
Ax + By-\-C = + C' =
* ii r or u the straight line,
are pevpondicular
1
}
^jty
if
^t.l'-f 7?/^'
=
0\ of 0.
should be noted, as in 2 of this chapter, that tho dhection of the lines depends only on the ratios of the coefficients of .r and y. It
II b.
Examples
Find the equation of a straight line which passes through the origin and makes an angle (i) 45, (ii) 30, (iii) 120, with the axis of x. 1.
2.
Draw
the straight lines
y
=
I,
l t
x-\y^
In what points do they intersect ? H. What angles do the straight lines x + y axis of .r mul with each other? 4,
2,
-1,
x
\i
4
make with the
Find the lengths of the perpendiculars from the origin on the straight
THE EQUATION OF THE FIRST DEGREE
36 lines
Draw
=
3x + 4// = 5, 12:r-5?/ the line in each case.
.r4i/44=0, 3r-
10,
=
0,
2y-Kb =
Q
t
5. Find the equation of the straight line which makes intercepts on the axes twice as long as those made by ly ^x 63.
Find the equations of the sides of an equilateral triangle referred to a pair of sides, (ii) to the bisectors of one angle as axes. 7. Two straight lines make intercepts a, b and ka, kb respectively on the 6.
(i)
coordinate axes.
Show
that they are parallel.
Two
a on the axes. Show straight lines make intercepts a, I and 2>, that they are perpendicular. 9. If the axes are inclined at 120, what angle does x 2y-f 3 make with the a?-axis ? __ 8.
=
A straight line
10.
it
The
11.
BC
base
makes intercepts
make with
^3 and 1 on the axes of x and y
;
what
the intercepts are 5 \/3 and 5? A of is a the 75, angle perpendicular from A on the triangle of length 3, and divides the angle A in the ratio 2 3. Find
angle does is
the axes? also
if
:
the equation of BC referred to AB, AC as axes. 12. Prove that the straight lines ;/ = 3^-f7,
5//
= 8# + 35,
and x
=
are concurrent. 13.
Put the equations
3#-f 4?y
(i)
into each of the standard forms
For what value of
m
12
;
2y-# + 4 =
(ii)
14.
=
IT,
I,
and
aro the
III.
straight lines y
3#47, y
mx+ 5
(ii) perpendicular? Find the equations of the straight lines making an intercept^ on the axis of y and inclined at an angle 30 to y \/3s 4 4. (i)
parallel,
15.
16.
Draw (i) (ii)
(iii)
17.
the straight lines
x cos ^ TT 4 y sin J x cos TT 4 y sin $ TT x cos^ Tr + ysin $n
a
n-
Draw
the lines
;
a
;
a,
(i)
y
(ii)
y
(iii)
.r
when
= =
= 3 and -3 respectively. What intercepts does each make
when 18.
2 and
a
>rtan $7r4 c
j:tan^7r
ytan
-'/TT
4c 4r
2 respectively.
;
;
;
c
Show
on the axes?
that the straight line x 4 // tan
whatever value
touches a fixed circle, 19. If the straight lines
# ft
=
8
^ =
a
have.
may
3x + my + l 3wy-9o;-f
3 sec
0,
0,
are perpendicular, find the value of m. 20. Find the equation of the straight line perpendicular to
which
(i)
(ii)
(iii)
cuts off a length 4 units frcffn the #-axis 4 units from the y-axis
cuts off is
distant 5 units from the origin.
;
;
THE EQUATION OF THE FIRST DEGREE 4.
The
of a
equation
f
A, B, and G
coefficients or constants,
straight
two
viz.
independent quantities,
Ax + By-\- C =
line
of the
ratios
equation represents the same straight line or divided by the same quantity.
+5 = 15# + 20#+25 =
Thus
3tf+4?/
if
in the general
,
involve
0,
37
A B :
each term
:
is
two
only C,
the
for
multiplied
0,
represent the same straight line. have seen that when certain pairs of conditions are given the Thus equation of the straight line can be formed.
We (a)
If
two points equation
(/>)
(.r t
,
yj, (r 2
,
the straight line are given,
its
1 \
and the direction of the line, i.e. the angle it makes with some given straight line such as the axis of x
If one point
are given, the equation (c)
yj on
y y
is
is
of the form y
the line and
If the direction of
=
HUC
+ C.
distance from the origin are
its
given, the equation is then of the form r cos 3 + y sin 3 = p. In each of these cases there are two independent constants in the
equation of the straight lino corresponding to the two given conditions.
The form Ax + lly + C
Any
straight line.
=
is
used for a general discussion of the
one of the equations
lx+my+\ = .r+my+n = fo'+
0,
(i)
0,
(ii)
#+w =
(iii)
represents a straight line and contains two independent constants. No one of them, however, is suitable for a general discussion the first cannot represent a straight line through the origin, the second ;
cannot represent a straight line parallel to the axis of #, the third cannot represent a straight line parallel to the axis of //. It is evident then that two conditions are necessary to fix a
and further that in the cases above given these two conditions are sufficient: we shall see later that two conditions are straight line,
not always sufficient. When one condition
the line)
Thus have
if
is
we
(e. g.
a point on the line, or the direction of
given, a relation between the constants can be found. know that the point (a, ?>) lies on the straight line, we
Aa + Bb + C =
0.
THE EQUATION OF THE FIRST DEGREE
38
one of the independent terms of the other, and the equation of the line can be found in a form which involves only one unknown or It is then possible to find the value of
ratios or constants in
undetermined constant. f
C
=
Aa
Thus, in the example given above, since
Bb, the equation
can be written
x
or
a
+
-j
(y
1)
=
0, 7>
which involves only the one undetermined constant
A
.
second
jA.
necessary in order to find the value of this constant and the equation of the straight line.
condition so fix
We
is
have seen that the general equation
f
yLt'-f/ty
+C
~
can be
made
to represent any straight line by giving suitable values to the When one condition is given, and therefore only one constants.
in the equation, the equation can no longer be made to represent any straight line, but only one of a definite group of straight lines. Thus, in the example given
undetermined constant
is left
=
A
must represent one of a (x I) a) + B(y above, the equation group or system of straight lines all of which pass through the point
By
(a, I).
giving
definite
the
values to
remaining constant
,
.
the
can be made to represent any one of this system of
equation
Hence we see that if an equation of the first lines. contains degree only one undetermined constant, this equation for different values of the constant represents some definite group of
straight
straight lines; a further condition, if sufficient, will determine any In the following work we shall [ar particular line of the group.
convenience use the Greek letter
Two line.
p,
for the
undetermined constant
\
conditions are not always sufficient to determine a straight If a point on the line be given together with the condition
that the line touches a given circle, there are two straight lines which fulfil both conditions. If a single insufficient condition like is given, it will be found that the corresponding equation in the constants representing this condition algebraically contains one constant in the second or a higher power, so that more than one
the latter
value of
it
in
terms of the other constant can be found
will be fully discussed at a later stage,
:
such cases
THE EQUATION OF THE FIRST DEGREE
39
Groups or Systems of Straight Lines. (i)
Straight lines ivhieh $>ass through a given 2*0 'ml
(h, k).
The equation y k = /^(r h) represents a straight line, and is satisfied by the coordinates (h, A) of the given point, whatever the fx may be: it therefore represents for different values of // the system of straight lines which meet at the point (//, A). Such an equation is said to be of a particular form thus, for
value of
:
instance,
an equation of the form y
of straight lines through the point (ii)
The
Straight lines inclined at
==
3) represents a
/jt(^
system
(3, 0).
a given angle
(y) to
t/ie
x-(u:/<.
direction of a straight line depends only on the ratio of the
coefficients of
x and
different values of
an angle
oc
p.
The equation y
y.
tan
jo
Ot-f-ju-
represents for
the system of parallel straight lines which
make
with Ox.
In particular, i/=-mx + n represents a system of straight lines mx. parallel to a given straight line y
=
(iii)
Straight lines perpendicular to a given straight
line.
y~mx
Let the given straight line be + c, then (p. 35) the equa= line perpendicular to it: a x 44n straight wy represents
tion
represents a system of parallel the given straight line. straight lines all perpendicular to the required form of the If the given line is a r 4 by 4- c 0, = 0. is bx + equation ay n it
therefore for different values of
(iv)
System of straight
two tjioen straight lines
The coordinates
IJL
lines gassing through the intersection
of the
Ar-|-Zty + C=0; A'x + li'y + C' = Q.
of the intersection of these straight lines satisfy
simultaneously both the equations
Ax 4- By 4- C = Hence they
0,
A'x + B'y + C'
Ax 4- By 4- (74- M (A'x + This equation a straight line
=
0.
satisfy also the equation
;
is it
/>'// 4-
C')
=
0.
of the first degree in x and y and represents therefore represents a straight line through the
=
intersection of the given lines e.g. .r + p. y of straight lines through the intersection of ;
represents a system
x
=
~
and y
0,
i.
e.
through the origin. (v)
S^emjtfiAtXQight
The equation x
cos
ot
lines distant
is
j?,
the origin.
=p
represents a straight line whose whatever value a may have OK is in
+ y sin a
distance from the origin
p from
;
THE EQUATION OF THE FIRST DEGREE
40
The equation can be written
this case the undetermined constant.
x (cos 2 \ a
sin 2 \ a)
4- y 2 sin
a (1 - tan 2
or
Put
for tan i
ju
oc
.
a)
+ 2#
a cos
I
a
tan
a
=
j>
2
e,
p (cos
(1
+
=
0,
2
1 ot -f
sin 2
J
a)
2
tan i
a).
and we obtain
M (+p)-2|Liy+jp-i.
=
the undetermined constant appears in the second degree. illustrated in this section is generally convenient for
The method
the use of the single constant ^ does not, numerical examples For example, we have however, give quite satisfactory results, ;
shown
in (iv) that
Ax + By+C+i*.(A'x + Vy+V) = represents, for different values of
^
Q
(i)
a pencil of lines passing through and of the the point of intersection straight lines f / If we give jm increasingly large values, the A'x + y + C 0.
Ax+Hy+C =
=
S
straight line represented by (i) approaches the position of the straight It is sometimes said that for an infinite line A'x+By + C'= 0. It coincides with this straight line. correct, however, to write equation (i) in the form
value of
/JL
it
\(Ax + By+C) + n(A'x + B'y+C')
is
=
safer
and more
Q.
(ii)
This equation contains only one independent constant, viz. the ratio but the equations of every straight line in the A//x or the ratio ///A When A is pencil can be found by giving finite values to A and /x, ,'
zero, the equation reduces to it
is
reduces to Ajc + preferable for
Sy+ C =
A'x + lf'y+C' = 0, and when /* is zero This method of writing the equation
0.
two reasons
:
we
avoid the- necessity of giving an
infinitely large value to the constant,
and we get a more symmetrical
equation.
Examples
II
Write down the equation of the straight
c.
line in the
form
in
which
it
represents (i) (ii) (iii)
A system of straight lines intersecting at (2, I). A system of straight lines parallel to 3aH-4y 6. A system of straight lines perpendicular to (a) 7^~;/ =
a?-3 (iv)
A
=
system of
st might lines
concurrent with 3.r -J-y-
r>
=
und x
Straight lines touching a circle of unit radius whose centre the origin.
(v)
(vi)
A
(b)
6,
0.
~ is
0.
at
group of parallel straight lines making an angle of 30 with the
straight line y
==
\/3jc
-f
12.
THE EQUATION OF THE FIRST DEGREE 5.
41
Determination of the equation of a straight line ivhen two conditions
are qiven, I.
T/ie straight line
passing through the two jtoints (x^
(x^
i/ A ),
,
This equation has Been already found in the form x y I
is given again as an example of the present method. Since the line passes through (x^ y ) its equation is of the form
but
}
The second condition, viz. that Thus we have x 2 x /x.
value of
l
Substituting for
II.
The straight
(:r.2
,
the
is
a* -f by -f
line parallel to
line, fixes
f/j).
the required equation
/x,
on the
lies
y2)
= M^
=
c
and
jtassiny through
the point (x^ y^.
Since
The point line of the
With
=
+ /A
(j.\,
j/^ lies
vlue of
c
=
0,
the straight line
one of the
is
0.
on
group required
this
+ ly +
parallel to ax
it is
group at + Z#
/jt,
it
is
;
hence the value of
given by aj\ .r^-f
rt(,r
?>(//
Ihis equation is instructive: it form implies that it passes through
-f
by^ -f
=
/x
for the particular
~
JJL
0.
the required equation! should be noted that ( 4. i) its //i)
(.r^ #,)
is
and
4. ii) its
(
form also
~
0. implies that it is parallel to .r+ />//-fc When the student is familiar with the forms of the equation corresponding to given geometrical conditions, such results can be at
once written down. III,
The
straight line prrpendicuhir to
through the intersection of the straight line* l>x
The second
+
ax + by +
c
-f
r
= 0,
-f-
by
-f
c
=
ami passing
\
y.r 4-
-f r'
= 0.
condition gives that the required equation
form (px + qy + r)-hM to
W
ax
=
0,
(/>'^4-'///
+
r')
=
0,
and since
the ratio of the coefficients of
,r
it is
and
is b
/^
and the required equation (a/ +
6 flO (jw: +
is
^+
r)
in the
perpendicular //
hence or
is
- (op 4- &j) (/J? + q'y -f r') =
0.
:
- a
;
THE EQUATION OF THE FIRST DEGREE
42
IV. To^find the equation of the straight line through the point jhjjc 0. which makes an angle a with the straight line ax + by -f c
=
The
straight line ax + by + c
=Q
makes an angle tan"
1
the #-axis, hence the required line makes an angle tan"
with the o>axis
=
y
a?
tan
7
~~ -x
.e.
(6
And
r
+a
tan
oi)
;>/
+
('(
since the line passes through
(#-) (ba
tan a)
There are evidently two
%
a
~
1 -f y o.
1
i
-+
tan {tair 1 a
i.
w
J
therefore parallel to
it is
;
~T
(
-f (jc
tan y
b
=
+ &tana) (h< k) its ft)
(a
strait at lines
+
b
0. is (vide II)
equation
tan a)
which
=
0.
satisfy the conditions
V. The straight line throuy' y ) distant p from the origin. Since the distance of the line from the origin is j>, the equation of t
the straight line
is
in the 2 l*
Since
(# 1?
T/J)
lies
on J
form
(
,
{
4. v)
(x+l>)-2yH+l>-JC
=
0.
this, to determine the value of
JJL
we have
=
b
0. + ^'i)-2^ 1 fz+ j [;-^1 M' a quadratic equation in p, hence there are two straight these conditions are lines which satisfy both the given conditions
This
is
:
therefore not sufficient to determine one definite straight line.
Note. The above results are not to be regarded as formulae numerical example should be worked out on similar lines.
Examples
;
each
II d.
Write down the equations of the straight
lines
which
satisfy the following
conditions:
The
1.
straight line passes through the point (-2, 3)
(b) perpendicular to
3#-y =
The straight line passes through + ly = 8. 3. The straight line joining the points
2.
and
is
fa) parallel,
6.
the origin and
perpendicular to
is
4a:
(a) (3, 2), (1, 1), (b)
(
-4,
5), (
-2, - 3), 2
(,&), (a 6, fc-6), (d) (acos<9, asintf), (acos0, asin0), (an 2 an), (f) (a, 6), (a + rcos0, &-frsin#).
(c)
(e)
(aw
,
2am),
9
,
4.
The
straight line cuts off a length 5 from the axis of
an angle of 75 with
it.
a?
and makes
THE EQUATION OF THE FIRST DEGREE
43
5. The straight line cuts off a length 10 from the t/-axis and is perpendicular to 11//-10.Z = 21. 6. Find the equation of the straight line drawn through the point (x^ y^
perpendicular to the straight line joining the points (a" 2 ,y 2 ), ^3 Va)7. Find the equation of a straight line passing through the intersection of which is parallel to px + qy + r = 0. cur 4 by -i c = 0, ax -f b'y -f c = 8.
What
the length of the =2?
is
perpendicular from the origin on the
3# + 4y
straight line
Find the tangent of the angle this perpendicular makes with the axis of x. The centre of an equilateral triangle is at the origin, and one side of the = 2. Find the equations of the triangle lies in the straight line 3# + 4y other two sides. 9.
What
is
angle
formed by the straight Of. -f-
10. j'
whose equations are
lines
x cos y sin 3 x cos |3 -f y sin ft
= p, = q?
Find the equation of a straight line inclined to the straight line y sin p at an angle 3, whose distance from the origin is 2j).
cos y
ex.
-4
Find the equation of the straight line joining the point of intersection of the lines ax + by 4 c c' = to the origin. '.r-f b'y 0, 12. A square has its centre at the origin one side of the square is the 11.
-}
:
7
=
0; find the equations of the other three sides. 13. Find the equation of the straight line which passes through the points ot intersection of each of the pairs of straight lines
straight line
\ l3//
.r
4
ax -f by -f ax - by i
c c
= =
a.r -f
i
- ax
'
by 4
4
=
d
^
j
by -i 14. The perpendicular from the origin to a straight line meets the point (h,k) show that the equation of the straight line is (
j
it
at
:
15.
Show
that the straight lines joining the points
(1, 1), (2, 2)
respectively
to the point of intersection of the straight lines
19.r
+ 3f/-29
=
0,
13#+ lly-27 =
are at right angles. 16.
(2,1), 17.
Find the equations of the medians of the triangle whose vertices are (3, -3),(-5,2i. The sides of a quadrilateral taken in order are a?
+ 2y
=
5,
3* + y
=
10,
.r
+ 6y + 8
=
0,
4#-y + 7 =
;
find the equations of its diagonals.
Find the equation of a straight line through the intersection of 3#-f 5t/-7 = and 4#-f 6y-5 = parallel (i) to the axis of x, (ii) to the 18.
axis of
y.
are that the two straight lines #-fy-fc = 0, \/3x + y-f d = inclined at an angle of 15. 20. Find the condition that the straight lines joining the origin to the points of intersection of the pairs of straight lines 19.
Show
to-f wr/
0| should be
(i)
perpendicular,
(ii)
coincident.
+1
=
0)
THE EQUATION OF THE FIRST DEGREE
44
To find the coordinates of a point whose distance from a fixed and which lies on a straight line through (ex, ft) inclined
6.
point
(a, ft) is r
an angle 6
at
Let
.4
be the point
Draw AN, Then i.
e.
r
is
to the x-axis.
(a, ft)
PN parallel
the coordinates of
and
P (#,
the required point.
y)
to the axes.
AX= x-% =
rcostf
PN
rsintf,
P
are
y-ft (r
cos
+ a,
The length r is measured from negative P lies on the other side of A. This result can be written Note.
r sin
A
-f
/:*).
in the positive direction
;
if
sin B
cos~
if P be regarded as any point (oc,y) on the straight line Al\ this may be regarded as the equation of the straight line. The equation in this form is often very useful, arid the meaning of the constants should be carefully
and
noted,
(ft,
straight line
ft)
'is
a fixed point on the straight line,
makes with the
and the distance of
(j?,
#-axis,
y) from
(%
(., y) is
ft)
Two examples
is r.
is
the angle the
any point on the straight
type of problems in which this form of the equation is useful refer to it again when equations of a higher degree are discussed.
Ex.
A
i.
straight line
the straight line
point Q.
Find
y
=
\/3#
the length
is
to
drawn through
line,
will illustrate the :
we
shall
the point P('2, 3) parallel to
meet the straight line 2x
+ iy
=
27 in
the
PQ.
= \/3x makes an angle J n with the .r-axis straight line y of the equation straight line through (2, 3) parallel to it is The
:
hence the
.
COS JTT
=
or
where r
is
r
'
the distance between the point
(a?,
y)
and the
fixed
point
THE EQUATION OF THE FIRST DEGREE Now
(2, 3).
on
lies
if
the
u* ^ 2
r
equal to PQ, then (.r, y) must be the point Q which the coordinates of Q are therefore 4y = 27
is
line
2.r-f
;
r
Hence
2
+4
2 r-f
\
^
J
^
/
r+3 =
hence
-
r is
Ex.
-
^
-
11,
^3 - 1 -
2
27,
/
(2\/3 + l)r=
i.e.
which
45
246,
the required length of PQ.
ii.
To find
the perpendicular distance
Ax + By + C =
the straight line
of the point
(a, I)
from
0.
The equation of a straight line through - A (y - b) is B (.r 0, a)
I)
(Yr,
perpendicular to the given
line
x
.
((
A
where
r
If
in
;
is.
Hence
1.
3j?
2
a
\/(x
~
-f
)
b)*
(y
^
and
(a, b)
(*r, //).
(jr,
y]
ia
0.
-
/
7/
_?!!
__
Examples
II
e.
^--^-. 4 a
)
-f
Find the perpendicular distance of
(2, 5)
-f
ft
)
-f
C=
from 12# +
0,
5?/
= 40
and from
16.
D is a lines x-y
A, B, C\
straight the parallelogram. 3.
B
the distance between
A(
+ 4f/=
2.
b
y
equal to the length of the required perpendicular, the point
Ax -f By + C =
on
~
parallelogram 7
and
3:r
+
A
;
is
the point ( -5, 2) J3C, CD are the Find the lengths of the sides of ;
== 28. 4//
Through a point P(h, k) a straight line is drawn, making an angle fr + fy/ = 1 at Q. .r, to meet the straight line
6
with the axis of
Find the coordinates of a point on
Hence
PQ
find the locus of such points
distant
when
PQ/n from
P.
6 varies.
4. A straight line is drawn through the point A (a, b) making an angle 6 c*. Show with the axis of x to meet the locus represented by x? + y* that it meets it in two points P,
independent of B. If AP, AQ are equal, prove that
(a sin 6
- b cos 6)* =
c
2 .
THE EQUATION OF THE FIRST DEGREE
46
line
7. To find the perpendicular distance of ant/ point from a straight whose equation is given.
Oblique Coordinates.
(a)
Let the equation of the straight line be given in the form
=p
XCOSOL + ycosfi
and
let
P (x^
the point be
Through
P draw
(i)
y^).
a -f y cos /3 == p. draw OLM per-
a straight line parallel to x cos
And from
Draw pendicular to these lines. PN perpendicular to the given line: then is the required
PN
length, call
f(*..yj
Now OL =
^\~
x cos a
But
(a?!,
y^
lies
on
^ cos a -f y Hence,
:r
# if
-f
?/
cos
fi
=p+#
and
.
hence
this,
therefore
chapter
OM = p + q of PM is
hence
the equation
*\~
q.
3 of this
by
j),
~7
it
t
cos
!
cos
ft
= jp + ^
a + ^ cos /i
the equation of a straight line
x cos a
+ # cos
j>
/:*
;
p.
given in the form
is
=
0,
the length of the perpendicular from any point on it is found by It substituting its coordinates in the left-hand side of the equation. follows from this result that
if
we can put
the equation of any straight
line in this particular form, the length of the perpendicular any point can be at once written down.
on
it
from
Compare the equations
Ax + By+C=Q and
,rcos(X4- //cos/3-
same
If they represent the
cos
Hence
cos 2
a -
A^
-
= cos -
__
2
&
cos 2
(ii)
straight line,
a _ ^ /3 '
~
cos/3 __
B
~~A~ TT
(i)
p = 0. p '
~C
cos ---= cos An
OL -f
a.
cos 2
/3
fi '-
2 cos a cos
+ &-2AB cos
(a
+ /J)
/3
cos (a -f ~
/^)
But
OL
THE EQUATION OF THE FIRST DEGREE therefore + = ft
cos
co,
(\
__
Hence equation
cos
ft
^
p
sin
__
written in the form
(i)
A sin
co
x -f
.
B sin
co
.
(ii) is
C sin
-f
?/
Therefore the perpendicular from (.r^ h By, -f C
//^
-_
(i) oc
cos (ii)
:
co
on the lino Ax-}.
sin
By + C
is
co.
Rectangular Coordinates.
The student should work out case
o>
2A/*cosco
l*~+i/2
(h)
47
same way the more simple
in the
the results are
The length of the perpendicular from (.i\, //j) on the is .^cossc-f ^ sin A' 3 -h?/ sin (\-p ~ j>. The length
Ax + By+C=Q
from
of the perpendicular is
(.c lt j/ T
)
P is placed on
equation of the straight line through
Fig.
(ii)
x cc)SLX+#sin^
Fig.
cos
,7-
and as above, since Fig.
line
(ii)
In the two figures given
(i)
on the straight
^T
(i)
Fig.
straight line
(.r 1(
Ci-f?/ sin
?/j)
cos \
on the
is
+
:\
jr
(ii)
.^ cosot-f //i sin \
Hence the length
:
(/>
0,
Fig.
(i)
Fig.
(ii)
(/
=r^
{
cos
3: -f //!
j^cos ^
;
line, ( /)
(ji)
+ 7)
=. 0,
7)
0.
q of the perpendicular
~x q
and these results are
?/,
opposite sides of the line th* parallel to the given line is
(jp+ q)
sin >
(i)
l
P
sin ?/ l
of different sign.
en
is
p,
^iiiLX,
THE EQUATION OF THE FIRST DEGREE
48
of the perpendicular changes as the point P crosses the being of zero length when P is on the line. When dealing with more than one perpendicular we must call those drawn from points on one side of the line positive and those from points on the other negative. It follows similarly that all points whose
The sign
line,
coordinates
make Ax + By+G
Ax + By + C~
0,
positive lie on one side of the line and those whose coordinates make Ax+By+C
negative lie on the other side. This can also be shown independently
The
means of discovering whether two on the same or opposite sides of a straight line this question of sign arises in such problems as/ find a point equidistant from three intersecting straight lines*; there are four such points, the in-centre and the three ex-centres of the triangle formed by the result only gives us a
points are
lines.
:
Thus,
+b
a1 x
l
if
the sides are 0,
y-\-c l
a.2
x + b 2 y-}-c.2 =^0, a^x +
the four points are given by the equation
_ ~~ To determine example
in any special case which solution corresponds for to the in-centre. the student should draw the graph of
the lines.
Example. lines
(i)
3a?
To find
the in-centre
+ 40-12=0,
(ii)
of the triangle formed by the straight
5a-12y-20=0,
(Hi)
24^-7^-72=0.
35
40
\ We
see
from the
sides of the line
and the origin are on opposite and hence the perpendiculars from these points must
figure that the in-centre
(i),
be taken of opposite sign. Since the substitution of
#
= 0,
y
=
in 3#-f
4y
12 gives us a negative
THE EQUATION OF THE FIRST DEGREE
49
value, the substitution of the coordinates (x' y \j] of the in-centre will give a positive result: hence the perpendicular fiom the in-centre (x',y) on
3#-f
4^-12 =
numerically equal to
is
The origin and the makes 5.r- 12 y y it
same
in-centre are on the
=
20 negative, x
x
y
,
side of the line
= \j
(ii)
:
will therefore also
x = 0, make
negative.
The numerical value of the perpendicular from x y on
this line
',
-
& (5^ - I2y' - 20') = ^ (12 y
Similarly, the perpendicular from
-i
(24y'-7a?'-72)
Hence the in-centre
and the in-centre
1.
is
(4
=
1-
2
therefore
- 5x' 4- 20). 7#
on 24y
(7*'-24*/'
-f
72
=
is
72).
will be given by
=
8jc-y
i.e.
(#', y')
is
a
%,
and 2#-f
11
Examples II
f.
32,
1/
=
33,
2$).
In the above example work out similarly the coordinates of each of
the ex-centres, giving a reason for the signs chosen in each case. 2. Find the centre of the circle inscribed in the triangle whose sides are
x-y + l = 3.
The equations
0,
x + y-1
=
0,
.r-3y + 5
=
0.
=
12. 0.
Relations between two straight lines whose equations are given.
8.
To find
(i)
=
of the sides of a triangle are x 0, 3j? + 4y 0, y circle escribed to the side y =
Find the coordinates of the centre of the
the coordinates of the point of intersection
of two straight
lines.
This
is
equivalent to finding a pair of values of the coordinates satisfy simultaneously the equations of both the
x and y which straight lines e. g.
;
in other words, to solving the equations simultaneously.
to find the point of intersection of
= 0, 3^-7^ + 6 = 0,
5* + 4y- 7 by cross multiplication the point of intersection
is (fy, l\).
THE EQUATION OF THE FIRST DEGREE
50 (ii)
To find
the angle leticeen
two straight lines whose equations are
given in rectangular coordinates.
Let the equations be
these straight lines
^r
make
Q
.
jty
Q
angles
A with the axis of x
;
the angle between
them
is
A
.,/
,
therefore
A' i
tan
_.
~* * i L
1
+
VB'
If the equations are givon in the
y
=
m,r
the angle between them 6
is
-f-
;/
r,
tan
form
one value
is
positive
-I-
c
,
+
,
1+wm'
is
we cannot
tan" 1
and corresponds
negative and corresponds
x
wi'-r
~
l
It is clear that in the general case
acute angle between the lines
=
m 1 -f mm
decide whether the
m
z
-,
or tan"
to the acute
1
7-, 1 H-
w mm,,
angle, one
is
to the obtuse angle.
Corresponding results can be obtained for oblique axes in the same
way
(
2)
;
these results are / A T>/
an
,
(AB
A
/
A / T>\
xjjsmco
_!
/\
THE EQUATION OF THE FIRST DEGREE (iii)
The condition
that the straight lines should be perpendicular
Since the tangent of a right angle
follows.
A A' + BB' =
coordinates
51
or
1
is infinite,
+ mm' =
in rectangular
,
and in oblique coordinates
AA' + BB' (AB' + A'B)cosu-(} To find
(iv)
the equations
or
the bisectors
of
the angles between the
of
straight lines whose equations in rectangular coordinates are
C=
A's + Ify +
0,
C'
=
0.
any point on either of the bisectors, the perpendiculars to the straight lines are of equal length. It is evident (see figure) that for points on one bisector these perpendiculars If
from
(-
ri
IJ\) is
it
are of the
same
sign,
and
for points
on the other bisector they are
of opposite sign.
AJT
TT Hence
-f /j'//i -f LI. LT I
and any point on a bisector
AJT+
BI/
C
A'JC l
= +
lies
_
}
~ -f 7/>/i -f G Ji^^ (
'
^
on one of the lines
+ C _ ^IV + B^+C'
-4.r-f
By+ C __
A'j + B'y+C'
which are consequently the equations of the bisectors. 4 (iv) that the equation of any straight (v) It has been shown in line through the point of intersection of the straight lines is
of the
form ')
whatever coordinate axes are employed. i)
3
=
0,
THE EQUATION OF THE FIRST DEGREE
52
convenient in order to save the constant repetition of to refer to it in the following such an equation as Ax + By + C It is often
=
way:
let
u stand
for the expression
Ax + By + C,
i.e.
u = Ax + By +
C,
=
represents the straight line, and for the sake of brevity can refer to the straight line u \ This notation (usually called
then u
we
l
abridged) is of much greater importance than appears at first sight. There is a large class of problems in geometry of a purely descriptive character (i.e. not dealing with magnitudes such as length, or size of
which can be solved in a quite general manner in such work = represents a straight line quite sufficient to recognize that u
angle) it is
:
where u
an abbreviation for any expression which, equated to zero, represents a straight line without any regard to the actual system of is
coordinates employed. This idea is introduced early in the when occasion offers, and it is hoped
student at the later stages to pass to
work and will be developed by this means to help the generalized coordinates more
readily.
If
u
= 0,
v
=
represent two straight lines, any line through
their point of intersection If u' EE (#',
y'}
Ax' + By' +
should
interpret v
f
lie
=
C,
is
u + \v
then w'
=
=
0.
is
the condition that the point
on the straight line u
=
;
and similarly we
0.
+ Xv = passes through some given point u'v At/ 0, hence w/ represents the straight (#'> y'\ then tt'-f 0, line joining the point (/, y') to the intersection of u 0, v r a w line represents straight through (/, y'} parallel Again, u If the straight line u
=
=
=
=
=
to
u
=
0.
In the particular case when the equations of straight lines are in
+ y sin a p = 0, x cos a + y sin a p, tfcos/S + i/sin/:} &
the form
x cos a
if
u = v =E
then u' and
are the actual values of the perpendiculars from
v'
to the straight lines
u
=
0, v
Hence the equations of the = 0, v = are straight lines u
A
bisectors of the angle between the
uv =
point can be determined as the
then u
=
(#', y')
= 0.
=
and u + v=Q. intersection of
two given
represent straight lines, their point of intersection can be referred to as the point (w, v).
straight lines
:
if
0,
v
THE EQUATION OF THE FIRST DEGREE
53
Example. If two trianyks AHC, A'B'C' are so placed that the points of intersection ofAB, A' B' ; HC, B'C' ; CA, C'A' lie on a straight line, then the joins of corresponding vertices AA', BB', CC' will meet in (Coaxal triangles are also copolar.)
a point.
Let CB,
CB' meet
at
I':
A
CA,
at
a straight line, Let the sides of the triangle
ABC
w
be z
~
0,
Now = 0.
ami the straight
line
PQR
;
All A'B'
at
tt,
where 1>QR
be the straight lines n
=
=
0, v
is
0,
0.
a straight line through the point of intersection of u 0, is a conIts equation is therefore of the form (tu-\ z - 0, where
B'C'
is
which can be determined
stant
y
for
any chosen system of coordinates;
for,
the axes are fixed, since BC, B'C are given straight lines their equations are completely known. Without reference to any particular co-
when
ordinate axes, therefore, we c vn consider the constant a as known. Similarly, C'A' passes through the intersection of the given straight lines CM, P(J and its equation is + ^ = 0, where b is a known constant. A'B' in the
^
0. Hence the sides of the triangle represented by cir \ c A'B'C' are the straight lines whose equations are
same way
is
an + z
Now
=
0,
bv
-f
z
0,
av -f 2?
=
0.
the equation bv 4 z~(cw f xr) represents a straight line through the point of intersection of the straight lines J>tM 2 = 0, c/r-f^ = 0, i.e. -
through the point
A'.
=
cw is equivalent to bv 0, and this represents (cw + z) = a straight line through the point of intersection of the straight lines v 0, w =0, i.e.- through the point A. Hence A and A' both lie on bv cw =
But
(bv + z)
=
:
then
the equation of the straight line AA'. In exactly the same way ctv-au Q represents the straight line and an-bv the straight line CC'.
this
is
-
BB
r ,
THE EQUATION OF THE FIRST DEGREE
54
Hence the equations of the three straight vertices, via. AA', BB', CC\ are
cw nu
Iv
cw au
-
bv
lines joining
corresponding (i)
= =
(ii) (iii)
=
0, which represents a straight line through the of 0, (i) and (ii), is the same equation as <m-&r point of intersection i. e the straight line CC' (Hi) passes through the intersection of A A' and CC' are concurrent. BE' in other words, AA',
But (bv-cw) 4 (cw-au)
=
BB
:
1
,
Examples
II
g.
1. Find the length of the perpendicular drawn from the point (2, 4) to the straight line joining the points (3, 1), (7, 4). 2. Find the equation of the straight line which joins the intersection of
=
5#4 ;/-! =
to the point (1, 3). Draw the tangent of the angle between them. 3. Find the coordinates of the feet of the perpendiculars drawn from the point (1, 1) to the straight lines x-2 y + 2 = 0, Zx-y + \ = 0. Find also the length of the perpendicular drawn from the point (1, 1) to the
the lines lines
and
3#-4t/4l
and
find the
straight line joining these feet. 4. Find the length of the perpendicular from the origin on (x cos6)/a 4 (y sin 6)/b
1.
5. Find the equations of the straight lines through the intersection of 6 == the straight lines 2,r 1/45 0, #4 3y respectively perpendicular
and parallel to the straight
line 5
#4 8//-10 =
0.
(2, -1), (0, 2), (3, 0), (-1, 1) are the angular points of a parallelogram, and find the angle between the diagonals. 7. Find the condition that the straight line 1x-\ w/i/4 n =0 should touch 6.
Prove that
the circle whose centre 8.
as
Find
coordinates
diagonal
is
is (a, b)
and radius
r.
which has (1, 2), (4, 3) one diagonal, and whose other
the^ equations of the sides of a rectangle
of
the
parallel to
a-
extremities of 4
3/y
=
0.
Find the coordinates of the orthocentre of the triangle whose sides = 9. are y = 2 x, 2y x, x 4 y 10. Show that the triangle formed by the straight lines whose equations 9.
are
4^~3y~8 =
0,
3#-4?/4 6
= 0,
x + y-9
=
is
isosceles:
length of the equal sides. 1 1. One side of an equilateral triangle is the straight line and the opposite vertex is the point (2, 1).
find
_
x-
\/3y
the
44 =
0,
Find the equations of the other sides. 12. Find the a?4& a y4r 1 = 0, conditions that the straight lines = = a a x4&2y4-c 2 should form an isosceles right-angled 0, a8o:4fe 8t/4c 3 1
triangle, the last line being the hypotenuse.
Give, with numerical coefficients, the equations of three such lines.
THE EQUATION OF THE FIRST DEGREE (
55
13. The algebraical sum of the perpendiculars from the points (xl} ;/ t ), X 2> Ui\ ( X ^y^> (#4> I/*) on a straight line is zero show that the line must pass :
through the
mean
centre of the four points. Extend your proof to the case of any n fixed points. 14. A straight line is drawn through (5, 9) inclined at 45 to the axis of x.
The straight
line
is
cut in P,
Show
intersect at T.
Q by x + 3y =
120, which pair the isosceles; give lengths of the equal
PQT ia
that
=
Ix + y
20,
and the tangent of the angle at the veitex. 15. If the straight line PQ drawn through the point
sides
to the line joining
of the triangle 16.
POQ, and the distance of
Find the coordinates of the centre of the
triangle formed by the
x
lines
=
1,
3# + 4y
5
inscribed in the
circle
= 0;
5.r- 12//+ 16
0,
those of the centre of the circle escribed to the side x 17.
area
Q, find the
from PQ.
2)
(4,
at right angles
(8, 4)
P and
inteisects the axes in
(6, 2) (4, 1)
=
also
1.
Find the coordinates of the foot of the perpendicular from (a, 0) on = j'-fa\ 2 and show that when A varies all these feet lie on
the line Xt/
,
a fixed straight line. 18. Show that the two straight lines (x cos
6)'/a f (y sin
6)/b
=
(x sin 6)/b
1,
- (y cos 6) /a =
(ae sin 6)/b
are perpendicular, and that the distance of their point of intersection from 1 V". the origin is independent of B if rr (1 e }
=
19. Straight lines arc
line tixi
ly
=
30,
drawn from the point
and these straight
of the locus of the points of bisection. 20. The product of the perpendiculars from
21.
lr
prose
The equations of the 2
.r-fty-J
=
Find the coordinates of 22.
The connector
straight line
+ byl
~
a 2 (l-e
=
=
1
,
find the
(-(/,
0)
=
2
).
>r
0, its
+ nnj
m~
0,
1
.r-h
ny-ii'
=- 0.
orthocentre.
in
Q
:
y')
with the origin
show that PQ OQ :
=
find the equations of
;>, (i) the diagonals, intersection of the diagonals parallel to the sides. t;
,
on
1
=
ax
-}
(ii)
meets tho
by'+l.
23. If the sides of a parallelogram arc the straight lines
u
equation
sides of a triangle arc
of any point P(jt'\ 1
O
(ae,
(# cos #)/-{ (ysin0)/& is b'\
meet the straight
(3, 2) to
lines are bisected:
lines
w=
0,
through the
Find the equation of the straight line which joins the point of section of u = 0, v = 0, to that of u -f ic = 0, r + ww = 0. 24.
inter-
four sides of a quadrilateral have equations ui v = 0, r-i w> = 0, -u = 0, t;~ w = 0. Find the equations of its diagonals. 26. Find the locus of the middle point of that portion of a straight line the fixed point (ft, k) which is intercepted between the 25.
//
r =- 0,
The
passing through axes of coordinates.
THE EQUATION OF THE FIRST DEGREE
56
Relations between three straight lines whose equations
9.
are given.
To find
(i)
the condition that the three straight lines
Ax + By+C=0,
A'x +
'!/
+ &=<), A"x + B"y + C" =
should be concurrent.
have a
common
+ Bj/
+C = 0,
If they are concurrent they
be
(#!
,
f/i).
Then
Ax
l
l
A'x +B'y l + l
C'
point
;
let its
coordinates
=0,
Eliminating (# 1? yj we find the required condition to be
ABC
=
\A' B' C'
A"
Ii"
0.
C"
Since any two straight lines meet in a point, one condition is necessary and sufficient that a third line should pass through the same point. (ii)
To find
the area of the triangle enclosed by
tlte
three straight lines
Method i. The equations can be solved in pairs and the coordinates of the vertices of the triangle found, and hence, Chapter I, This method is often useful in the case of numerical 7, the area. examples.
Method ii. The
method obtains the
following
result in determinate
form.
Let the equations of the sides of the triangle be
a.iX
and
+ b.iy +
1'j
=
0,
=
0.
intersect at (<,,f/j).
(iii) Suppose (ii) Then (i) and (ii) at fa, yj.
or, let
us say
AJ
I
We
have similarly J
=
^ = --G! f
l
= y~ =
(i)
(iii) (iii)
and
(i)
at (r^,
y2 \
THE EQUATION OF THE FIRST DEGREE where
^4 2
A
A%, &c., are formed from
,
by permuting the
}
57
suffixes in
cyclic order.
Then the
The product
C,
y! 2 7? 2 C 2
-
. j
determinant and of the determinant
of this
which we denote by A,
M^ + ^/Jj +
A, B,
l
=
area of the triangle
a2 12
c>>
',
is
+
&2
B a + c2 C
value
is
A2
a 2 -4 3
CjG'j,
a^a+bjBj + CaCs
,
tJ
j
A I
j
A
or
so that
;
its
and we have
,
for the area the
j
A
i
(/
A*
-
expression
-
-
^
ij
i 1
G-> 2
C
A
where
--,
cl
,
c2
,
r
i
i
i
(
a/v
If A
iii)
///rev
t'jr
^
,
C
2
,
tj are the
ft,
'
>5
in the determinant A.
5
we
+ lli y -f
straight
C\
,
Here tho coordinates have been taken oblique coordinates
and
a 2 b^" c 2 j
minors of
r
?y i
lines
multiply by sin
L\
=
0,
which
be
i
For
octangular.
(o.
J 2 x -f 7> 2 //
w
to
-f C',
=
?
A ,x -f J^ :
;/
-f
Cj
not concurrent, the equation of
= any
straight tine can be crjwessed in the form
l(A
l
y+1^y^C }^w(A^ + ^i + C }^n(A^x + ^^^ 1
l
Let ^i.r4-7?//4-C'=
be the equation of the straight line which
is
to be put into the required form.
Compare
the coefficients with those of the above equation, then
U\ + m(
r
2 -f
nCj
=
C.
This gives three equations to determine the three constants/, they are in general sufficient to find these constants. straight lines are concurrent we have the relation
A,A
2
It,
R
z
c
c, c,
A,\ B.
-
?n,
n:
If the given
0,
|
l
in
which case we do not is no solution.
there
\
find finite the values of
/,
m, and
n,
and
THE EQUATION OF THE FIRST DEGREE
58 Cor.
If values of
(Ax -f B$
I
is identically true,
/,
-f
m, and n can be found so that the relation
Cj)
+ m(A 2x + B 2 y + C2 )
-f
n (u4 8 x 4-
3 ?/
+ C3 ) =
we must have /
A l -f m^ 2 -f w^4 3 =
=
0,
0;
and consequently
=
B,
0,
the straight lines are concurrent. Thus, for instance, we can see from their form that the straight lines are concurrent. (8) v v~ 0, v w = 0, i.e.
Proposition
iii
of this section can also be proved in the following
manner.
Let the throe equations in abridged notation be iv
= 0, If
and
PQ
Since its
is
CA, any other line,
PC is a straight
equation
Hence lines lu
AH represent these
let J9(7,
PQ
is
is
+ mv =
of the form (p. 52) lu
=
+ mv
0,
P and
in
=
r= 0,
0, v
=
0,
0.
;
= Zw -f wv -H w?; =
or
of
it
e;
a straight line through the intersection of the straight its equation is therefore of the form and w (lu -f
hence if w =
0,
join PC. through the intersection of u ~
let
line
A B meet
u
lines.
v =
0,
w=
mv)
are
-f
nw
0, ;
any three straight
lines,
any other straight line can be expressed in the form
=
0.
the equation
THE EQUATION OF THE FIRST DEGREE Cor. If M multiply the then
U=
0,
=
0,
first
i?
=
w=
0,
0, lu
mv 4 nw
-f
three equations by u EE fae, F
=
F=
case the fourth line
is
J,
TF
wit?;
are
four straight lines,
respectively,
and
let
= nw,
=
represent the three straight lines, and in this represented by the equation
TF
0,
=
m, and n
59
tf+F-f JF=0.
considered
:
when
descriptive properties of a quadrilateral are the following example illustrates the method, which will be
This result
is
useful
referred to again later.
four sides of a complete quadrilateral are represented by the equations u = 0, v =- 0, ir = 0, ?H r-f ic = 0, to find the equations of its three diagonals. If the
Let ABA'B' be the quadrilateral.
Now
the diagonal
A A'
straight lines v
passes through the intersection of the pairs of
=
but the equation can also be written
n
,
i
=
)
[straight line thiough A'\ [straight line through JJ
r-f /r (it
-f
r4
and therefore represents the straight
>M
Similarly the equation can be written
line
u
(?<-! r-f
?<
?r)
AA
=
/r)~r
[straight line through C'] [straight line through 6')
=
and represents CC'.
tH
Also the equation can be written
(n
and represents DB'. Hence f?-fw=0,
=
0,
-{
v
-f
r rr)
-fr
= -
[straight line through #'] ?/'
=
[straight line
through B}
are the three diagonals of the
quadrilateral.
Examples II 1.
What
is
the
value
of
a
if
h.
the three straight lines
a-
4 y -4
3#-j2=:0, x i/-f3a=0 are concurrent? 2. Find the area of the triangle formed by the three straight X-5 = 0, y + 2x-l = 0, #-t/4 1 = 0.
=
0,
lines
THE EQUATION OF TH
GO
FIRST DEGREE
For what values of a are the three straight lines concurrent? 3#4-2ay-f 4 = 0, x + y 3a = 3.
4: The equations of the sides of a triangle are 24t/-7o? = 72 find its area. 5. Prove that the straight lines
3# + 4*/ =
12,
2a;
+y+
l
=0,
5#- 12 = //
20,
:
(6-f (c -f
c)x-lcy
2 a(& 4-6c +
3
c ),
= 6 (c -f ca + a2 a? - dby = c (a -f db )
a)#
2
cay
)
2
are concurrent. 6.
What
,
2
(a
-f fc)
is
the point of intersection ?
-f
fc
Find the condition that the three straight lines
=
2aX-f a\ 3
=
2av-f a>
,
should be concurrent. 7.
\ 2x
Find the area of the triangle formed by the axes and the line = cX, when the axes are inclined at an angle o>. Prove that the perpendiculars of a triangle are concurrent.
+y
8.
Show, using abridged notation, that the three bisectors of the angles of a triangle are concurrent. 10. Prove analytically that the bisectors of the vertical angle and the bisectors of the exterior base angles of a triangle are concurrent. 9.
11.
The equations of three M = 3a?-h4y-7 =
and that of a fourth
straight lines are
t?~4#-f5*/-6 = 59 x + 1 y 21 = 0.
0,
line is
Express the last equation in the form hi
-f
0,
^
#-# + 1=0,
mv + nw =
0.
Find the equations of the three diagonals of the quadrilateral formed by these four lines. 12. The six vertices of a complete quadrilateral are AA', BB', CO', and the diagonals form the triangle PQR. If the equations of the &ides are and u = 0, v = 0, w = 0, w-f v + w = 0, find the equations of , QB\ RC\ of PA, QB, EC. (See figure, p. 59.)
PA
Show 13.
u -f
that PA', QB', KG' are concurrent. a quadrilateral are the four straight lines u + v 4 w = 0, find the equations of its v -f w i;-H0 = 0, u 4- v w 0, 0, If the sides of
=
=
diagonals.
10.
Relations between four straight lines whose equations
are given.
Anharmonic
Ratios.
D
are four points on a straight line, Definition. If A, B, C, then the value of the expression AC.DB/CB.AD, in which the signs as well as the magnitudes of the segments are taken into
THE EQUATION OP THE FIRST DEGREE Anharmonic Ratio of the four points A, B, denoted by (AB, CD).*
consideration, is an It is frequently
There are as
many anharmonic
61 C,
D.
ratios of four points as there are
A, B, C, D, that is 24 ; but if one ratio the others can be completely and uniquely determined For the sake (vide Russell's Treatise on Pure Geometry, Chap. IX). of preciseness we shall refer to the above expression as the anharmonic
permutations of the letters is
known
all
A, J?, (7, D. be noted that, if C divides the segment AB in the ratio Tt should divides the segment A B in the ratio V m', these ratios / in, and ' having sign as well as magnitude as in Chapter I, 6, then the
ratio of the points
D
:
:
'
'
anharmonic
of
'
ratio of
A> B.
C.
D is
I
I'
mmv r
In Chapter I a harmonic range was defined as four points, two which divide the distance between the other two internally
and externally in the same ratio
AC " AD CB BD
:
i. e.
AC.BD_ 1
which, with due regard to sign, gives -
CB. thus,
when
ratio is
the range
AD is
-
'
~ -I
harmonic, the value of the anharmonic
1.
I. If any four concurrent straight lines OP, OQ, OR, OS (called a pencil of rays, vertex 0) are cut 'by any straight line (called a trans-
versal) is the
*
in four points
same for
This notation
anharmonic
A, B,
all positions
is
C,
not universal.
ratio that
D,
the
anharmonic ratio \AB, CD}
of tlw transversal.
Many writers use (ABCD) we should denote by (AC, JBD\
to denote the
THE EQUATION OF THE FIRST DEGREE
62
The
ratio
AC.BJ) _ &OAC.&OBD AD.BC~ AOAD.&OBC
(Euc
*
VL
*'>
_OA.OC. sin LAOG .OB. OP. sin LEOD OA OD sin LAOD .OB.OC. sin LBQC
""
.
.
sn and this quantity
is
independent of the position of the transversal
ABCD. is
This proves the proposition so far as the magnitude of the ratio concerned it can easily be verified, by drawing transversals in ;
different positions, that 0, 2, or 4 of the segments AC, CB, AD, change their sign ; the sign of the ratio is therefore unaltered.
DB The
constant anharmonic ratio determined on any transversal by the pencil is called the anharmonic ratio of the pencil. Cor. i. Since the anharmonic ratio of a pencil of four rays depends only on the sines of the angles between the rays, the anharmonic ratio of any parallel pencil is the same.
Hence the anharmonic ratio of any pencil of four rays is equal to that of a pencil formed by four parallel straight lines through the origin consequently we need only consider pencils with vertices at the origin. :
Cor.
The
ii.
pencil formed by joining any point to four points forming
a harmonic range
a harmonic pencil.
To find the anharmonic ratio of y = px, y~qx> y~ rx, y = sx.
II. (a)
four
is
lines
the pencil
(b)
Ifu of
= 0, v =
the pencil
are
h in the points
coordinates of these points are
ph),
(h,
any two
formed ly
=
The
qh\
AD. BC
ratio
the
*
Let the pencil be cut by a transversal x A, B, C, D.
(h,
formed by
(h,
rh\
(h, sh)
lines
hence
^ (rh-ph) (sh-qh) (sh-ph) (rh-qh)
anharmonic w-fc 2 v = 0,
straight lines, to find the
the
:
u-Jc^^O,
THE EQUATION OF THE FIRST DEGREE
63
Let the lines cut any transversal in the points A, B, C, D let be the point (j^, y t 7^ the point (#2 y 2 ), and let w lf t/ 2 and ^, 2 be the values of w and v when the coordinates of these points are ;
A
),
t;
,
substituted in them.
Let
^=
--
A,
=
IJL
;
then the coordinates of
since the points A, B,
Now,
respectively,
on the four given
lie
lines
we have HI
Mi =
Substituting from
(i)
and
(ii)
(fc,
=
w 2 -fc 2 t'2
(i)>
in
(^ * 3 ^ + A - *4 t^ -f
(iii)
)
(ft a
)
^x (fc 2
and
-*3 - *4
t;
)
we have
(iv)
V2
)
2
(ii),
= 0, =
.
A^ft-jy^...*^
Hence -L-LtJllv/"
-
/7
/-
v
= 77K -r ^rv; = the anharmonic
But -
This result section
D
(7,
is
,
.
ratio of the four lines.
a complete analytical proof of proposition I of this A/^ found is independent of the position
for the value of
;
This proof, though more difficult, is preferable deals simultaneously with the magnitude and the sign
of the transversal
because
it
of the ratio.
Cor.
i.
An important
result follows
the pencil formed by the four lines
it
from this
=
0, v
=
0,
the anharmonic ratio of
:
M
+ Xf
~
0,
=
u + \'v
is
= -1, hence the four straight lines u = 0, For a harmonic pencil A = 0, tt-f At? = 0, M \v = form a harmonic pencil. But given any four are two of them, the equations concurrent straight lines, if n == 0, v = of others can be put in the form lu -f mv = 0, I'u -f m'v = 0. Hence the equations of the rays of any harmonic pencil are of the form mv = 0. u a* 0, v 0, lu 0, lu + mv
A
,.
-,-,
Cor.
y =
or
rx,
ii.
y
The condition that the four straight
= sx
should form a harmonic pencil
lines
is
(p-r)(q-s) + (p-s)(q-r) =0, = 2 (pq -f rs). (p + q) (r s) -f.
y
= px
y
y
= qx,
64
Ex. it
i.
THE EQUATION OF THE FIRST DEGREE Find the locus of a point P which is such that the lines joining
to the
joining
Let
points
it
P be
0),
(a,
to the
(
points
the point (,
#
~ __
(0,
(0, &),
77),
harmonic conjugates of the
0) are
a,
lines
--6).
then the equations of the four straight lines are
y
These are parallel to rj
Hence, Cor.
ii
above,
-+-
i.
e.
the locus of
Ex.
ii.
(
Each
,
q) is
e r
6V + aV = a*6
2 .
diagonal of a complete quadrilateral
divided har-
is
monically by the other two.
AB\
Let the sides
B'A', A'B,
BA
=
u
*be
v
0,
=
0,
w = 0,
w+
r
+w
=
0.
was shown in 9 that the equations of A A', BB', CC' are
Then
it
Now
C'R
is
intersection of
and
w=
a straight line through the AB' and BA', i. e. of u =
0.
It also passes
A A'' and
BB',
i.
through the intersection of e. of v -f w = 0, w v =
equation is therefore w can also be written (t* -f t>)
its
-I-
w
= 0,
(t? -f
w)
;
for this
=
0.
Hence the pencil ffA, C'A',C'R, CC' is - 1^ = 0, w -f ^ = 0, which u = 0, w = 0, from its form is seen to be harmonic.
B
Examples
II
i.
1. Find the equation of a straight line which is the harmonic conjugate with respect to the axes. of o? + 2. Find the anharmonic ratio of the pencil #-fJy=0, wye=0,
y~mx =
= 0, # + jpy = 0. Find the condition that the four straight lines joining the point (#, y) to the four corners of a square whose centre is the origin, and whose sides are parallel to the axes, should form a harmonic pencil. 4. Find the fourth harmonic of the pencil y px = 0, y qx = 0, y rx = 0. = 0, Y = 5. The equations of two intersecting straight lines are find the value of the anharmonic ratio of the pencil formed by Jj-XT-f m^Y^ 0,
x -f ny 3.
X
2
r= o, x+ w r= o, *+ w r= o. f
8
s
*
4
4
:
THE EQUATION OF THE FIRST DEGREE
65
Polar Coordinates.
11.
We
have shown in Chapter I that if the initial line and a perpendicular through the pole are taken as axes of Cartesian coordinates, then (x, y) and (r, Q) being the same point x
= r cos
and
= r sin 0.
y
Since the equation of a straight line is linear in x and general equation of a straight line in polar coordinates is
y,
4rcos0-f J3rsin0 + 0=0. To find
(i)
tfo length
p
the equation
this perpendicular
Let P(r,
of a straight
of the perpendicular to
6)
makes with
it
the
(i)
line in
from
polar coordinates, given tJiepole and the angle a which
the initial line.
be any point on the
ON
line,
the perpendicular to
the line.
Then
POZ=0, hence in the triangle rcos(0
OPN
a) =jp,
(ii)
and
this equation, being true for the coordinates of any point on the line, is
the equation of the
Expanding cos
Cor.
Cor. form since
Of.
is
of the form
.
(i)
cos a -f r sin
= 0,
3, II.)
(ii) is
of the form
straight line perpendicular to
r sin (0 is
(Cf.
jp
straight line parallel to
The equation of any
ii.
a
sin
.
found above.
The equation of any
i.
becomes
a) the equation
(0
r cos
which
line.
-)
(ii) is
of the
= $,
in this case increased or decreased
by \v.
The equation of any straight line through the pole is of the form B =a ot, for any point (r, a) lies on it whatever value r may have. Let PQ be any straight line Cor.
iii.
Ar cos + Br sin and
OR a straight line through
Since the equation
(i) is
-f
C
the origin parallel to
(i) it.
equivalent to
AX+BV+ c the straight line
0,
PQ makes an
o,
angle tan- ( -j^\ with the initial line OZ. 1
THE EQUATION OF THE FIRST DEGKEE
66 The
straight line Off, parallel to
consequently
its
equation
it,
makes the same angle with
also be written ^4
04 B&inQ
cos
=
0.
way we can show that the equation through the pole, making an angle 3 with ^rcos^-f #rsin04 C 0, In the same
is
according as
/ Acos(0 X) + Bain(0X) = the angle is made towards the initial line
Example.
To find
;
(ii)
tf-tair^-^J, which may
OZ
is
(iii)
of the straight line
(iv)
or
away from
it.
the angle between the two straight lines I
- == I
-
and
cos0 + cos0
a
= cos0 + cos0
(i)
/J.
(ii)
These straight lines are parallel to the lines
which pass through the These can be written
cos 6
+ cos 6 - a.
= 0,
cos
+ cos 0^3
=
0,
pole.
2cos}3.c6s(0-iCX) = 0, 2co8}0.co8(0-Jj3) =0,
and i.e.
and 6 The angle between these straight lines is i(3-|3), ami angle between the given lines which are parallel to them. 12.
this
is
also the
Envelopes.
The equation
+ my + n =
can represent any chosen straight line provided that the values which c&n be assigned to 7, m, and n Ix
If, however, there is some given relation between = m 2n) the equation Ix + my + n can represent n and I wij (e.g. one or other of a group of straight lines. only In the simple case Z = w 2, we can replace I by its value in terms of m and n and the equation then becomes
are unrestricted. Z,
m(x+y)
= w(2#
1),
m and n any one of a pencil x+y~Q and 2 #1 = 0,
and
this represents for different values of of lines passing through the intersection of
i.e.
the point ($,
The equation stants (vide
J),
+ my + n=sO
contains only two independent con4); for the sake of simplicity we will discuss the lx
THE EQUATION OF THE FIRST DEGREE equation in the form Ix + my = 1 which can represent any ,
which does not pass through the origin. A relation then between I and m means that
67 straight
line
Ix
+ my =
1 represents
a definite system of lines if we take a series of values for Z, gradually increasing by small quantities, and find the corresponding value or values of m, the line represented by :
lx+my
=
1
and m will take up a series of different positions, each differing slightly from the one before. for these values of
I
Compare the case of a point whose coordinates (x y) are connected by some given equation (e. g. 8# + 4#=l); if series of values are given to x, increasing by very small quantities, and. the corresponding values of y are found, the corresponding point will take up a series of positions each differing slightly from the one before. If a point continuously changes its position, a curve is formed on which all the points lie the more points we plot the more clearly So if the line continuously changes its this curve is indicated. is which all the lines envelope or touch a curve formed position, the more lines we draw the more clearly this curve is indicated. The curve thus formed by a moving line is called an envelope, 9
;
;
It should the curve formed by a moving point is called a locus. be clearly understood that the envelope of a line moving under a constraint or condition is as simple an idea as the locus of a point
moving under a
constraint or condition.
y
(i)
The
figures illustrate
the locus of a point whose distance from of a line whose distance (ii) the envelope
(i)
the origin is constant, from the origin is constant.
THE EQUATION OP THE FIRST DEGREE
68
In (i) two points of the series lie on any given straight line, and in (ii) two lines of the series pass through any given point. In (i) the line joining two consecutive and very near points is said to touch the locus, and in (ii) the intersection of two consecutive and very near lines
To
is
said to lie
on the envelope.
consider this algebraically
connected by any relation, substitute in lx + my
=
l
:
and in terms of
the coefficients
if
I
we can in general find I we thus obtain an equation
;
m are m and
containing
only one arbitrary constant
We
shall consider only the two cases, where the undetermined quantity occurs in the equation (a) in the first degree only, (b) in
the second degree at most. For case (a) we have already
shown
at a point,
a point.
i.
the envelope
e.
is
The most general equation
that all the straight lines
in case (b) is
\\ax + by + c) + K(afx + Vy + e') + (o!'x + V'y + c") where
all
meet
^ 0,
the coefficients except A are supposed known.
Now
through any point (x l9 yj two lines of the system pass, the values of A being given by the equation
A 2 (ax l + fyi + c) + A (aX + Vy v + c') + (a'X + fr'Vi +
These two lines will be coincident (a'^
+ &>! + c')
2
f
l
(a'x is satisfied
+ Vy + c') 2
t/ x )
is
0.
if
= 4 (ax + by, + c) (a' x
but in this case the point (x l9
=
l
+ 6"yi + c")
;
on the envelope; thus the equation
= 4 (ax + by + c) (a"# + b"y + c")
by the coordinates of
all
points on the envelope,
i.
e. is
the
equation of the envelope.
is
Note. In abridged notation the envelope of the line A3 + Av-f w v a = 4uw.
Ex.
i.
To find
the envelope
of a
line
whose distance from the origin
constant.
The equation
of the line can be written
#cosO(-f ysinCX-i*
where p is constant. This can be written
X 1 (x 4- p)
where
Henca the envelope or
is
1
y
=
- 2 X y + (p - x)
A=tan|3(. (x + p) (p x),
0,
0,
is
THE EQUATION OF THE FIBRT DEGREE Ex. the
To find
ii.
(he envelope
of the straight line
are connected by the condition a
coefficients
+b=
_ + y1 = c,
69
1
where
when c
is
a constant. Since a 4 b
the equation of the straight line can be written
aa -a(#-y-f c) + r#
or
where a
is
an undetermined constant,
The values point
(#',
This point
is
(a?',
of
a
for lines of the
y] are given
system which pass through a particular
by a a (x' y' -f c) + ea?' 2
;
for this
is
(x'
- y' + r) a
Hence the equation of the locus of given system of lines, is
0.
quadratic in a hence two lines of the system pass through any If these are coincident, (#', #') lies on the envelope. y').
The condition
This
*= 0,
shown
is
in the figure.
leaf.
(#', y'),
i.e.
the envelope of the
THE EQUATION OF THE FIRST DEGREE
70
Illustrative (i)
AC,
A straight line parallel to the AB in P and Q respectively.
formed by
ABC
the straight lines
the ratio
Take AB,
\2
:
(X+
AC for axes
2
1)
of
Examples.
base
BC of a triangle meets
Show
the sides
that the area of the triangle
BP, CQ,
PQ
(2A + 1),
where X
bears to the area of the
Mangle
of AP to PC.
is the ratio
x and y and let B, C be the points (6, 0), (0, Then the coordinates of P and Q are
Hence
BP is
c).
the line
x
(X
4- 1
(i)
and
B
CQ
is
the line
^ The coordinates
of their point of
X equations
:
by subtraction A
Substitute in
(i)
to find
l>
-
I/ --
=
Ac
tJ
,.
iu
.
are
intersection I
0,
~
1
2\-f
I
1
.
the point
The area of the
triangle
PKQ is then 1
xTi X-fl
X6 2X +
Xc 2X +
1
1
1
X-fl
g
X 5c8Jn^
i
2
X-fl
1
'*(X4-l) (2X-fl) 1 a
X fccsinJ. *
(X
APKQ:
+ 1) 2 (2X+T)'
1
given
(say).
or
X Jf
c
hence
X-fl
i.e.
v
Xb
2X+1
"'
by these
THE EQUATION OF THE FIRST DEGREE (ii)
The area included by the lines bx cos Qi + ay sin a bx cos Z>#
at tan \
The
area =
\
(/3
ay sin
j3 -f
= = =
71
a&, a&,
cos y + ay sin y 6 tan tan i i y) (a (y a)
6 cos
a a sin a
b cos
3 asin/3 a&
3
a&
t
i
asiny & cos
ft
a sin
ft
6cos|9 rrsin/3!
I cos
/3
a sin
j3
b cos y a sin y
cos
ft
b cosy
asiny
b cos
a sin
i
l
ft
ft
9)
(r.
sin
cos/3 sin
cos y
a 1 W sin (/3-
-f*
ft)
(>-) bin (ft-y)
sin
sin y 1
cos 2 sin | (|9 - ft) 2 sin J (y
sin
ft
sin
\ (ft + ft)
- ft) sin
~
(y
+ ft)
2 sin J
1
ft
- <3) cos J (ft
2 sin j (a
- y)
cos J
cos %(y-fi). sin 1
(ft
sin J (y-/3) cos J (y-|9) cos J
(ft
-
2
+ a) (ft + y)
(j3
y) cos }
-y) cos
(ft
-y)
(/3
- ft)
0-y)tan J(y~ft)tani(ft-^), for sin J(/3-f
(iii)
J/
ft)
cos|(ft 4-y) ~-sin|(y-f ft)cos|(/3-fft) = sini(/3-y).
straight
lines
through
through the points A, B, C parallel concurrent, then straight lints
draicn
MN, NL,
respectively to the lines
LM are
LMN parallel respectively
to JSC,
C4,
AB are concurrent.
This is a good example of the advantage of using quite general coordinate axes in dealing with a general proposition. Let A, J9, C be the points (a? y x ), (x^ y 2 )> fe* #s) an(l ^ ^> A' the points t ,
The line 3fJV is *(78-%) Hence the line through A (xlt
yj) parallel to .this is
From symmetry the equations of NL, LM,
the lines through
B and^ C parallel
are
The condition
that
I
(i), (ii),
7i-
|
7a
and
fi-fa
(iii)
^s
should be concurrent,
is
-
0.
" (*?i
" ya (f i - f a) '/a)
to
THE EQUATION OF THE FIRST DEGREE
72
Add
the second and third row to the topj then 2*i(i?-i7*i 0.
- 93) - Sy t ( 3 - 8 ) = 0, - - ~ (% - fc) (& &) (ij! * t ) (& &) 2 a?!
Hence either or
which cannot be condition
(/ 2
true unless L,
M,
N are
collinear
;
therefore the required
:
is
ZSifoi-'hJ-Syitf.-W-O.
(iv)
From symmetry to
BC, CA,
AB
the condition that the lines through L, 3f, should be concurrent is
2&(y8 -y.)-2i7 X (^-^ = But the conditions
(iv)
and
(v)
are identical
;
^
parallel
0.
(v)
this can be at
once seen by
expanding the terms.
Hence the required proposition
(iv)
A
point
P
divides
established.
is
a fixed
line
straight
AB
in
the
ratio
i:p m, and Q divides another fixed line CD in the ratio p + and n are constants and p is variable. Find the locus of the where middle point ofPQ.
n:pn
m
P
Take AB, (a, 0),
(&, 0),
CD
for coordinate axes
(0, c),
(0, d).
and
2p if
Jf?,
J'
the mid-point of PQ,
m(a By eliminating p we
is
of
A, B, C,
i' (a?,
2p
#),
~~ V)
__c + d
n(c
get
a~b m(a-b) the equation of a straight
line.
D
P and Q are
c(p-n
I
>U \ >
Hence
let
The coordinates
B
f
~^T(c^d)
d) ~'
be the points
THE EQUATION OF THE FIRST DEGREE (v)
A
sum of
straight line moves so that the
on the axes
Show
is
and equal
constant
that the locus
lines at
=
c cos w,
of coordinates containing the angle
(x+y)
1
o)o:
(c
""""
*
-L
(c-a)a? +
or
+cos
cu)
_i
"
"""*
(1
o>.
Let the equations of the two straight lines be * y _y 1.5, ~~ ^ t a c-a c ,
it
of intersection of two such
right angles to each other is the straight line the axes
made by
to c.
the point
of
the intercepts
73
0y-a(c-a) + 0y o(c 0)
>
0,
(i)
= 0.
(ii)
Cross multiplying to find the coordinates of their point of intersection
x
we get
f/
ao(o-a)
ab (i)
and
(ii)
(c
a)(cb)
c
are perpendicular
a) (c
(c
1
y
or
But since
c(6-a)'
(c-a) (c-0)(o-a)
x
b)
+ ab
*=
{o(c
a)-f-a
(c
[Ch. { (c -a)(c-b) + ab}\ (x + y) (1 + cos a>)
i.e.
Hence (iii) which is the required (vi)
A
A
(h,
k)
c
cos
8
(ii).]
a>,
locus.
straight line
with the axis of x
=
II,
;
OP is
drawn through
a straight
making an angle 26
tlie
origin
making an angle
AP is drawn through the point same sense as 6) with OA produced.
line
(in the
Prove that the locus ofP, the intersection of these straight
h
lines, is
k
.
the axes being rectangular.
Let the angle so that tan 6
#
x
LOA .
so that tan<
Then the angle
OPL
=r
,
and
let
Pbe
the point
(a?,
y)
;
THE EQUATION OF THE FIRST DEGREE
74
and
OA
sin^OPA
OP
sin
/.
CAP
_
i.e.
8*
""
sin/GPL
n # cos
0P~~ o
i.e.
^ O^P
sin
-f
^* _ ~~
'
8in~<9
*ft
=
or
OP sin
~~
Piwd
cos
COB 6
^ coa OP cos ^
_
OP1
(vii)
sin
2 sin (9 cos
OP
with the lines
cos
sin20
t
x
y
the equations of two straight lines which
y
= x tan
them ivhose areas arc
/3 -f
c
;
y
c 2 , f/^e 00:^5
-==
x tan y -f c
awrZ
make equal angles
/om triangles with
of coordinates leing rectangular.
pass through the point (0, c), and make angles a?-axis Let AB, A'B' be the required lines with the y respectively inclined at an angle 6 to the #-axis and equally inclined to AA' BB'. If p be the perpendicular from the vertex C on these lines, the area
The given
lines
AA\ BB'
,
y
of the triangle
CAB
or
CAB'
is
Now
Hence
if
AB is the straight line
the perpendicular from
on
it is
--y)
or
(0, 0)
equal to
3,
i.e.
But
c
2
p=
c-Y/cot$(/3--y).
THE EQUATION OF THE FIRST DEGREE = c sin $ (ft + y) + c Hence q and the required equations are There are two other straight
-y),
75
.
which
lines perpendicular to these,
satisfy
the conditions of the problem. (viii)
//
P
and
l
P2
are the feet of the perpendiculars drawn
+ w y -f w a =
(x', y') to the straight lines liX
1
from the x + + m^y w 2 = 0, 2
0, t point 2 2 2 lies in the ; hence show that find the value of OPt + OP2 obtuse or acute angles between the straight lines according as the ex-
P^
pressions (Z 1 #' + tn^'-f same or opposite signs.
n^ x
(Z 2
#
/
+ w 2 ?/ + n 2 and )
Z
1
7
2
-l-m rn 2 have the ]
Since the question is one of sign, it is better to proceed in the most straightforward manner possible, and thus avoid any unnecessary complication of signs. Let the straight lines y intersect at A.
The equation
of
AP
l is
so that the equation of OP, is
The coordinates
of
the form x' +
l^
AP
we
Pj
on
lies
We
l
P
l
are therefore of
y' + w^fc, and
find that
k
since
is
shall use the following abbreviations
:
jy'-HJj, ^t/' -f
Then the coordinates
P
of
l
LL*
x -
1
,
n,
,
are ,
1
Similarly, the coordinates of
,
and
g
-,
P
a
y
-
are
Hence *
-f m
2 i
.
(
*A
IV-fWj
1
_
*A r 2
V-fma
)
mL >
(
i
8
U^H-wii
L _ W ]* * a J2 + m a )
Now the angle Pi#P2 Is obtuse or acute, and consequently P^Pj is P9* is negative or positive, acute or obtuse according as OP18 -fOP,f 1 i, e. according as LjL, and J^-f t^m, have different or the same signs.
-P
THE EQUATION OF THE FIRST DEGREE
76
A straight line moves so that the product of the perpendiculars on it
(ix)
from two fixed points
constant
is
Let the two fixed points be
The condition
(-a,
0),
(a,
Ix +
straight line
find the equation of its envelope.
:
my +
1
0),
and the equation of the
= 0.
(i)
gives us that
(-/a-fl)
(fo.fl) L \-=
v
'
- constant - c2/(say), .
.
,
2
.e. a s a a P(c -fa )-fc m
or
Now
the values of
I
w
and
*l.
(ii)
which
for those straight lines of the system
pass through some particular point (x^
,
t/,)
are given
by
+ l =0 Q 2 2 a J*(c + a ) + c w =l.
(iii)
/ai+itift
and The values of
(and note that from (iii) to any one value of one and corresponds only one value of m) are thus given by i.
I
(iv) I
there
P {(c2 -f a8 ) y* -f e
e.
If the two values of I given by this equation are coincident, two coincident straight lines of the system intersect at telf yt ), and (xl y,) is on the envelope. ,
The condition
^
for this is
-
i.e.
or
_
+ aa ) yi -f c^,8 } {c2 -^},. 2 * a 2 c (c + a ) yt - (c -f a ) y t fxftf, 2 f 2 1 2 1 2 2 c ^ -f (c -f a ) y! = c (c + a ). 2
a
{(c
2
9
Hence the equation of the envelope
is
(x) ABC is a triangle, and any straight line cuts the sides JSC, CA, AB in the points P, Q, my other point, and OP cuts AB, AC in D and JB, OQ cuts BC, BA in F and G, OR cuts CA, CB in J, H. The straight lines JG, HD intersect in
is
A
No is
we
special details of any of the lines in this question are given there this property of the triangle is purely descriptive ; ;
nothing metrical
shall therefore use the abridged notation.
:
THE EQUATION OP THE FIRST DEGHIEE PQR be
Let
triangle be
the straight line
=
u
0, v
=
0,
Now DE, GF> JH are of lines
(u, x),
(t>,
#),
w=
let
the sides J5C, CA,
AB of the
:
+ x=
0,
mv 4- x =
0,
lu
= 0. = mt> -f nw -f + a?
nt0
Now
and
0,
0.
straight lines through the intersection of the pairs let their equations be
(w x) t
x
77
the equation
a:
can be written in either of the forms
(nw + x) v
= 0,
+ x) = 0;
therefore represents a straight line through the points of intersection of and nw + x = 0, and also of w and wt?4 x 0, i. e. through the points J and G.
it
v
=
=
=
Hence the equation of JG is mv + nw + x = 0. In the same way the equation of HD is lu 4 nw + rr
= 0.
=
Now
the equation (/ti-fnw-f a?)-(wi?-f nw-f a;) represents a straight line through the intersection of ,76? and HD, i.e. through 0'. But it reduces to
lu-mv~
of u
=
0,
and
?
i.e. it
=
0,
i.
represents a straight line through the intersection e. through C. Hence the equation of CO' is
mv
lu
But i.e.
is
the point of intersection of
of
Ju
and therefore is
+#
=
4-
x)
(7u
a straight line through 0. But this equation reduces to
CO'.
In other words
COO
1
is
0,
=
ft
DE and GF, wt>4a? = 0,
- (mi? 4
lu-mv
a?)
= 0, i.e.
the straight line lu
Examples II 1.
lines
= lies
mv
on the straight
line
= 0.
j.
Show that the feet of the perpendiculars from the = # + y-4 = 0, a? + 5y-26 0, 15tf-27y-424
origin to the straight arc collinear.
line is drawn making an angle of 30 second straight line is drawn making intercepts 3 and 5 on the -positive directions of the axes of x and y respectively. Determine the distance of their point of intersection from the origin. 2.
Through the origin a straight
with the axis of
a?.
A
3. Find the locus of a point which moves in such a way that its distances from the straight lines 2o:-y-f5 = 0, 4a;-2y-8=0 are equal. 4. Find the acute angle between the straight lines 12a?-~5y-5 0, 6-3# 4y=0 and the equation of the straight line which bisects the obtuse angle between them. In which angle does the origin lie ?
Through the origin three
straight lines are drawn, making angles 30, the of Find the coordinates of with axis 150 x, of length 4 units. 120, their extremities and those of the centroid of the triangle of which these 5.
extremities are the verticei.
THE EQUATION OP THE FIRST DEGREE
78
6. Find the length of the perpendicular drawn from the point (2, 2) to the straight line joining (3, 1), (7, 4). 7. Show that the straight lines joining (3, 0), (3, 4) to the point of are intersection of the straight lines 19# + 3y-29 = 0, 13#+ lly-27 =
at right angles.
Find the equation of the two straight lines which pass through the = 0, 2a?-/ + 3 = 0, and are such that the perpendicular from the point ( 1, -1) on each is of length 6/5. 9. Prove that the line joining (1, 1) and (23, 4) passes through the 8.
intersection of the lines a;-f/-f 2
intersection of #4- y
7
=
and 2#
6
3y
=
0.
Find in what ratio the straight line 3#-f 2y between the points (6, 5), ( -3, 2). 10.
=
7 divides the distance
11. The coordinates of ABCD are (3, 4), (6, 3), (5, 7), and (4, 6). Find the equation of the straight line which joins the mid-points of AC and BD. Show that it cuts AD, CB in P, Q respectively such that
CQ/QB =
AP/PD
5.
Find the area of the quadrilateral whose vertices are (2, 1), (4, - 3), (2, -5), (-1, 4), the axes being inclined at 60. 13. Find the polar coordinates of the foot of the perpendicular from 12.
(3, 0)
14.
on 2x~
the line
T
makes on the axes
Ox
in such a
being the
initial line.
way that the sum of the
intercepts
Find the locus of either of the points of trisection of the portion intercepted between the axes. 15. Find in their simplest form the equations of the straight lines joining the following pairs of points it
is
4.
:
(i)
(ii)
(aw
2 ,
2 aw), (aw 2
,
2 an);
(am, a/m), (an, a/n)
;
(acostf, &sin0), (acos, fcsin0). the equations of the straight lines from the vertices of a triangle (iii)
16.
Find
perpendicular to the opposite sides, *given the vertices
(3, 4),
(1, 5),
(6, 7).
At what point do they intersect ? 17. Find the equations of the medians of a triangle whose vertices are (3, -2), (-6,5), (4, -7), and find the point where they intersect. 18. A straight line is drawn from the point P(a, b) in a direction inclined at an angle Of to the axis of x to meet the straight line x/a + y/b = 1 at Q. Find the length of PQ. 19. Show that the straight line 6sin(a-f $)x = a cos ^ (a 4- ) y bisects the distance between the points (a cos a, b sin a), (acosft b sin 3). 2 ; 20; The equations of three lines are 5x-y = 7, y = 7#-5, y-f 4x
find the length intercepted 21.
xx^ + yyi 22.
on the third by the other two. t#i y\) which is such that the straight
Find the locus of a point
A
angles
;
=a
-passes through the fixed point (h> k) for all values of (x l9 y,). straight line of given length slides between two lines at right find the locus of a point dividing it in a given ratio.
Find the coordinates of the centre of the triangle whose vertices are (2, 3), (3, 4), (6, 8). 23.
line
2
circle circumscribing the
THE EQUATION OF THE FIRST DEGREE 24. If
Ax + By =
1,
ax + by
=
1
79
are parallel, find the distance between
2 2 them, given a -f b = c* and -4 = ka. 25. Determine the locus of the intersections of the straight lines given by tx/a -y/b + t = Q, x/a + ty/b -1 = 0, t being a variable.
26. Two straight lines cut the axis of x at distances a and -a, and the axis of y at distances b and 6' from the origin. Find the point of intersection. 27. If a number of such pairs of lines be drawn with a the same for all
=
a 2 find the locus of their intersection. 28. Find the area of OLM, where are the feet of the perpendiculars let fall from the origin on x C08 = 4 y 8i n = jp a a? cos ft 4 y sin (X 2 2
and
bb'
,
LM
^ ^
^
,
.
29.
\y
=
30.
Find the locus of the orthocentre of the triangle whose sides are x + a\ 2 fty = x -f a/* 2 vy = x + at>2 for different values of A, p, Find the coordinates of the circumcentre of the triangle formed by *>.
,
,
,
the lines x + 2y + 3 = 0, 2a? + 8y + 1 = 0, 3#-f 5y + 2 = 0. 31. Find the coordinates of a point which is such that the line joining it to the point (7, 4) is bisected at right angles by the line 3-r-y = 7. 32.
Find the orthocentre of the triangle whose sides are
x/m+y/p-l =
0, a?/n-f
y/p-l =0, y
=
0.
33. Two straight lines have equal angular velocities <> in opposite Find the polar equation of the locus directions about two points 0, A. is pole, OA initial line, inclined at JTT to OA. 34. Find the angles which the straight lines
of their intersection
when
and
OA =
a,
and each
line is initially
make with
(i)
rsi
(ii)
//r
(iii)
J/r
=
cos 8
-f
e
cos (B + CX)
the initial line.
Find the polar equation of the locus of the feet of the perpendiculars from the pole on a straight line which passes through the fixed point (p, (X). 36. Find the polar coordinates of a point P which is distant d from a point Q (p, a) when PQ makes an angle 3 with the initial line. 35.
37. Find the polar equations of the straight between the following pairs of lines
lines bisecting the angles
:
(i)
(ii) (iii)
(iv)
38.
Thd
e
=
a, 6
rcos0
=
-
ft
;
sin0-8in(0 + a) r cos (6
= 2J = 0, cos + cos (0-0) =
rsin^
f>,
a)
=
jp,
r cos
p cos
;
(X,
from a point P on the = is x to the line joining the feet y parallel 0. P on 3# + 4y-12 and 4#-3y + 8
line joining the feet of the perpendiculars
y = 0,
two
fixed straight lines of the perpendiculars from Find the equation of the locus of P. 39.
Find the area of the triangle formed by the straight rcos (4-3)
=j>,
rcostf
=^aec
ft,
lines
rsin(0 -a) =j>.
THE EQUATION OP THE FIRST DEGREE
80
x tan tf,, y = x tan a a triangle lie on the lines y the at circumcentre the t/ being origin: prove that the locus of the orthocentre is the right line a?2sin0-yJBcos0 ~ 0. The
40.
=
vertices of
a?tan0s
,
,
Pbe two.points on Ox, and JB, Q two points on Oy, ^4 and B being P and and varying in such manner that 1/04 - 1/0P= I /OB - 1/0 Q, fixed, show that PQ passes through a fixed point. 42. The lines l(l-a)x + bly = l-a, 41. If A,
m (m
a)
x + &wy
=m
a
make a constant angle ft with one another when I and m Show that the locus of their intersection is 2 3 (bx + ay) tan (ax by + 1 / 4 ax.
vary.
Of.
Find the perpendicular distance of the point
43.
rcos(0-(X)
(i)
SB
p,
(ii)
= l/r
(/, ff)
from
cos0-f ecos(0-a).
a straight line moves so that the sum of the perpendiculars let fall on it from two fixed points (8, 4), (7, 2) is equal to three times the perpendicular from a third fixed point (1, 3), show that this line passes 44. If
through one of four fixed points.
AB
45. A straight line makes intercepts OA, OB on the axes of x and y of lengths 4, 3 respectively if P is the mid-point of AB, find the equation of OP referred to BA as axis of x, and the perpendicular from (7, 8) upon ;
BA
as axis of y. The vertices A,
46.
B
of a given isosceles triangle
ABC, right-angled at C, perpendicular straight lines OA, OB. feet of the perpendiculars from A, B on OC are NC. CX is constant ; (ii)
move one on each of two OC meets AB at X, and the
MN.
Prove that
P
47.
is
and -4 2
l
of
A
l
B^
48.
CO
point,
.
OM
t
f y * n -#i on the straight
an(* the &x * 8
A% B
l
lies
The ends
whose equation is y = mx, and are drawn meeting the axis of x in points
on the straight
P any two straight lines
through
A
any
(i)
fixed
BC of the base
^* line
line
Prove that (he point of intersection
whose equation
of a triangle are
Find the locus of the vertex when (i) 49. If the coordinates of two points
(a, 0),
AB -A&=*c>\
is t/-f
- a, (
be
(ii)
(2, 3)
0.
0).
A& + A(?
1
A and B
w# =
and ( - 1,
2
c
a .
4), find
P and Q in AB and in AB produced respectively which AP/PB = AQ/QB = X. Hence find the ratio in which AB is divided by x + y = 6.
the coordinates of the points for
50.
If
cx
= 0,
/3
= 0,
y
=
represent in Cartesian coordinates three + ny holds for
straight lines such that an identical relation Zot + m/3 all values of (x, y), then these lines meet in a point.
Hence show that the
bisectors of the angles of a triangle
~
meet by threes
in four points. 51. (i)
Find the equations to
The two
angle 45 with (ii)
two
The
lines
straight lines through the point (-4) 3) which
make an
Sx-y + 5 ~ 0.
straight lines through the origin, each of
#+y * 0, 2#-3y
which forms with the
4 a triangle whose area
is 5.
THE EQUATION OF THE FIRST DEGREE 52.
Prove that
(1, 2) is
81
the centre of one of the four circles touching the 4a? + 8y-15 0, 0, 5a?-12y + 6 0,
three straight lines 8a? + 4y-16 and find which of the four circles it
is.
Given two straight lines 8a?-4y-f 5 ==0, 3a?-7y-80, determine the equation to the straight line through their point of intersection making the same angle with the first straight line as the second does on 58.
the opposite side of 54.
A
55.
AB,
CD
Prove that
PQ
it.
variable straight line through the fixed point (/, g) meets the axes of coordinates in P, Q. Prove that the points of trisection of PQ lie on one or other of the loci whose equations are Sxy - 2gx -fy = 0, 8 xy gx - 2/y 0.
56.
are two finite straight lines P, Q are their middle points. in the same ratio. and :
divides
AC
BD
Find the conditions that the straight line joining the origin to the = aor-f fcy-f c should 0, a'x + b'y + c'
intersection of the straight lines bisect the angle between them. 57.
ABCD
(\/5, ITT),
is
a rhombus, and the polar coordinates of ABC are (4, JTT), Find the coordinates of the remaining corner, and the
(4, |TT).
equations of the sides and diagonals. 58. Find the locus of the intersection of two straight lines which pass through (a, 0), ( a, 0) respectively and include an angle of 45. 59.
Prove that if the three straight lines
ax sec 6 by cosec ax sec by cosec ax sec ^-6y cosec ^
are concurrent, then sin (6
-f
^)
-t-
sin
-f
(<
6)
-f
= c*, = c1 ,
sin (^
4-
0.
<)
The equations of two parallel straight lines are 4a?+8y12, 4# + 8y = 8 obtain the equations to the straight lines which pass through the point (-2, -7) and have a length 8 intercepted on them between the 60.
;
parallel straight lines.
Find the coordinates of that point on the straight line 2a?-y-5 = the sums of whose distances from the points (19, 18) and (9, 8) is least. 62. The vertices of a triangle lie on three fixed concurrent straight lines, and two sides pass each through a fixed point prove that the third side 61.
:
passes through a fixed point. 68. Show that the lines
10*-y-301 are concurrent, and find the anharmonic ratio of the pencil they form. 64. A triangle is formed by the axis of x and by the straight lines whose
l0;
find tne equation of equations areitf+^y^l and i#-iVy + the locus of the centre of the rectangle inscribed in the triangle and having one side on the axis of x. 65. (xl9 y x \ (a?a ya ), (*ff y8 ) are three points A, B, C, and are such that ,
the straight line BC is that the equation of AB 197
a5 and CA xxl + yy l xxl 4 yys * a*. ,
is
F
is
ff;ra
4 yya
a*.
Prove
THE EQUATION OF THE FIRST DEGREE
82 66. 67.
What curve do all lines of the form aX 2 -Ay + # * touch ? touch ? What curve do all lines of the form \*x-a\ + y What curve do all lines of the form (x cos X)/a + (y sin X)/6 = 1
touch ? Find the envelope of a straight line PQ which meets the axis of y in Q, and is always perpendicular to SQ, where S is the point (a, 0). 70. u4 JBC is a triangle, D, #, F the feet of the perpendiculars from A B,C respectively on any straight line. Prove algebraically that the perpendicular from Z), E, F on EC, CA, AB respectively meet in a point. 71. The sides of a triangle ABC are u find the 0, v = 0, w = 68.
69.
y
;
0, equation of the straight line joining A to the intersection of lu + mv lu + nw~ 0, and the equation of the harmonic conjugate of this line with
= and w = 0. Find the equation of the straight intersection of the two pairs of lines respect to v 72.
tj
=
where
w-ff-ft^
) r
line
which passes through the
e=0) *
' I
r, w are abridged When the axes are
forms of the equations of straight lines. oblique (o>) find the equations of lines through (P>
,
from the point
y) to the straight line lx+my + n h x (tx' + my' + n) k~y'
(#',
75. If a triangle
A
0,
then
ABC remains similar to a given triangle, B move along a straight line, find the
and
if
the
and
equation of fixed, point the locus of C (polars). 76. Find the lengths of the sides of the triangle formed by the lines x cos 0< -f y sin Oi = p, x cos ft + y sin /9 = q, x cos y + y sin y = r. Prove that the area is
be
2 sin (0 - ?) sin (y .
and
B are
ZJf a
)
sin (a
-
)
are Points P, such that PQ is of constant length, AP, BQ meet in K. taken in Find the locus of R as P<) moves along LM. 78. P, Q are two points one on each of two fixed straight lines at right angles, such that PQ subtends a right angle at a fixed point. Find (a) the locus of the middle point of PQ; (b) the envelope of PQ. 79. If the coordinates of the vertices of the triangle ABC are (xl9 y1 \ and those of the triangle (&, ft), (#a> V*)> ( x*> ys) respectively, w e tliat h 8 perpendiculars from A, B, C to B'C, C'A' (* *!*)> (fs> I*)* 77. .4
fixed points,
fixed straight line.
LM
ABC
^
1
A'B' are concurrent if
What
1
t
1
1
1
1
0.
can we infer from the symmetry of this result in the two sets
of coordinates ?
THE EQUATION OF THE FIRST DEGREE
83
80. Pour straight lines a, b, r, d being given, show that in general one and only one straight line can be drawn meeting them respectively in = BC = CD. points A, B, C D (in this order), so that AB 9
Discuss exceptional cases. 81. P, Q, R are three points in the sides
ABC such
BC, CA,
AB
of the triangle
that
BP:PC=l:m; CQ QA = m
_ :
AP, BQ, CR
are joined
:
n
;
AR RB = :
n
:p.
show that the area of the triangle formed by
:
these lines
___
82.
(mn + mp -f np) (nl + mn + mp) (nl + ml + mn) The triangle ABC is formed by the lines a^-f btf + Ct = 0, a 2x + 6 2y -f c2 = 0, a 3 a? -f 63 -f c s = 0, t/
lines joining the vertices to the origin meet the opposite sides in D, E, F. Show that the sides of the triangle intersect the on the straight line corresponding sides of
and the
DEF
ABC
x
(ajc,
+ a 2 /c.a -f a s /c 3 ) + ij (bJ Cl + 62 /c 2 -f 6 3 /c 3 ) -f 3
=
0.
83. PQ, a bar of fixed length, slides between two bars intersecting in 0, and from any one point in PQ lines are drawn in all directions and move in rigid connexion with PQ: show that there is a point in each of these lines which will trace out a straight line as PQ moves. 84. One side of a quadrilateral is fixed and its length is 2k, the adjacent sides are each of length a, and the opposite side is of length 2c. Prove that the equation of the locus of the middle point of the last side may be written in the form J 0' 2 + & 2 ) 8 -4fcs a (r> + & 2 ) = 4c 2 fcy -4fcV, where 2 = aa #a + y2 and the fixed side is taken as axis of x, and b a = c a 4-A; a *
,
a-
,
a perpendicular line through its mid-point as axis of y. 85. The triangle AOB has the angle at equal to
o> and JT for its O y show that the equation of AB referred to OA, OB as axes of x and y is #/(# -f # seco>) +y/(y + ^o sec ) = 1> an(* * ne further end of
orthocentre
the diameter through
where (#
,
:
-f IJ Q
cos w)/(y l
of the circumcircle of the triangle -f
x
v
cos
)
=
(y
+ .r
F 2
cos <>)/(#!
4- //!
cos
CAB = o>)
is (x^y^) cos .
CHAPTER
III
EQUATIONS OF HIGHER DEGREES. CHANGE OP AXES 1. THE equations of several straight lines can be combined into a single equation : thus
(Sx + 2y-l) or is
= 0, 7y-2 = 0,
(5*-3y + 2)
15x* + xy-6y* + x+
evidently satisfied by the coordinates of any point on either of the
Htraight lines
% x + 2y -
and
5#-3# + 2 =
1
= 0, 0,
and conversely the coordinates of no point can unless it is on one of these lines. Any equation in the variables x and y,
/(*
y)
satisfy the equation
= o,
will then represent straight lines, into factors of the first degree.
if,
and only
if,
/(#, y) breaks
up
A
single equation may represent partly straight lines and partly is drawn, there will be some curve. If the graph of /(#, y) a straight line in the figure corresponding to every linear factor of
=
2.
A
th Iwmogeneous equation of the n degree represents n straight
lines through the origin.
Let
w a # -f- ajaf -ty + a a s n
-
V
be any homogeneous equation of the w Divide the equation by t/ thus
-f
nth
. . .
-f
any*
=
degree.
,
this is
an equation of the nth degree in
roots, real or imaginary.
(
) and consequently has
\y/
n
CHANGE OP AXES a2
If the roots of the equation are o^,
86 then the equation can
*>
written
hence the original equation represents the n straight lines
0^ = 0,
x e.
i.
n straight lines,
#-0(2 y
= 0,
= 0,
...a-o^y
real or imaginary, all passing
through the
origin.
=
If a 0, r = represent two straight lines, a homogeneous tb of the n degree in u and v must represent n straight lines through equation the point of intersection of these two straight lines.
Cor.
i.
In particular the equation
a,(x-ay ^a l (x^a)^\y-l) + ^(x-a)^(^b)^...+a n (y-b) i
represents n straight lines through the point
Cor.
ii.
An important method
arises
n
^Q
(a, b).
from the
result of this paragraph.
Consider the equation
ax + 2hxy + by* + 2gx + 2fy + c t
whatever locus
it
it
Q;
(i)
represents, the straight line to +
cuts
=
in two points, which
my + n =
(ii)
be found by solving the equations (i) and can at once write down the equation of the
may
Without thus solving we (ii). straight lines joining the origin to these points of intersection ; forthe equation (Hi) is
homogeneous
in
x and
//,
and
is
of the second degree:
represents two straight lines through the origin. But the coordinates of any point common to the loci
(i)
therefore
it
and
(ii)
satisfy
the equation (iii): hence this equation represents the straight lines joining the origin to the points of intersection of (i) and (ii). The same method can be used with equations of a higher degree.
3. (<*!
x + ty +
when (i.
Consider the equation
e.
(a 2x
+ 6 2y + c 2
)
(a^x + 6 3 y + c3
)
.
. .
(ax + b n y + c n
)
=
;
(i)
the factors are multiplied together the terms of the highest
nth ) degree are obtained from the product
Now a^+biy = is a straight line through the origin parallel to &iX + b l y + cl = 0. Hence the equation (ii) represents n straight through the origin parallel to the n straight lines represented by equation (i). This result is of great importance : we see that if any equation represents straight lines, the terms of the highest degree lines
EQUATIONS OF HIGHER DEGKEES
86
equated to zero represent a system of straight lines parallel to them through the origin. Any question dealing only with the directions
by a single equation can thus be by considering parallel straight lines through the
of straight lines given
at once sim-
plified
origin.
Example. To
find the equation oftivo straight lines through the point
3) parallel to the straight lines
(2,
-2 = The equation
15.
2
-f
0.
(i)
represents two straight lines through the
xy-bif
origin parallel to those represented
by equation
(i)
hence
;
represents straight lines through the point (2, -3) parallel to
(i).
4. Most of the properties of equations representing straight lines can be investigated by comparing the equations with the product of linear factors, and the majority of problems on them are little more than algebraical exercises in the comparison of coefficients.
When
the axes are rectangular the directions of the lines can be x tan 9 in the equation to parallel straight lines through the origin this gives an equation for tan 0, and the
found by substituting y
= ;
values of 6 thus found give the angles which the straight lines make with the axis of ,7. give some illustrations of these
We
points.
Ex.
i.
What
is the
equation
perpendicular respectively
to the
o/n strand n
through the point
lines
k)
(It,
straight lines given by the equation
= 0.
w
Let
= (a^ %) -f
The
n -f
;>!#"-
(a^c
V iw n -y + -f
+ brf) (a& + 68 y).
straight line perpendicular to a^x
-f
1$ =
.
.
,
.
.(a n
x -f bn y).
a$
is
equation of n straight lines parallel to those required
which is obtained from the right-hand side of and -a? for y.
(i)
bfl
(ii)
=
hence the
is
by substituting y
(ii)
for
a',
Since the right-hand side is identically equal to the left, the equation of lines through the origin perpendicular to the given straight lines obtained by making the flame substitution in (i) this gives
n straight is
;
and straight
;
lines
through the point (h k) parallel to these are }
y^^^^
CHANGE OF AXES Ex.
Find
ii.
the conditions that two
ay?
should be perpendicular
+ cxy
bx* y
-f
87
of the straight
lines
-f dy* =
2
one another.
to
The equation of the three straight lines perpendicular respectively to the a? for y, given straight lines is found by substituting y for x and i.
dy?
e.
- cx*y -f bxy* - ay3 =
Hence the two expressions ax* + bx*y 4- cxy 2 -f a*y 3
(to
,
8
cx*y
0.
-f
&#y
a
3
at/
common
quadratic factor, since each of the two perpendicular becomes the other in the equation of the perpendiculars. Add a times the first to d times the second, and take a times the second from d times the first then the common quadratic factor is also a factor
have a
straight lines
:
+ d' )x'> + (ab-cd)xy + (ca + 6d)y 2 } 2 - cd)xy + (a 2 + a?)y2 (bd + ac)o? (ab l
of both
x{(a*
and
y\
)
;
consequently the quadratic factors in these are identical, a&
cd
- (a6- cd) a
i.e.
2
ac ~~
W
+ dP + &rt-f ac
=
0. :
+ ctan 2 + 6tan^-f a
o*tan s
a2 -f cP
a 2 + d*
The solution can also be obtained as follows put y ** #tan0 in the equation, then
i.
-f
if
=
0.
This must give two values of which differ by a right angle, e. two values of tan B whose product is 1.
Hence
the roots of the cubic
if
dt*
are
t
ly
t
But
2
*3
,
W
and since
or a8
4-
t
l
* * 2 3
= -
/.
is
?.
=
iii.
;
^1
=
Put y
=
^;
=
;
or a2
-f
ac
we have
+
W
4-
d2
0,
this special case is left for the reader's consideration.
Find
the axis
= #tan0
which the straight lines
the angles
5^(^ + make with
+ ct* -f bt 4- a
- 1.
a root of the equation,
a2 c 4-a6d + ad 2
unless a
Ex.
8
we have
2 2 )
-20y 3 (rc2 4.y 2 )-f 16t/
5
=
of x.
in the equation, then
5 a 8 2 5tan^(l-ftan^) ~20tan ^(l + tan ^)-fl6tan ^
Multiply throughout by cos
5
0,
then
5 sin 6 -20 sin 3
6+ 16 sin 6
=
0,
0.
0,
EQUATIONS OF HIGHER DEGREES
88
1 4 sin 6 {16 sin 6 - 20 sin 6 4 5}
hence
sintf-f
.*.
0,
2coB20sin0+2cos40sin00;
sin04 8in80-sin0 + sin50-8in3tf
.-.
0,
sin50=0,
i.e.
hence 50
(n being an integer), B are 0, 86, 72, 144,
180n and the values of
288.
should be carefully noted that any other value of n gives a value of 6 corresponding to one of these directions.
N.8.
It
Examples III 1.
=
xy
;
(ii)
a?'-4y*
-
0;
(v)a?'-2a?y + y' 2.
Show
that
(iii)
= 0;
:
#8 -7a?y + 10y* (vi) x*-2^/8xy + y*
;
2#7 + 3#y-2y*-a; + 8y-l =
lines at right angles, 3.
a figure
loci in
Represent the following
(i)
a.
(iv)
x*-xy
;
6.
represents a pair of straight
and draw them.
Find one equation representing the diagonals and sides of a square
referred to convenient axes. 4. is
The centre of a square
parallel to the axis of x,
6. 7.
What
does the equation
Draw the locus Draw the locus
What
the point
(a, a),
one of
its
diagonals
side is of length 2 a. its and diagonals. sides equation representing
Find an 5.
lies at
and each
afy*-aV-aV
y-aV-6V
+a*62 8 = 0. (ff-a) -(y-&)*
-f
a4
=
represent ?
0.
the equation of parallel straight lines through the origin ? lines through the origin perpendicular to the lines #* + 2ta?y y* = 0. 8.
is
Find the equation of a pair of straight
What do you
conclude from your result ? Find the equations of pairs of straight lines through the origin ~ 0. 0, (b) x-y making an angle 6 with (a) x + y 10. Find the angles which the straight lines given by the following equations make with the #-axis. Hence write down the separate equation 9.
of each line: (i)
tf'-aty-Say' + Sy^O; s
(ii)
a
11. If
ti
- mV
>
=
2
a?
sin3^~8a:Vco83a-8^y
0,
t>*=0 represent straight
5 8iii8^-fy cos3aif
lines,
= 0.
what does the equation
represent ? 12. Find the equation of two straight lines (a) through the origin, 8 a 0. (b) through the point (3, 0), parallel to a? -8ay + 6y t
13.
Find the equation of a pair of straight lines perpendicular to the + 2hxy 4-fcy2 = 0, and passing through the point (6, a).
lines ax* 14.
What
lines
the angle between
are represented
them?
by or-2a?ysec0+y
?
?
What
is
CHANGE OP AXES
89
tfEEffCOsa + ysina-^, u=E#cos0+yBhrj3-g, express in a single equation the bisectors of the angles between u = 0, v =- 0. 16. Find the coordinates of the points where the straight lines 15. If
8*8 +sy- 4y-6a;-22y~24 Can you draw the lines 0. cut the axes of coordinates and the line a?+y from these data ? If so, find their separate equations. should be 17. Find the condition that the two lines ax' + 2hxy + fcy* l
perpendicular. 18. Find the conditions that the equation ax* should represent,
+ $bx*'y + $cxy* + dy*
three coincident straight lines (b) two coincident lines and another ; (c) three lines equally inclined to each other. (a)
;
Find the equation of three straight
19.
make angles with
the axis of (i)
(ii)
x
lines
through the origin which
:
+ 120;
+ 60,
6,
0-45,
When does as8 + bx*y + cory2 +
0, 8 dfy
0+45. =
represent a pair of perpendicular and a straight line bisecting the angles between them ? 21. Find the equation of the straight lines joining the origin to the
20.
lines
points of intersection of the lines ,2# +
8y
=
1
and
a
3a?
+ 2#y-y'-7a;-8y + 3
0.
Find the condition that the straight lines joining the origin to the 8 2 2 points common to the loci # +y = a and Ix + my = 1 should be coincident. 22.
23. Form the equation of four + ^Tr, + f TT with the axis of x.
u = values of u
=
straight lines
which make angles
0,
+
n,
w=
represent straight lines, and u, t/, w are the h and y = k, show that a uvw*(uv uv) + v>u' (t?M>' wv') + icuv*(wu' uw) = represents the joins of (h, /) to the vertices of the triangle formed by 24. If
t
u,
and
t>,
0, v
t>,
0,
w when x =
u\
Show
that the equation of any pair of perpendicular straight lines 2 = 0. the through origin can be put in the form or y + 2Jw?y Hence show that 25.
#*+ Bx*y +
Cky + Dxtf + Ey*
represents two pairs of perpendicular straight lines if
B+D =
and
E=
1.
The greater part of elementary algebraical geometry is occuwith the properties of the locus represented by the most general pied 5.
equation of the second degree, viz.
When
the left-hand side of this equation has two linear factors the which are parallel to the straight
locus represents two straight lines lines through the origin
uyuATiuiNS ur
90
We propose here to discuss the properties of this latter equation. To find the nature of the lines ax2 + 2hxy+ ty* = 0. (i)
If the equation
and
is solved,
m w 1}
2
are the
y are the
two values of - so found, then x
m^ = 0,
w2 # =
y
two
straight lines. These are therefore real, coincident,
m w 2 are t
real, coincident, or
,
or imaginary according as
imaginary.
Hence, according as ft
the
lines
2
ab
is
positive, zero, or negative,
the
represented by
are
equation
coincident,
real,
or
imaginary.
The
idea of an imaginary line has been adopted in order to preserve continuity thus, for instance, we say that two tangents can be drawn :
to a circle
from a point which are
or imaginary according as the point lies outside, on, or inside the circumference. Imaginary points and lines cannot be represented in the same way as real their existence is indicated by the algebraical "consideration of real, coincident,
:
geometry and has been accepted in Pure Geometry with fruitful results. They offer an explanation of many facts, and their recognition saves the necessity of a long series of exceptions.
(ii)
To find
When
(See Chap. IV).
the angle between the straight lines ax*
-f
2hxy
the two values of ( 2,
where
=
2
lf
j
given by the equation are the values of tan 6 l9
are the angles
which the
straight lines
the #-axis.
Put therefore
-
x
= tan 6 in
the equation and
we
= 0. Hence and
= 0.
the axes are rectangular, since the equation of a straight line with the #-axis is y x tan 0,
through the origin making an angle
tan
+ by 2
tan
{
+
tan
tan O l tan
2
= 2
=
r-
-
>
get
make with
CHANGE OP AXES The angle between the ,
V
/a2 _ 2
^
reader should consider the case
Cor.
i.
a
-f
&
Cor. h*
e.
hence
4a b
ft
2 ),
tan 0,
.
The
when
If the lines are at right angles, 6 l
hence
infinite,
i.
.
,.,
_
is
lines is (6 l
the
condition
91
that
b is zero.
-6 2 =
and tan(^~^2 )
^7r
is
be perpendicular
should
the lines
= 0. If the lines
ii.
=
Note.
are
coincident, B l
=
2,
and tan(^ 1 -02 )
is
zero,
ab.
When
the coordinates are oblique,
angle between the lines
it
may
be shown that the
is
tan- 1 -^-V a+b
2 h cos o>
which (iii) To find the equation of the straight lines between ax* + 2hxy + by 2 0.
bisect
the angles
=
If
#!,
2
are the angles
#-axis, the bisectors
make
which these straight angles %(0i
+
2 ),
lines
make with
^+^^ + $2)
axis.
Hence the equation
of the bisectors is .
~x tan i.
e.
i.
e.
?
)
-
tan
= o, ^
/y.2
A u,
the
with this
EQUATIONS OP HIGHER DEGREES
92
2 (^-y tan ($ + 2) = 2xy, a 2 - tan Ol tan (# -y ) (tan Ol + tan 2 = 2xy (1
Hence i.
e.
i.
e.
)
l
2
)
A (a;2 -ya )
Hence the required equation
Note
= (a-b)sy.
is
The condition that these lines should be should be positive, which is always the case. i.
1
real is that (a-fc)
-4
4/t
8
Note li. The equation satisfies the condition that the bisectors should be perpendicular to each other. (iv) To find the condition that the straight lines ax* + 2hxy + by* should be harmonically conjugate tvith respect to the straight lines >'
=
= 0.
If neither 6 nor b' is zero, let
= 6 (y-px) (y-ff) - A). a V + 2h'xy + Vy* = 6' (y (y = 0, y go; = straight lines # ax* + 2Jixy + by*
and
rA')
Now
the
jpjc
conjugates with respect
to the straight lines
y
ro;
are
= 0,
(p-r)(-) = -(|>-)(-rX 2(pa + rs) = (l> + a)(r + s).
if i.e. if
harmonic
y (
5#
=
10. II)
Expressing this relation in terms of the coefficients in the given equations,
we
find
ab'+a'b
The reader should prove that b or
V
=
this condition holds
when
either
is zero.
The converse can be easily proved by reversing the steps in the work above. The equation ax 2 + 2hxy+ty2 = contains only two independent constants, viz. the mutual ratios a:h:b.
put ax* + 2hxy+ by*
If a
^ 0, we can therefore
= a(x+,py) (x+qy),
since the right-hand side contains the two independent constants JP, This is a useful comparison to make in many special cases.
q.
CHANGE OP AXES Ex.
To find
i.
on the straight
the product
lines
of the perpendiculars from ax 2 + 2hxy+ty* = 0.
03? + 2hxy + by9
Let
Comparing
93
coefficients
=a
p + q*=
(a:
+ j>y)
(*
-f
the point
(f, r>)
gy).
,
^ a
The product
of the perpendiculars
V {a
Ex.
ii.
To find
(lie
8
4-
aa (l> + q)* -2a*pq + a
anharmonic ratio of (lie pencil formed
pairs of straight lines whose equations are
ax* + 2hxy+by* Let
(w?*
a
V
=
and
a'x* +
2h'xy+Vy*
+ 2hxy + by* = a(x+py)
-f
2 fc'a?y 4-
= a' Vy*
(a: H-
ry)
(x
+ qy),
(a: -f
sy ),
.,
then
The anharmonic
ratio
=
Now Hence
j>s-f^r-. \{ (p + q)(8 + r) +
and similarly pr+g
-
f ,
{hh
-
(j>
-g)(- r)}
= 0.
'by
the two
EQUATIONS OP HIGHER DEGREES
94
the anharmonic ratio
therefore equal to
is
ab' -f a'b
Ex.
iii.
To find
- 2 hV + 2
the area of the triangle
= 0,
2 equations ax + 2hxy -f by*
Ix
=
Let the points of intersection of Ix -f my act*
+ 2hxy + &y
=
3
whose sides are given by the
+ my = be
a
and
of 2/1
(x l y x ), ,
is
(a?,,
ya )
;
at the origin,
is
are given by the equation
2/2
+ 2h(a(} I \ y
/v
\
-2
4-
)
+
6
=
0.
\i>
'
/*
-1
Hence
with the lines
1
then the area of the triangle, since one vertex
The values
1.
=
2/2
2/i
Again, the values of y l and ya are given by (1
+(!-
-,,,)
y)
+
V-0,
y'(am'-2Wff) + &f) + 2y(W-am) +
i.e.
The area of the
triangle
is
=
0;
therefore
Examples Illb. Prove that the equation 4a^~12opy + 9y 8
represents two coinci1 two parallel straight lines. dent straight lines, and 4a?2 -12rry-f9ya Find the equation of coincident lines perpendicular to them and passing 1.
1). through the point (1, 3 2. Find the angle between the lines 60*' - 103 art/- 72 y
3.
Show that the equation
lines,
and
Draw
:r
3
-2crycot2a-y
find their equations.
the locus
when a
30.
l
0.
represents two straight
CHANGE OF AXES Show
4.
95
that the two pairs of lines
^Q
and
5.r2 -12:rt/
=
+ 6t/ 2
contain the same angle. 5. Find the area of the figure enclosed by the
lines
x
= 3,
y
=
2,
Prove that the two pairs of lines
6.
ax^ + acxy +
cy*^
0,
(3
+ (T 1 ) a?2 + a;y -f (3 + a" 1 )^
=
have the same bisectors of the angles between them. 7. Find the length of the intercept cut off on
x + y-\
=
by x* + 4xy + y'*=*
0.
Find the separate equations of the bisectors of the angles between
8.
2 (a-f8)# +
xy + ay*
=
0.
x-y =
bisects the angle between 4 x* - xy -f 4 y 2 = 0. 2 that the 10. have the straight lines (a -f X) a? -f 2 hxy + (/; -f X) y* = same bisectors whatever value X may have.
Prove that
9.
Show
Find the anharmonic ratio of the pencil formed by the two pairs of == and x 2 -f 7 xy -f 9 y 2 = 0. 12. Find the condition that one of the lines ax^ + Zhxy + lnf = should be to one lines of the coincide with, (a) (b) perpendicular 11.
lines
3# 2 -5#i/-ff/ 2
aV-f ZKxy + Vy*
=
0.
Find the equation of a pair of straight conjugates with respect to each of the pairs 13.
and 3#* +
which are harmonic
lines
7.ry-f
4y
2
=
0.
14. Find the equation of a pair of straight lines which are at right angles and have the same bisectors of the angle between them as the straight
lines
y-2x Show
== = 1. 0, ?/-3rr 1 that 11 y* + \bxy-x
=
represents a pair of lines through the I. origin inclined at 30 to the line #-f 2y 16. Find the condition that # cos a + y sin ft p should be parallel to 15.
=
=
1 one of the straight lines ax -f 2 "hxy + by = 0. Find the between 17. the angle straight lines
1
x
1
-f
y
2
=
4 (x cos 6 -f y sin
2
0)
.
2 1 2 form that the straight lines a cos a x + dbxy -f b sin a y & pairs of straight lines which are harmonic conju2 = 0. gates of ax* + 2 xy -f 6y 19. Prove that the straight lines joining the origin to the points common
18.
1
Show
for different values of
to
(x-HY + (y-fc)
2
=
c
2
and kx + hy
=
2hk
will
be at right angles
if
denotes a 20. Prove that (a + 2fc + b)x* + 2 (a-b)xy + (a-2h + b)y* pair of straight lines each inclined at 45 to one or other of the lines given
by
EQUATIONS OF HIGHER DEGREES
06 21.
Find the length of the intercept cut
tfcosa-fysina Interpret the results when atanf a-2&tan<X-fc0.
aa? 4-2fajy-l-
by
j> f
when &
ac or
5
on
off
?
7
0.
the equation
satisfies
and Find the condition that the straight lines aa?7 -f2fcry4&y2 to-f wy+n should form a right-angled isosceles triangle. 23. Find the equation of the straight lines which are harmonic conjugates 22.
1
of both the pairs ax* + 2 hxy -f 6y9 0, 24. One of the lines ax^+bxy+cy*
aV + Ztey-ff'y
1
other two 25.
is
Two
Show
a*0.
{oo'-a'o} / faa' (6c'-i'c)
fixed straight lines
$ by a
=
0. + 2 Vary + coincides with one of the lines
&y
that the tangent of the angle between the
1
are intersected in P,
a'a?
+ <*'
(a&'-a'ft)}.
whose equation
is
wy =
variable line ZaH
Find the area of the triangle formed, and,
ff
2
1.
the area
if
+ 2:rycotO--i/8 is
constant, find the
equation of the locus of the mid-point of PQ. 26. Find the condition that the two pairs of straight lines oa?
8
+2fc*jy-f&y'
=
0,
aV-f 2h'xy + Vy*
=
pencil (i) when lines of the same pair are conjugate, when lines of different pairs are conjugate. 2 27. Show that the conditions that the straight lines ax*-2hxy + by' = should form an equilateral triangle with a? cos a + ysin a p are = = a/(l 2 cos 2 a) */(2 sin 2 a) &/(! -f 2 cos 2 a).
may form a harmonic (ii)
28. A straight line of constant length 2 ? has of the straight lines o#? 4- 2 fca?y + 6y3 0. Show that the locus of its middle point is
extremities one on each
its
=
(ax -f Ay)' 29. If
tfcosa + y
sin
a =^> makes
then the values of tan Sl n 2 80.
.
+ (hx -f fcy) 8 -f (aft - ft2 ) J a
-
-f
tan
0,
#,,
0.
^ 2 with the lines
and tan ^ tan ^ are
^-(a-6)^-ft and -oT7Ta~ 4
The diagonals
angles
-
,
o*
f
-2M + n
,-
, -
fc
,o a
.
,_.tan
(^ v
=
x
a). '
c and y = c, and a pair x Show that the other two sides intersect
of a quadrilateral are
opposite sides are oa?* +
V
0.
of at
the point {26c/(6-a), 2cw/(a-&)} and are parallel to
81.
lines 82.
Find the length of the intercept on the line y 2 fcry -f &y2 when the axes are oblique.
oa1
Show
lines ax*
tw# +
c
made by
the
-f-
that,
when
+ 2 hxy + by1
the axes are oblique and inclined at an angle also include an angle o>, if
4 ab cos1
o>
- 4 h (a
-I-
ft)
cos a
+ (a f ft) 2
0.
CD,
the
CHANGE OF AXES
97
CHANGE OF AXES.
6.
change the position of the axes of coordithen necessary to find what the equation of any locus referred to the original axes becomes when referred to the new axes. If the axes are changed from Ox, Oy in the figure to (XX, 0' Y, the transformation can be made in two stages. It is often convenient to
nates
(i)
:
it is
We
can transfer to a pair of axes parallel to
through the point (ii)
We can
Ox and Oy drawn
0'.
then change the directions of these axes to O'X, O'Y.
The two
the first step is simple stages can be examined separately in all cases the second, however, in the case of oblique coordinates, :
;
is
very involved and
ever, for future
work
is
to
We
need, howrarely required in practice. of axes alters the degree
show that no change
any locus to do this it is necessary to show that, the coordinates of a point referred to the original axes are (x, y) and referred to the new axes (A", Y), then the new equation is obtained of the equation of
:
if
from the old by some linear substitution such as x = lX + mY+n, y = l'X + m'Y+n'.
We
shall prove this, but otherwise confine our attention to special
which experience shows are required in the processes of analysis. This part of the work is often omitted by the student. This is a mistake, as many elementary properties become clear if the results cases
of transformation are understood
a
considerable
of the work.
number
:
we
intend
of easy exercises
to
therefore
to
give
emphasize this part
EQUATIONS OP HIGHER DEGBEES
98
To change the origin without changing the direction of the axes. Let the new origin be the point 0\h, k). Suppose the coordinates of any point Pare (, y) referred to th axes Ox, Oy and (X, T) referred I,
to parallel axes
O'X, O'Y.
Draw PLM parallel new axes meet the Then
the
if
Hence,
to
meet O'X,
to
Oy
original axes in A,
in the equation of
QJC in
L
and
M
and
let
to the axes Otf,
Oy
:
JS.
any locus referred
we
substitute (h + X) for x and (k+Y) for y, the equation obtained is that of the same locus referred to O'X, O'Y.
Note. of axes
:
The point P has been taken in the positive quadrant for both sets the student should draw other figures and see that this method
gives results true for all points.
Examples IIIc. 1.
What
does the equation
through the point
(1, 1)
3# + 4#
=
become when
7
parallel to the original axes ?
drawing the graph of the locus. 2. Find what the equation 4 x
1
-f
8
xy
-f-
3 y1
=
referred to axes
Verify the result by
becomes when referred
to parallel axes through the point (-2, 8). Verify, by finding their separate equations, that the
new equation
still
represents two straight lines. 3. Take any pair of coordinate axes and a pair of parallel axes through the point (-4, 5). Verify, by drawing, that the coordinates (x y) of the following points become (X, Y) when referred to the new axes, where a?=sX-4 and t
r+5.
t/
(i>(7,2). (iv)
(
(ii)
- 6, - 8).
(v>
(6,5). (
- 4, - 7).
(iii)
(-2, -8).
(vi)
(
- 6,
8).
Prove that, when the origin is changed but the directions of the axes are unchanged, the coefficients of the highest powers of x and y in an 0, or a, h, b in equation are not altered, e.g. a and b in ax + by + c 4.
a
xM
2 hxy
+ by
-
}
2 gx + 2 fy +
c
.
CHANGE OF AXES II.
To change from one
set
99
of rectangular axes
to
another
I
of
rectangular axes without changing the origin.
Let the new axes make an angle with the original axes, and let P be the point whose coordinates referred to Ox, Oy are (x, y), and referred to
OX,
OY are
(X, Y).
Draw PL perpendicular to OX and LN, LB perpendicular and parallel to 0,r. Then
PM
perpendicular to Ox,
= OM = ON-UL = OL cos 0-Psin = X cos 0- T sin # = MP= NL + BP = 0sin0 + iPcos0 = Xsin0-f- Tcos x
The equation
of
by substituting (X
0, 0.
any locus referred to the new axes is thus obtained Fsin 0), (X sin + cos 0) for x and y in the cos
Y
equation of the locus referred to the original axes.
Examples Hid. 1.
What
does the equation x* + y*
= a*
become when the axes are turned
through an angle 6 ?
does the equation # 2 -y 2 = /** become inclined at an angle \ir to the original axes? 2.
What
when
referred to axes
For the equation a?2 -ya -2aa? + 2&i/ +
EQUATIONS OP HIGHER DEGREES
100
Transform the equation x1 -f 4xy + y2 4 6 #-3 = by turning the axes through 60 and changing the origin to (1, -2). What do you conclude from your result ? 8. If the new axes are inclined at an angle. 0, taken in the positive sense, 7.
to the old, find the equations of the new axes referred to the old. Hence find the values of in terms of x, y> and 6 discuss the signs of the and expressions so found.
X
III.
f
l\)
Y
;
change from a pair of oblique axes
rectangular axes, retaining
lite
same
to
a convenient fair of
origin.
Y
L
*
Retain the original axis of x and take a line perpendicular to
it
for
axis of y.
P be (a y] referred to the original axes and (X, Y) to the new. Draw PL perpendicular to Ox, and PM parallel to Oy. Then x = OM = OL-ML = OL-PL cote* = X-Tcotco, y = PM = Ycosecco. and Y cosec Hence for x and y we substitute (X Y cot Let
1
,
co)
Y
To
transfer back to the original axes the expressions x + y cos to, y sin to.
N.B.
we must put
co.
for
X and
^
Examples III If the axes be inclined at
1.
axes
flie
v
an angle
o>,
e.
find
by changing to rectangular
conditions that the lines
should be perpendicular. 7 2, Referred to oblique axes inclined at an angle o> the point I is (x^ yj, and the line AB Ix -f my -f n = 0. Change to rectangular axes and hence find the length of the perpendicular from
P on AB. 2
does the expression # -f-# changed to rectangular axes ? Interpret the result geometrically. 3.
What
2 -f
2a?ycos
become when the axes
are
CHANGE OF AXES
101
When
the axes are inclined at <> the equation of a \ air of straight lines 2 hxy + ty* 0. Change the axes to rectangular axes, form the of the bisectors of the angles between them, and find what the equaequation 4.
is
ax*
=
-f
becomes when the axes are changed back
tion of the bisectors
to the
original axes.
IV. To change a pair of rectangular axes to another pair of angular axes whose equations refcned to the original axes are given.
rect-
\ new axes #cosa+y sin a p = 0,
Let the equation of the the form,
referred to the old be reduced to
#sina + ?/ cosa
q
= 0.
P be
any point whose coordinates referred to the original axes are (r, y) and to the new axes (X, Y). Draw PM perpendicular to = perpendicular from P on 0' Y O'X, then X = 0' Let
M
= #sin a + y cosa Y = MP = perpendicular from P on = x cos a + # sin a p.
(i)
<7,
Note. Tn the figure
O'X (ii)
X
and Y placed in the positive quadrant XO'Y\ In the forms chosen the substitution of the coor-
P is
are therefore positive. in the equations of the lines gives dinates of and jj, q, hence, since P is ou the opposite side of the lines to the origin the substitution of the
coordinates of P will give positive results as required. Note that p and q were considered positive in the figure. In any special case under consideration draw a rough figure and determine the signs.
Equations (i) and (ii) give us two linear simultaneous equations from which to find x and y in terms of X and Y the results are ;
(X
-I-
3) sin a,
sin a.
Cor. In the case of oblique coordinates the expression for the perpendicular from any point on a straight line contains the coordinates only in the
EQUATIONS OF HIGHER DEGREES
102
hence, if the method just explained is used when transforming to any other pair;, we shall get two simultaneous first equations of the degree between the old and new coordinates, and the change is effected by a substitution of the form consequently
first
degree
:
from any pair of axes
Hence, however the axes
may
be changed the degree of any equation
unaltered.
is
Example.
What
does the equation of the straight lines 7x-
come when
+ txy + 4# 2 =
the axes are the bisectors of the angles between
The equation
of the bisectors
le-
them ?
is
.e.
Now we know that the equations of two straight lines equally inclined to the avaxis are of the forms y-irr = 0, y-fwitf 0; hence the single equation representing the two lines referred to the bisectors of the angles between them as axes contains only the a?a and \f terras: suppose it is rt#
2
+ fy2 =
0.
(i)
The coordinates lines 2a?
+y
=
in this case are the perpendiculars from any point on the and 0, i.e. in terras of the old coordinates are
x-2y =
Change equation
(i)
or
back
to the old axes
a(2a?
this
is
4a
which are consistent and give Hence the required equation
4-
=
0;
=7,
b
4a-46 = 4. a 4 = 4, a = J, I = ft
.
is
8#2 + 3y2 =
is
become*
therefore equivalent to
-f
7.
it
+ y) 9 -f&(;r-2//) 2
Hence
The
;
0.
lines are evidently imaginary.
When any equation of the second ax* + 2hxy+ly* + 2fjx + 2fy+c =
Invariants.
degree
transformed by any change of axes to another equation of the
second degree, such as /
c
= 0,
CHANGE OP AXES certain relations
To show
(i)
between the constants are unaltered by the change
we
these relations
103 ;
call Invariants.
that
when we transform from one set of rectangular and a& h* are unaltered.
co-
ordinates to another the quantities a + b Firstly.
We have seen (Ex. Ill
c.
4) that
changing the origin without
changing the direction of the axes does not affect the coefficients have only then to deal with a change in direction a, h, and b.
We
of the axes.
No change
Secondly.
of axes affects the degree of the terms
2#r + 2/# + c; hence these terms do not
We have then only to when
the coefficients
affect
a',
ft',
and
I'.
what the terms ax + 2hxy + by2 become 2
find
the direction of the axes
is
changed.
Suppose the axes turned
through an angle 0. Then ax2 + 2hxy + by2 becomes a
(X cos 0-
T sin 0)* + 27* (X cos 0- Tain 6)(X sin + Y cos 6) f*(Xsin0+rcos0)
=X
2
[a cos
2
+ 2h cos
+ b sin
sin
+ 2XY [(6 -a) if
0]
cos0 + Mcos 2 0-sin 2 0)]
sin
f
Hence,
Y 2 [a sin2 0-2ft sin0 cos 0+ b cos 2 0].
ax2 + 2hxy+by 2 becomes a'x 2 -f 2h'xy + a' It'
= a cos + 2h cos = a sin 2 2h cos 2
sin sin
a'+ft'ssa +
i.e.
(cos
2
b'y
,
we have
+ 6 sin 0, 0+ & cos 2 2
tf,
ft;
2
0-sin 2 0) + (&-a) sin = ft cos 20 + 1 (6- a) sin 20 2&'=:27icos20- (a- 6) sin 20.
and further, fc'= A
2
2
cos
;
/.
Also we can write
2a
or
/
= 2ftsin20 + a + 6 + (a-6)cos20,
a-~==a-r. the proposition states and tbe are invariants if a'X? + 2h'XY+VY*
Note. One point needs careful notice proof implies that is
a + 6, a&-/t a
obtained from a# 8 -f 2kry-f
does not follow
when we
3
fcy
by the processes of transformation.
are told that c.
:
and
It
EQUATIONS OF HIGHER DEGREES
104
represent the same locus referred to different rectangular axes that these relations are true for either equation may have been simplified by :
some constants.
multiplication or division by 2 2 Thus, for example, ff ~t/
=
X
and
2
- Y z - 2 */3XY=
represent the
same pair of straight lines (the angle XOx being 30), but the values of a&-fc2 in these two equations are -1 and -4 respectively: the fact is that
when
the process of transformation is completed the second equation 2/3 .XY= 0. appears in the form All we can say, then, when we know that
J^-jr
2
+ by*
4-2fo?t/
flrtf
=
and a'x* + 2h'xy +
represent the same locus referred to different axes,
where X
is
is
b'y*
=
that
a constant. .
T
In any case, i. however,
-
,
-'%
.
.
,
an invariant.
is
Example. WJiat does the equation of the pair of lines 7x* + 4xy + 4y 2 become when referred to the "bisectors of the angles "between them ?
We know that
the
new equation
is
and we suppose that this equation axes to the pair of bisectors.
=
of the form
is
the result obtained by changing the
a + fc=s7 + 4aall,
Then
afe=7.4-2 2 = .-.
a
=
8,
fe
24;
= 3 and 6 = 8, 3Z + 8 F = 0.
=
3 or a
3 8 or and the equation is 8X 2 + 3 T 2 = The two results correspond to two cases when a particular bisector
taken as axis of
is
X or as axis of Y.
The new axes being now called the axes 2 2 or 3*2 + 8i/ 2 = 0. 8tf + 3t/ =
of
x and y the
results
can be
written
A
proof of this invariant property due to Prof. Boole is Suppose that we transform applicable also to any change of axes. (ii)
an equation from axes inclined at
o>
to axes inclined at a/
;
and
that on making the substitutions for transformation the expression 2 b' the expression ax2 + 2hxy + by2 becomes a' 2 + 2 2 2 x + 2xy cos a>-fi/ represents the (distance) of the point (x, y) from
V X Y+
X
the origin, and
We
when
T
:
transformed must therefore become
suppose that the origin
is
unchanged, for
we have shown
CHANGE OF AXES
105
that a change of origin only does not alter the coefficients and therefore such a change need not be considered. It follows
will
then that the equation ax2 + 2hxy + by 2 + Jc (x*'+ 2 jcy cos
to
a,
//,
+ if) =
&,
(i)*
become a
Hence
X*+2h'XY+b'Y* + k(X* + 2XYcoBu'+Y 2 = )
if
the value of
7c
such that the
is
a pair of coincident straight
lines,
(ii)
equation represents the left-hand side of the
if
e.
i.
0.
first
a perfect square, the second equation must also represent equation coincident lines and the left-hand side of (ii) is also a perfect square. (i) is
The
conditions in these cases are
+ k) (b + k) = (h + + (' fy(V + 1) = (ft' + (a
Hence any value
which
of k
a+
2/i cos
Z>
sin
2
co
ft
satisfies
cos co) 2
a! __ ~~~
(ii)
(ii)
;
these
2 // cos a/
sin 2
-A
.
also satisfies
(i)
+V
co
aft
(i)
,
cos a/) 2
They may be written
equations are therefore identical.
Hence
Jc
to'
a'ft'-A' 2
2
The student should work out the case of rectangular axes in the same way. We again note the words in italics it is supposed that the second equation is in the form given by the process of transformation without :
subsequent simplification. If
we merely know
that ax 2 + 2 lixy +
2 &
=
aV + 2A jy + 6 i = /
and
same locus
represent the
/
8
referred to axes
we can say is that 2 (a + 6- 2 A cos co) _ (a' +b'
inclined
at
co
and
co'
respectively, all
"
(aft
-A
2 )
sin 2 co
"~
2 A' cos a/) 3
- 7/ (a' V
Examples III
2 )
sin 2
7
co
f.
In the following exercises 1-6 it is understood that the general equation of the second degree aa?a -f 2/*##-f fy/ 2 -f 2gx + 2fy + c 0. is transformed by a change of origin or a change in the direction of the axes from one set of
=
rectangular axes to another. 1.
Show
that
which contains
it is a?.
by a change of origin only to remove the term Find the equation which the coordinates of the new
possible
origin referred to the original axes satisfy. Can this always be done, and in how many ways ?
EQUATIONS OP HIGHER DEGREES
100 2.
(bj
Discuss the removal under the same conditions of (a) the y term,
the constant term.
What 3.
the geometrical significance of the transformation ? is it possible by a change of origin to remove both the x and y Examine the case when nb = ft*. In how many ways can it be
is
When
terms ?
Where is the new origin ? Can the y term and the constant be removed simultaneously
done ? 4.
When
?
impossible ? 8 5. Show that by changing the direction of the axes the term (a) a? or 2 Find the equation giving the (b) y or (c) xy can in general be removed. value of the angle through which the axes are turned.
is this
In what cases
is
the transformation impossible ? 2
If the equation can be transformed to y -f 27a? + 2my tions exist among the original constants of the equation ? 6.
8.
The most general equation
0,
what condi-
of the second degree in
or
this contains five independent constants, viz. the ratios of ff,
and y
a,
ft,
is
I,
f, c.
The which
greater part of analytical geometry is concerned with the loci this equation represents in the various special forms to which
can be reduced, and under the various conditions which may exist among the independent constants. The student will thus do well to
it
acquire early a knowledge of the notation by which the discussion of shall discuss in the next paragraph the equation is simplified. the properties of the equation when it represents a pair of straight lines
We
and include this notation. 9.
If the general equation
f(x, y)
=
ax*+2hxy + l)if + 2gx + 2/# + c
=
represents a pair of straight lines, then the expression /(.r, y) can be The condition for this is worked resolved into two linear factors.
out in most text-books on Algebra: we append here the most obvious method because it applies to any system of coordinates, and to equations of a higher degree.
Let
ax 2 + 2 hxy + 1)y* + 2yx+ 2fy + c
=
Then, comparing
r)
(p'x + cfy + /).
coefficients,
a=pp',
From
(px + qy +
b
=
qg.',
c
= r/, 2/= 3 r' +
the equations 2/=#r'-f2'f, 20
=
CHANGE OF AXES 2 (fp - gq)
107
= r (pg' and Z(fp'-94) = S( therefare 4 (Jp-
obtain
j
2
(ft
c/i
i.
This condition P,
,
*",
It includes all cases, whatever values
is
necessary. may have.
#', s', r'
Conversely, to
show that
it is sufficient,
e. if
i.
abc-ha/^-a/ -^ -^ = 0, then ax + 2 hxy + by 2 + 2gra; -f 2^ + c can be factorized 2
2
2
2
We
can always find
7
g
j), ^, j)',
,
.
so that
ax* + 2hxy+ ly* = (px + qy) ($'x+ q'y), $p' = a, jpg +jp g = 33 = b; and evidently /
/
where
/
r
2ft,
cannot both be zero, nor can j/ and
j>
and q
g'.
We are given that thus
4 (jfr- OT)
Now
c
put
= r/,
then
2
.
(//-^O = -c (f -i 2 t/p-0g)
=
(^- wO = S (p'g-wO-
Solving these equations for
/and
2/ = ^ + gV and p'qpq' is not zero.
If /and g have these values ax*
-f 2/Krj/
+ bv
The condition V
i.
( l'
e.
X2 =
2
4-
it is
2gx + 2fy+c
we
find
2g = jpr
y
provided that
r (pa'-jp'g)
r
-h
evident that
=
(pa +
/+ r)
proved to be sufficient I n this case, however, we have is
thus
fp-M =
or
fp'-gq'
=
Either of these conditions combined with
except
0.
ptfp'q = gives us since pq'p'q = 0, it
gx+fy is a multiple of px + qy ; further, evident that a#2 + 2/w^ + by 2 is a multiple of (p^f a^ -f 2hyy -f by2 + 2gx + 2fy 4- c is of the form
that is
and can therefore be written
in the
when
2
.
Hence
form l(px+qy+<x)(px + qy + p).
EQUATIONS OF HIGHER DEGREES
108
The
condition
is
therefore sufficient in this case also.
equated to zero represent parallel straight lines. The reader may examine the special cases q and 10. is
q'
The
factors
when p and
$
or
are both zero.
In discussion of the general equation the following notation
convenient
:
= a.s + 2/m/ + % + 2gx + 2fy + u' = ax'* + 2 My' + by'* + 2 gyf + 2fy' + X' = X = ax + hy + g, + hy' + g, = Y= hf + W+f, Y hx + ty+f, Z' = gx'+fy'+c, Z = gx+fy + 2
2
u
c,
c,
atxf
c,
C The
=
A2
aft
H = fg
,
ch.
can be remembered in the notation of the differential
latter
calculus, thus
u
" '
-
'
dc
~db
W
^"%'
It is evident that
and
u' 11.
Now
straight lines,
=
=
the equation u represents a jwrfr o/ non-parallel these must intersect at some point (#', 7/0* If> then,
if
the origin of coordinates
is
changed to the point
equation must represent a pair and is therefore of the form
i.e.
(#', y'), the resulting of straight lines through the origin,
the constant term and the terms containing
The transformed equation
Hence, equating the term to zero, we get
r,
y disappear.
is
coefficients of
x and
y,
and the independent
X = 0, = 0, i.e. r=0, +2gx' + 2fy' + c = 0, x
r:=0,
ax* + 2 hx'y'+
2
by'
i.e.
i.
e.
t'
= 0.
CHANGE OF AXES
wW
But, identically,
X' and Y'
hence, since
109
are zero, ax' +
i.e.
we
have Z' =
also
0,
hy'+g = 0,
=0. Eliminating a/ and /, a
A which
is
= =
the condition that u
hg
=
0,
should represent a pair of straight
lines.
Now X' =
and Y' =
point of intersection on each of the lines
(a/, t/')
;
but these are the conditions that the
(
referred to the original axes] should be
ax + hy+g This point
is
therefore given
x
i. e.
by
~ ""
y
__ "~
V^fy
0,
fl*^
/
" '
"l^ft*
the point of intersection of the straight lines
is
(
^ -A
referred
to the original axes.
We can obtain other forms by using either X = and Z = or Y = the results are identical because A = and Z =
0,
:
Since the point of intersection of the given straight lines lies on = 0, there must be some linear each of the lines 0, Y 0, Z I.
X=
relation
=
between X, Y, Z, such as IX + mY + nZ = 0. have by Algebra (or from the theory of Determinants)
Now we
the identities
a Q. + /JF+ 0(7
= 0,
gG+fF+cC= A = 0. Multiply these equations by then
#,
y and
1 respectively,
and add
;
Since the point of intersection of the straight lines lies on each = 0, Y = 0, i. e. each of the given lines is a straight of the lines II.
X
X=
=
0, their equations must 0, Y through the intersection of be of the forms ZX + mF=0, l'X + m'Y= 0, and consequently the must be of the form pX* + gXY+ rT 2 = 0. equation u =
line
HO
EQUATIONS OF HIGHER DEGREES
We proceed to obtain the equation in this form now bX^hY= b(ax+hy+g)-h(hx+by+f) = (ab-h*)x-(fh-bg) = Cx-G, -hX+aY = -h(ax+hy+g)+a(fus+l>y+f) = (ab-h*)y-(hg-af) = Cy-F. Hence bX?-2hXY+aY* - X(bX-hY) + Y(uY-hX) = X(Cx-6)+Y(Vy-F} = 0(Xx+Yy)-QX-FY = C(u-Z)-GX-FY = C-(GX+FY+CZ) = C, = for GX + FF+ CZ identically. Thus the equation M = can also be written ;
Note. This enables us to factorize the equation of a pair of straight with numerical coefficients.
u
lilies
=
represents straight lines, they must be parallel to the pair of straight lines through the origin which are III. If the equation
ax9 + 2 hxy + by z =
given by
The
0.
bisectors of the angles between the lines
straight lines
drawn through the
e.
=
are therefore
point f ~, ~j parallel to
h(x*-y*)-(a-l>)xy i.
M
= 0,
they are the lines
'[(")'-('-?)><-'(-*)(-')* or
h[(Cx-a?-(fy-F)*]-(a-b)(Cx-G-)(Cy-F) = Using the results in (II), we can write this equation
0.
Ji[(bX-hY)*-(aY-hX)*]-(a-1))(bX-hY)(aY-hX) which reduces, on our dividing by a6
/t
2 ,
=
0,
to
IV. To find the condition that the straight lines u
=
should be
parallel.
The angles between the angles between
=
are equal to the straight lines u the straight lines through the origin
CHANGE OF AXES
111
=
When
are parallel these straight lines the straight lines u are h2 the ab hence coincident, through origin
=
In this case we have
G=
F
and
Thus bX
A=
and
(b)
C=
;
it
.
follows at once that
0.
= hT
and
hX = aY;
parallel straight lines, (a)
and
the equation IX 2 the straight lines
consequently
if
u
=
2hXY+ aY2 = becomes an X = 0, Y = are identical.
represents
identity,
V. To find the product of the lengths of the perpendiculars from any point
= 0.
u
the straight lines
(#', y') to
Let
Then the product
of the perpendicular from Q, j/a + gfy + **' qy + r
=
s
(px'
The numerator ao;
/2
4-
We have
also
Hence
(p
equal to 2 *jy + Jy' 2 + is
pp
= a,
'
j^'
6, jp#'
2
+ (?-) (p'* + q'*) =r = PY2 +
Hence the required product yinci
on
+ gy'+r) (pV + cfy' + /)
p'^/2
(
VI. To
(#', y')
=
=
'
+p
+ c,
^ j/ry
=
that
is w'.
2h.
+ ^'2 + ^'-2 + /
s fl)
- 2JP/M'
y^
-
foows o/ the middle points of the intercepts
the straight lines
to
u
made by
= 0, on a system of straight lines parallel to j = OC
V -,
and
deduce the equation of the straight lines bisecting the angles between
the straight lines
u
=
0.
EQUATIONS OF HIGHER DEGREES
112
Let the point
P (a,
/?)
be the mid-point of any one intercept
AS.
Then the equation of the straight line AB which makes the f\ ti /? V tJ = ^ - for it is rparallel to 7 = - and passes intercept 7 r r is I m m I 1*
,
through (a, /3). This equation can be written
m
where r from the
the distance of any point (x, y) on the straight line fixed point (a, /?) and & is a constant put for convenience
is
If the value of r is either
given locus
on the
its
;
BP or PA,
then the point
(#,
y) is
on the
coordinates are then (klr + a, fcwr + /3), and, since
it is
locus,
Consequently this quadratic in r gives the values of PA and PB these are to be equal in magnitude and opposite in sign hence the This gives coefficient of r in the equation must be zero.
:
;
t
Hence
(a, /?) lies
on the
+ bmfi + gl+fm} =0,
line
l(ax + hy + g) + m(hx+by+f)=
0,
(i)
with our previous notation,
or,
ZX + wr=0,
(ii)
X=
Y=
0. a straight line through the intersection of 0, Now if this equation represented one of the bisectors of the angles between the lines it would be perpendicular to the intercept AB, and i.
e.
T therefore to
-,
=-
I
The condition
1J
m
.
for this is
(la+hm)m-~(lh + mb) I or
Zm(a-&)+7*(w
2
--Z
2 )
= 0, = 0.
But any point on the locus satisfies the equation
or
Y X - = in
I
CHANGE OF AXES
118
Hence, in the special case considered, a point on the one of the bisectors, satisfies
-
-XY(a-b) + h(X 2 or
i.
e.
on
r2 = 0,
-
x -r = XY 2
locus,
)
2
a-6
'
k
This represents a pair of perpendicular straight lines (for, if the equation be written in full, it will be seen that the sum of the coefficients of
Hence the
# 2 and y2
is zero).
locus of the middle points of the intercepts oc>
straight lines parallel to j
ax
2
=
made on
it
+ 2hxy + fa/ 2
by -f-
20& + 2/y + c
= 0,
when
this equation represents a pair of straight lines, i$ a straight their of intersection and when I and are such through point ;
m
line
that the locus is perpendicular to
-
=
I
m~,
the locus
is
one of the
perpendicular straight lines
a- 6
~~h'
J
which equation therefore represents the two bisectors of the angles between the straight lines u = 0. When the straight lines u = are parallel, and therefore ab = h 2 ,
Y= X=
X=
we have
seen that the straight lines 0, It follows from equation (i) that the straight line between the straight lines u 0.
=
Examples III
are identical lies
midway
g.
u ax -f 2 hxy -f 6t/ -f %gx + 2fy + c = is a of represent pair straight lines, and consequently the coefficients are connected by the relation A = 0. It must be carefully noted that the results given are, as a rule, only true in this special case. 1. Prove that Bu cX*-2gXZ+ aZ*. 2. Show that G/C = A/G = H/F, and that F/C = H/G = B/F. In
the
following
exercises
==
1
a
supposed to
=
3.
4.
AX+HY+
Prove that GZ=Q. Find for what values of X the following equations respectively represent
a pair of straight lines (a)
(b) (c)
(d) (e) 1267
:
114 If all
5.
EQUATIONS OF HIGHEE DEGKEES the coefficients in the equation u = are known except g
t
that the equation can represent real straight Examine the case when CA is zero. positive. 6.
If
u =
7.
= H/h
(a-f b) 8.
lines
(aP + 2Mm + bm*) =
Show
=
- B/a
-A/b,
that the necessary and sufficient condition that the triangle
formed by the straight is
is
the distance between them.
is
Show
show
CA
represents two parallel straight lines, show that (a-f V)d*
where 2d
provided that
lines,
u
=
and lx + my~
may
1
be right-angled
0.
that the equation
represents a pair of straight lines, and that they form a - b)fg + h (f - s ) 0. ax* + 2 hxy + by* 0, provided that (a
rhombus with
=
Find the condition that u should represent (a) two parallel, two lines. (b) perpendicular straight 10. Find the equation of the lines x* + S*/2xy + 5y* = referred to the bisectors of the angles between them as axes. 11. Find the equation of the straight lines x* + xy-y* $x~-y + \ =0 referred to the bisectors of the angles between them as axes. 12. Prove that a^-f 9y 2 -f 6#y + 4rr-f 12y~5 * represents two parallel straight lines, and indicate them in a figure. are quantities, the difference of whose reciprocals is constant, 13. If X, and p, q are constants, show that (kpx + pqy)* = (Xa^-f ^y 8) (Xp'-f fig2 -!) represents two straight lines equally inclined to each of two fixed straight 9.
fj.
lines. 14.
u
=*
15.
Show
that the area of the parallelogram formed by the straight lines
and ax* + 2hzy + by*
=
equal to c/(2
is
Prove that the two straight
lines
-
(x tan a y sin 6)* are inclined at the same angle whatever value 6 may have. Turn the axes through an angle tan" 1 (tan a cosec 6). (x*
16.
Show
+ y2) (COB* 6 sin* <X -f sin* B)
=
that the equation of the bisectors of the angles between the u can be written in the form
straight lines
=
x-2gy] + 17. If
the axes are oblique and inclined at an angle 30, sketch the locus
Show
that if the straight lines given by turned through an angle a, their equation in their 18.
ay? + 2hxy + ltf-2{(b-a)xy + h(x*-y*)} 19. If
the axes and two pairs of the
ax + 2 hxy + by* 1
new
are
tana + (^2 -2^y-f ay*)tan 8 0( =
five lines
contain right angles, prove that the equation of the %
-*/*
=
position will be
-// - (-)*/(-/)
fifth
can be written
0.
CHANGE OP AXES 20.
Show
115
that the coordinates of the orthocentre of the triangle formed + Zhxy + by 1 = and the straight line Ix+my = 1
>y the straight lines wxP
= y/m = (a -f 6)/(awif 2 him + K*). given by x/l s 4 21. Show that the four lines a? + 7a? y + 15tfV +
ire
t
harmonic pencil. 22. If the two straight
lines
u
7ay~6y
4
form
are equidistant from the origir, show
23. Show that the angle between one of the lines ax* + 2hxy + by* is equal to the angle Hid one of the lines ax* + 2hxy + by* + \(aP+y*) = between the other two lines of the set. 24. If p is one of the anharmonic ratios of the pencil formed by
ax* -f 2 hxy + by*
show that 25.
* 0,
a'x*
-f
2h'xy -f
b'y*
= 0,
(i
U-/o
Find an equation for X so that
X*+ Y* + \u
=0 may
of straight lines. 26. If the same straight line occurs in each of the
represent a pair
two pairs
ax* + 2 hxy + by* =
and d
is
0, 0, a'x* + 2 h'xy -f b'y* the angle between the other two, then + 2 cot 'Q as= aa'/(ha' h'a) + bb'/(h'b hb').
2a n b*x n yn + b**y* n 27. What is the meaning of the equation a* n x* n where x and y are coordinates with respect to oblique axes ? 28. The base of a triangle passes through a fixed point (/, g\ and sides are respectively bisected at right angles
Show
that the locus of the vertex
by the
its
lines
is
Find the condition that one of the lines ax* + 2hxy + by* = may 0. Jrr with one of the lines a'x* + 2h'xy + Vy* 30. Obtain the equation to the bisectors of the angles between the line? u = 0, in the form 29.
make an angle
31.
Prove that there
is
always one real value of fc, for which the equation
=0 A
is not zero. represents straight lines. In this question 32. Find the values of k for which the equation
(lx+my+ 1)
(r* + m'y +
1) -f kxy
represents pairs of straight lines.
Give a geometrical explanation. Showthat,if ax* + 2hxy + by*
33.
=
and a l x* + 2}i l xy + 6,y*
formed by any change of axes, the expression (a is unaltered.
H
2
=
are trans-
CHAPTER IV ANALYTICAL NOTATION, A REVISION AND EXTENSION THE
1,
geometrical ideas employed in the previous chapters and equations have been those of Euclidean
to obtain our formulae
geometry, with the addition of the sign convention used in Trigonometry.
The coordinates of a point, x and i/, are numbers which are the measures of the distances of the point from two fixed straight lines in terms of some chosen unit of length. Conversely, if any real numbers are chosen for x and ?/, we can, having chosen a unit, plot a point of which they are the coordinates.
We
have here implicitly a number which is of there is assumed units, any system the measure of any such distance (e.g. the diagonal of a unit square), and thus the ide& of number has been used in a wider sense than that, in
beyond the scope of this book The reader is referred to G. H. Hardy's
that of a rational number. to dwell
on this
idea.
It is
Course of Mathematics. have shown that a geometrical property of a point can be expressed by a relation between its coordinates, and, conversely, that
We
a relation between the coordinates of a point expresses the fact that it lies on some locus. Thus, if a point moves on a straight line, a relation of the form Ix + my +
n
between its coordinates. Conversely, if any set of numbers be assigned to Z, wi, and n (excluding the case when I and m are both zero), any points whose coordinates = lie on a certain straight line. satisfy the relation Ix + my + n there
2.
is
The
te=
points of intersection of two loci.
two loci, we obtain their equations and find sets of value of x and y which satisfy these Each set of values gives the coordinates equations simultaneously. If
we wish
to discuss the intersection of
of one point of intersection. For example, the point of intersection of the
two straight lines is the point whose coordinates and l'x+m'y+n' = are (ww'-tn'w)/(Jw' I'm) and (Z'-w'/)/(?w' I'm). If, however, the straight lines are parallel, their equations are of the form
lx +
my+n =
ANALYTICAL NOTATION and for
we cannot
find
fo-f wy
any
+ n'=
117
0; in this case the method fails, x and y which will satisfy This result is to be expected, for
set of values of
these equations simultaneously. in Euclidean geometry parallel straight lines are straight lines in the same plane, which, being produced ever so far in either direction,
never meet.
We
proceed to investigate the intersections of a straight line with a locus whose equation is of the second degree. The nature of the locus is immaterial for our present purpose we wish to discover whether the method of solving the equations of two loci always :
gives satisfactory results. Consider then the points of intersection of the locus
x 2 + 4xy + 3y 2 -2x-2y +
1
=
(i)
Ix + my + n = 0. with the straight line (ii) The equation can be solved by substituting y (lx + n)/m or x = (my + n)/l, obtained from the equation of the straight line, in the equation of the locus (i). This substitution evidently gives
=
us in general a quadratic equation in either x or
Lx* + the equation in x is Three cases may occur
Mx + N =
y.
Suppose that
0.
:
(i)
(ii)
L is not zero the equation is quadratic. L is zero, M is not zero the equation is the ;
;
simple equation
L and M are zero the solution fails entirely. Note. If m = 0, we substitute for x and get a quadratic L'y -f Mfy + N* = exactly similar cases may then occur. (Hi)
;
in y,
2
;
Case
i.
If the roots of the quadratic in x are real
and
distinct,
we
have two distinct real values of x and one value of y corresponding There are therefore two points to each satisfying both equations. Let us examine special
which the straight line meets the locus. cases illustrating the possible results.
in
(a)
The
straight line
whose coordinates are
(
4y
=9
2f, 2J),
meets the locus in the two points
(-4J,
2}).
#=
2 gives us the quadratic # this gives us only one point of intersection (1, 0). 2 1 the equation for x is x -f (c) For the straight line y
(b)
The
straight line
=
the sets of values of x and y are then 1 + i, or, in the usual notation, (
1
(
1),
impossible for us to plot any points either of these sets of values.
(
+
\/
1
^, t,
1),
1).
(1
2# +
l
2x + 2
=
;
J~l,
1),
it
is
Evidently
whose coordinates
= 0;
are given
by
ANALYTICAL NOTATION
118
meets the In the Euclidean sense therefore the straight line y = locus in one point only, and the straight line y = 1 does not meet If
it at all.
we
investigate similarly, the intersections of the locus
with the straight line y = ft, we find that, unless k lies between and 2, there are two real points of intersection, and the distance between them is 2 VA; 2 2i; this distance becomes smaller and smaller as k approaches one of the values 0, 2. It is clear that == is the limiting position of a straight line which meets the
y
two
locus in
points.
Instead then of saying that the straight line
y=
meets the locus in the single point (1, 0), we say that it meets it in two coincident points (1, 0), (1, 0). In the case of the straight line y = 1 we found two distinct sets of values of x and y satisfying the equation, but we cannot plot any We say that this straight line meets points to correspond to them. the locus in two imaginary points.
Thus by adopting the
ideas of
'
coincident points
'
and
'
imaginary
we
are able to say that (for all straight lines which come points' under Case i) a straight line meets the locus in two points which may be real and distinct, real and coincident, or imaginary and distinct.
Note. When the coefficients of the equations are real we obtain a quadratic equation with real coefficients the imaginaries so found are called 'conjugate'; that is to say, if (a + bi, c + di) are the ;
coordinates of one point,
(a
Thus one cannot have
other.
cdi)
li,
are the coordinates of the
a real straight line meeting the locus
in coincident imaginary points. If the straight line is # + y in the equation of the locus and obtain
Case ii
= 2, 4#
we 9
substitute
= 0.
This straight line then meets the locus in the single point This
y
y
=
(2 J,
2
x
J).
a single point in a totally different sense to that in which meets it in a single point. get simply one point, not two
is
=
We
coincident points. Case ill. If we take the straight line x+y 0, we cannot find and which of x This straight y satisfy both equations. any values line does not meet the locus at all. This is a totally different result
=
to that
x and
for y = 1 there we found sets of values for but could not plot corresponding points ; here we find no
which we found y,
;
values for x and y at all. It is convenient and important in Analytical Geometry to be able to assign complete generality to our results ; to say that 'Every two straight lines meet at a single point', 'Every straight line meets
A REVISION AND EXTENSION
119
* every locus of the second degree in two points', Every equation of the first degree represents a straight line ', and so on.
To
effect this
above,
and to include the second and third cases
we require, '
illustrated '
in addition to the non-Euclidean ideas of coincident
'
'
points and imaginary points ', the ideas of points at infinity and ' the straight line at infinity '. proceed to develop these ideas. *
We
3.
Homogeneous
Coordinates.
The
straight line.
The
general equation of a straight line contains only two independent constants, but we found that in order to represent every straight line by a general equation we had to adopt the form
lx+my + n
=
we saw
further that although the equation in this form apparently contains three constants, in reality it is given by two independent ones ; one of the constants, though not always any one, ;
can have a purely arbitrary value, other than zero, assigned to it. constants I, m, n have no absolute values and no geometrical meaning in themselves, though the ratios of two of them to the
The
third are perfectly determined for
any particular
straight line,
and
have precise geometrical meanings. If we give any set of values to Z, in, n (except simultaneous zero values to I and m, a restriction we shall practically remove later) we have an equation, the locus of which is a straight line we may ;
refer to it as the straight line
(I,
m,
n),
and, since a set of values
we may
call Z, w, n Such coordinates have no the coordinates of the straight line. The set of coordinates absolute values, although their ratios have.
of
I,
m, n completely
fixes the straight line,
km, kn (where k is any number) determines the same straight and 15 line as the set I, m, n\ e.g. the equations 5#-f 10^ the same # + 2y 8 line. Such obviously represent straight
Jd,
=
=
coordinates are said to be homogeneous. Any relation between the coordinates
I, m, n, expressing some property of the straight line, must be homogeneous in those coordinates. For example, the fact that the straight line passes through the point
expressed equally well by the relations kla + kmb + kn 0, where k is any number. (a, 6) is
=
la
A
+ wH- n
=
and
non-homogeneous
bm + en2 = 0, cannot indicate any property relation, of the straight line. The coordinates of an arbitrary straight line, # + 2# + 8 = for instance, can be made to satisfy this relation by such as al +
choosing them to be equation a + 26 + $ck
(*,
2fe,
8 k) where k
is
determined by the
= 0.
Unless we have to deal with a straight
line passing
through the
ANALYTICAL NOTATION
120
origin we can put n to be lx + my + 1
=
1 and take the equation of a straight line and we can call it the straight line (, w). By giving arbitrary values to I and m we can obtain the equation of any straight line, except a straight liw through the origin.
=
To
;
obtain complete generality,
we
require the
homogeneous system
of coordinates.
Homogeneous Coordinates. The Point. may now seem natural to inquire whether we cannot
4.
It
obtain
Cartesian coordinates by adopting some complete generality system of homogeneous coordinates which we may use when the ordinary coordinates appear to involve a loss of generality, as in for
and
We shall
see later that there are systems of coordinates, Areal and Trilinear, in which a point is determined uniquely by a set of numbers, the absolute values of which need cases
(ii)
(iii)
above.
not be fixed although their ratios are, and that Cartesians may be Let us take regarded as a special or rather limiting case of these.
we will call f, ?j, f, which have themselves no absolute values but are such that, for any particular How can set, the ratios of two of them to the third are fixed. a set of three numbers which
whose Cartesian coordinates
these represent the point
are
(x,
#)?
Let the ratios There is one quite simple way of effecting this. x be and The and ij/f /C y. respectively point (x, y) will then is be defined by the set of numbers (#f, #, ), where arbitrary* if ( is not zero, the set of numbers (, 77, Q define the whose Cartesian coordinates are /f and rj/. Our equations point in x and y now become homogeneous in We write f/ 77, for #, and multiply by the power of for x and ij/ necessary to
Conversely,
.
,
clear the equation of fractions.
The general equation
of the first
degree then becomes l + m*i + n = 0, and the general equation of the second degree becomes a 2 + &Tj 2 -hcC 2 -f 2/Yj+ 2&rj = 0.*
2g+
Any two
equations of the
are satisfied
first
degree
l+mr) + n = Q by a common set
and
l'
+ m'tj + w'
of values of
,
TJ,
= viz.
,
ww'
m'n,
nl'n'i, Im'l'm. have seen that any set of numbers (, TJ, Q define a point is not zero. in the Euclidean sense if We shall now say that
We
such a set continue
*
We may
to define
a point even when
Such
is zero.
a point
notice that this form of the equation explains 'the conventional
distribution of the coefficients in the expression
ac
a
-f
a
2/Mcy4-&j/
+20-f 2/y-f-c.
A REVISION AND EXTENSION is
not a point in the Euclidean sense
;
it
121
cannot be plotted.
It
may,
however, be regarded as the limit of a sequence of points which can 0. be plotted. Let us consider a straight line l(x a) + m(y b)
=
Any point on this straight line has Cartesian coordinates of the In homogeneous coordinates we may take form (a + mt, b lt). the coordinates of this point to be (w + a, Z-f &(, 0, ( being the same
Now
as 1/t.
as the point
and further from the point
moves along the t
(a, 6),
straight line further increases in absolute magnitude,
being positive for points moving in one direction and So that, as the point moves for those moving in another. negative its
sign
further and further from
(a,
its
6),
homogeneous coordinates take
Z+b(> (), when: ( is continually diminishing. (m-fa, We as a limit. These coordinates tend to the numbers (w, Z, 0) is at line the on the then that I, 0) 'point infinity' straight (m, say
the form
= 0.
Note two things about this point at infinity'. Firstly, its coordinates are independent of a and b ; secondly, we arrive at the same point at infinity in whichever direction we proceed
l(xa) + m(y
b)
'
'
'
So that a straight line has only one point and a set of parallel straight lines have the same
along the straight at
*
infinity',
line.
Thus we may now say that parallel straight point at infinity'. lines meet at a point at infinity instead of saying, with Euclidean 0, Geometry, that they do not meet. The equations Ix + my + n (
'
'
=
n'=Q
common set of solutions, the equations + Zf+WT) w'C=0 have, however, the common set
have no
= 0, (m, -1,0).
(
We = 0.
see that all points at infinity possess the Now is a form of the equation
=
which
is
that
=
common property Z + mi] + n = 0,
We say then We call it the = as the
the general equation to a straight line. is the equation to a straight line. will now arrive at straight line at infinity '. equation of a straight line from other considerations.
We
1
k(l
-f
mr)) + n
=
represents for all values of
k,
The equation other than zero,
=
and
The 0. parallel to the straight line Ix + my by this straight line on the axes of coordinates n/km. Therefore as k diminishes, the straight line
recedes further
and further from the origin in one sense or other
a straight line intercepts are n/Jd
made
Now as k diminishes the equation according to the sign of k. tends to the form n k(l +mrj) + n 0, or, what is the sa'me
=
thing,
=
C = 0.
We
see then that the equations of all straight lines, as these 0. straight lines recede from the origin, tend to the same form
=
ANALYTICAL NOTATION
122
There
in the plane one 'line at infinity',
is therefore
on
'points at infinity' lie
and
all
it.
The homogeneous coordinates of a point dividing in the ratio X 1, the distance between the points whose Cartesian coordinates are So that the point (^iiVi), (x*,V& are (xl + \x^ Vi + ^y^ 1 + A). :
'
9
at infinity" on the straight line joining two points may be said 1. to be the point dividing the distance between them in the ratio
We
may
notice also that the
two points
point at infinity
'
on the straight
the harmonic conjugate with respect to of the middle point of the segment joining them.
line joining
them
'
5.
Cases
We may lines with
(ii)
and
is
(iii)
now resume
rediscussed.
the discussion of the intersection of straight
the locus # 2 + 4#y + 3# 2
and (iii). Take the straight line meet the locus in a single
The equations of this line +7)-2C = and
2x2y+l = 0,
in cases
(ii)
#+# =
2, which previously we found to use homogeneous coordinates. and point, and the locus now become 2
=2
+ 4r + 3T,2_2C-2r,C+C2 ?
= 0.
f in the equation of the locus, we obtain Substituting Tj or C=0; combining This (4^~90C=0. gives us these results with + TJ 2f=0, we get the two sets of values
4f-9C=0
1, 4)
(9,
and
*
point at infinity'. points, one of them a
Take now the
The former is the point whose Cartesian we found before the latter is a which J), So that x + y = 2 now meets the locus in two
1, 0).
(1,
coordinates are (2J,
;
'
point at infinity
straight line
x+y
'.
= 0,
which appeared
to
have no
points of intersections with the locus. Its equation in homogeneous 0. coordinates is f + ry TJ in the equation of the Substituting locus
we obtain
= f* =
=
;
so that
now this
straight line meets the locus
'
two coincident points at infinity ', the point (1, 1, 0) repeated. We now have a method of making all our results general. For example, the properties which we have proved for a system of straight lines passing through an ordinary point will be true for
in
a system of parallel straight lines ; for a system of parallel straight lines is a system of straight lines passing through a 'point at We can in future deal with the nature of a locus at infinity'. an infinite distance from the origin by making our equations homogeneous, and considering the intersections of the locus with the straight line
limiting values,
= 0,
instead of entering
upon an
investigation of
A REVISION AND EXTENSION 6.
The
third coordinate (
at first to the reader
owing
may
present
some
123
slight difficulty
no geometrical meaning
to the fact that
It is, however, Euclidean sense can be assigned to it. if with it our of Analytical we wish scheme to dispense impossible Geometry to be Projective \ The Cartesian system of two co-
in the
'
ordinates to
based on Euclidean notions, and
is
their restrictions.
It
is necessarily
subject
might be thought that we can avoid
a third coordinate by the use of the symbol oo (infinity) ; that we might in fact represent the 'point at infinity' on the straight line
Ix+my+n =
But while the two Zoo). by the coordinates (moo, coordinates m, 0), ( J, m, 0) define the (f, homogeneous And we same point, (-Moo, moo) and ( Zoo, + WQQ) do not. Our parallel straight cannot choose +00 in preference to oo. lines would now meet in two points, not in one. Apart from this, such as and it is undesirable 1, 2, 8, ., infinity is not a number to regard it as if it were, which we should be doing if we sets of
'
'
.
employed the
to
oo
symbol
It is also undesirable to
represent
think of the
'
.
a
Cartesian
coordinate. '
straight line at infinity
except being read as constant ', is sometimes employed. This is open to the very obvious objection that the equation '(7=0' habitually stands for the statement C is zero '. as
= 0.
The equation C =
0,
G
*
*
open to the much more serious objection that '(7=0' discrimidoes not discriminate between = 0, f2 = 0, * = 0, ., a It is also
.
.
occasionally extremely important to make. It is most desirable that the reader should understand clearly that 'points at infinity' and the 'straight line at infinity' are conventions of Analytical Geometry. They are not realities.
nation that
it is
Parallel straight lines do not actually meet They do not meet for all purposes of Mathematics. In the Integral Calculus and the
Theory of
Infinite Series there are
Geometry, and, what
is
Analytical Geometry. Polar Coordinates. It
no 'points
at infinity*
and no
They are conventions
of Analytical of coordinates in of more, particular types There is no 'straight line at infinity' in
'straight line at infinity'.
is possible to dispense with 'points at We could line and 'the at infinity' altogether. infinity' straight with their we can without their use that prove prove anything and we use, but we should only be taking unnecessary trouble, should miss many of the beauties of Analytical Geometry. We have already said that points at infinity are not real points '
'
in the Euclidean sense.
sense in which
we
They are not
'
'
imaginary points in the have already annexed that term to indicate
ANALYTICAL NOTATION
124 a point to which
we
its coordinates.
A
'
point
terms
can assign definite, but complex numbers, for 'point at infinity' may also be an imaginary i
in the sense that its coordinates are '
'
ideal points
and
'
'
fictitious points
are not particularly satisfactory. the reader to do is to think of
complex numbers. The are sometimes used, but
Perhaps the simplest thing
them
for
as
'points at infinity' iii inverted commas till he is sufficiently familiar with them to think of them as points without confusion with real Euclidean points. '
'
The notation
,
has been used for homogeneous coordinates any possible confusion with
?j,
to avoid, at their first introduction,
the x and y of the Cartesian coordinates that might arise from But x, y, and z are the symbols used calling them #, y, and z. writings for general trilinear coordinates, of which our homogeneous coordinates can be considered to be a special case
in English
;
and
x, y, 8 are generally employed for all types of homogeneous point coordinates. very little experience will enable the reader
A
to use x, y, % without confusion with
x and
y.
CHAPTER V THE CIRCLE A
1. Definition. circle is the locus of a point which moves so that distance from a fixed point is constant the fixed point is called the centre and the constant length is called the radius.
its
:
(A.)
To show
that the equation
+ 2hxy + by2 4- 2#z + 2/# + c =
ax 1
(i)
in rectangular coordinates does under certain conditions represent a circle, and to find the conditions.
Let
C (a,
/3)
be a fixed point; then any straight line through
- ^^
x-ot IS
cos
where r
y-fi
sm ^
2i
the distance of a point
is
(x,
=
,..,
()
*V
on the
y)
(a, /?)
line
from the fixed
point C.
Suppose that this points P and Q. Then, if r on the locus Thus,
if
is
line
meets the locus represented by
put equal to either
CP or
CQ, the point
(x,
(i)
y)
in
two
must be
(i).
r is equal to either
CP or
CQ, the coordinates of the point
{rcos0 + a, r sin0 + /3} satisfy equation
(i).
we
Hence, substituting,
find
+ 2r{(aa + ;*/? + #) cos + (fca +&+/) sin 0} + aa 2 + 2fta + & 2 + 20a + 2//3 + c This equation
is
quadratic in
r,
and
its
= 0.
(m)
roots are the lengths of
CP
and CQ.
The equation
(i)
can represent a
circle,
if
we can show
that the
point C (a, /i) can, subject to certain relations between the constants of this be selected so that as the line revolves about C equation, (i. e.
(a)
as
varies)
(7P, CQ shall be always equal and opposite values shall be independent of 0. locus will then be a circle whose centre is (7,
the values of
(b) these
The
:
THE CIRCLE
126
The
first
condition
(a) is fulfilled if (or, /3) is
for in this case the equation
will be of the form
(iii)
Ai* Let
a,
/3
The second
condition
a cos
(b)
a cos
Let
and h
'
then requires that 2 A sin
-f-
This
0.
2 +-& sin
cos
so in one and one case only.
is
= =
ft ft
2 (cos 6
+ sin 2 0) = tan + a
then (6 ft) tan 2 0-1-27* independent of values of tan ; hence 6 = ft, A 0, and a is
;
=
for all
we then have _ 0*
+ 2A sin0 cos0 + 6 sin 2
2
;
acos2 0+2Asin0cos0+&sin2
2
shall be independent of
ft
= B.
have the values given by these equations
_
where
chosen so that
ft
=
ft,
or a
=
6
= 0.
Hence the general equation of the second degree represents a = 6 and A = 0, and in this case only.
circle
when a
Put 6
= a and =
in equation
ft
(iv).
- ?,
Hence
a
'
a *
--
/*
i.
e.
the centre of the circle
From
equation
(v)
'
is ( V
we
-,'
a'
a
find
the value
of the square of the
radius.
Thus, putting 6
=a
and h
= 0,
a and
/3
r2
just found,
and substituting the values of /O
O
= ^2 + -2 - -.
We
thus conclude that the equation of the second degree represents a circle when it is of the form a(#2 -t-#2) + 2
=
and that
its
centre
is
the point (
-,
),
and
its
a, it
becomes
radius
a If the equation is divided throughout
a
by
is
THE CIRCLE or,
127
writing 0j, /i, andc t for ->-, ~,
which can be written where
Note
/i) is the centre
gly
(
(vi)
f
and VV^-f/i 2
If the quantity
the radius.
negative the radius
is imaginary, (g*+f*-ac) and the equation does not represent a real circle. Note ii. If r is the radius and the centre is at the origin the equation becomes a?2 + y2 = r2 which is the simplest form of the equation of a circle. i,
is
,
and radius of the circle are given we is always satisfied by the cowhich equation ordinates of any point on its circumference. For let C (a, /3) be the centre and r the radius, and suppose P (x, y) to be any point on the circle. (B.) Conversely, if the centre
can write
down an
the definition
By
CP 2
= f*
z
(^-a) +(y-^)
i.e.
9
2
= r2
,
which corresponds with the form found above. This
equation
is
(vi)
the
evidently
general equation of a circle, for the centre and radius have been chosen in
_
By comparing this equation with the general equation of the second = 6 and h 0. degree, we see that a
general.
=
The
discussion in (A) is given for two reasons: (i) it helps to prepare the way for a more general analysis of the general equation ; method is not always convincing to the student. (ii) the second
We
have used rectangular coordinates it is rarely necessary to If in (B) the axes of oblique coordinates in work on the circle. coordinates were inclined at an angle <*>, the equation would be
Note.
:
use
so that the
a
=
b
and h
more general conditions = a cos
for a circle referred to -any axes are
Examples 1.
V a.
Find the centre and radius of each of the following a = 0; (i) #' + y -2#-4y-fl (ii) (iii)
(iv)
aa + y*-6#- 0; (* + 12x* +
circles:
THE CIRCLE
128
(*-a)(a7#2 + 7y2 -3#-2y-3 = 0. Write down the equations of the (v)
(vi)
2.
(i) (ii) (iii)
(iv)
(3, 4),
circles
whose centres and radii are
5 units, 3), 1 unit,
(-2,
2 sin
(2 cos 6, (0, I*), 1
0),
2 units,
unit,
(v) (0, 0), a.
Find the
where they cut the line x = y. following circles and note any special points on them
real points
Draw the 2 (i) #
3.
:
-
(ii)
(iii)
(iv) 2
(v) 4a?
+ 4f/> + 12#-8y
=
Find the equation of a
4.
2 units; If
5.
11.
circle
when the axes are inclined #2 4-a?f/ + y2 -8#-7f/ + 6 =
whose centre
is
3)
1,
(
and radius
at (a) 60, (b) 120, (c) 45.
represents a circle,
find
the
between the coordinate axes, and the centre and the radius of the In Question 3 put each of the equations
6.
2 2 2 (#-a0 + (y- 9) t=r 7. Show that the equation
(i),
(ii),
and
(iii)
angle
circle.
into the form
.
j
of the circle
x 2 -f y 2
=r
2
is
unaltered
if
the
axes are turned through any angle 0. 8. Write down the equations of the circles given in Question 1 when the axes are changed to any pair of rectangular axes through their centres. 8.
The
general equation of the circle
contains fhree independent constants, g, f, and r. circle can therefore be drawn to satisfy three conditions, if these conditions give equations from which g, /, and c may be found.
A
Thus we may be given (a) The two coordinates
of the centre
(c)
Two points on the circle and Three points on the circle.
On
the other hand,
(b)
we
and the
radius.
its radius.
made to satisfy for example, a circle will not in general
see that a circle cannot be
more than three conditions
:
pass through four given points.
Example points (2,
i.
3), (6,
The equation
To find the equation ofa cirde which passes through 1) and whose radius is 4 units.
will
be of the form
(*-a where
(&,
/3) is
its centre.
the
THE CIKCLE The conditions that the given
-a)
(6
may
points
on
lie
=
2
129 are
it
16;
or
3,
/.
by subtraction
a-
or
8.
Hence
= 16, 6 = 0, 2/3* -4/3 -23-3 = 0. - 1 or 3 and CX = 2 or
substitution
by
(0
or
4- 1)
2
4
2
(3 -/3)
2
i.e.
/3
.-.
|3
There are consequently two
Example
ii.
three points (x,
2b find
y^
(a? 2
,
the equation
yj,
Suppose the equation of the a?
The conditions
2
which
circles
(.r 3
,
of
6.
satisfy the given conditions,
the circle
# 3 ).
circle is
+ y s + 2$w + 2/0-f
c'-=0.
that the three points should
lie
-f
2/y
2 ^rs
-f
2/y s + c -f o? 3
a -f
c
4- a; 2
2
4-
0,
tii)
= 0,
(iii)
2
= 0,
1/ 2
-f t/ 3
These three equations give the values of
#,/and
= i.
this circle are
2
t
2
2 gx 9
(i)
on
fyi + 2/2/1 + c + a^2 4- y =
We fail
passing through the
c,
(iv)
provided that they are
c if
0,
the three given points lie on a straight line. Hence, in order that may be finite the three given points must not be collinear. The equation of the required circle can be expressed in determinant
e. if
the circle
notation by eliminating g, f, and c from equations (i), (ii), (iii), and (iv) for equation (i) is satisfied (by hypothesis) by the coordinates of all points on the required circle. Thus, ;
x* -My2
x y
'.
'
v* V* -
is
#3 y*
*
-
the equation of the circle through the points (#lt 1267
I
,
(#a
>
^2)*
THE CIRCLE
ISO
Example
Two
iii.
variable straight lines are at right angles
such that the middle points of their intercepts on
Find
the locus
tlie
and are
axes are fixed.
of ttieir point of intersection. Suppose P any two perpendicular lines through this point have equations of the form
A (h, 0),
Let the fixed mid-points of the intercepts be is
the point (,
(0
/
B
-8(0, k).
17)
-?) +
wfy-i;)These meet the
^x^ -
A
:
(*-)-(>, (a? -<)=0. axis of
x
the
in
points
and the
axis of y in the points
Hence
thus
f(2A-2f) + (2A?-2r )
therefore
and (,
always
17)
which
is
circle
on
is
i
?
=
O
f
the equation
satisfies
a circle whose centre
AB as diameter
?
;
is
for its centre is
^
e and radius i(\/^2 + &"')> i* e the mid-point of AB and its radius
(\h> \k)
-
) AB.
Example
iv.
To find
the equation
triangle whose sides are
of the
=
circle
circumscribing the
0,
= 0, = 0. Consider the equation
=
0,
and C are arbitrary constants. The coordinates of any points which satisfy the equations of any two of
where
-4,
B,
the sriven lines also satisfy this equation
:
this equation therefore represents
THE CIRCLE some locus passing through the
181
vertices of the given triangle
this locus
:
is
a circle, provided that (i) (ii)
the coefficients of x 2 and y 2 are equal, the coefficient of xy is zero.
4ftZ8 -ma tw8 )-f J?(J8 Z1 -mjW l ) + C(J1 J2 -m 1 w a ) = 0, A (Ja w8 + Z 8 ro 2) + B(l 3 m l + ll m 9 ) + Oft m, + 1^) = 0. Cross multiplying to find the ratios A B: C, we have A B, C proportional to v 2 (Vi-wijWi) ft w + **!)- ftJj-wfjW,) ftm^ /iWg) = ft + Wi ) ftwf a -/a w8 \
Thus
:
t
and two symmetrical expressions. Hence the equation of the circumcircle
is
- Za w8 ) (I 2 x + m 2 y + 1) (f8 a; + wi 8 + 1) (I* + m ) (1 9 m a + (/2 2 4w^ftw 8 -~/8 m )(^-fw 8 y-fl)ft-|.m 2
i/
t
1
l
y-fl)
=
V b.
Examples
1. Find the equations of circles whose centres are (-6, which pass through the point (0, 1).
Show that the points (4, 3), (8, 3), Show that the circle whose centre
2. 3.
the point 4.
and
(0, b)
find
(4J,
2)
is (a, b)
cannot
5),
lie
(3,
on a
-4) and
circle.
and which passes through
also passes through the point (2 a, b).
Show that the point
(7,
- 5)
on the
lies
0.
circle
x 1 + y* - 6# 4- 4y - 12
= 0,
the coordinates of the other end of the diameter through this
point.
Show
also that the points (5cos0 + 3, 5sin0-2}, {5sin0-f 3, 5 cos0-2} on this circle whatever value 6 may have. 5. Find the equation of the circle which passes through the points (1, 5),
lie
What
(4, 6), (5, 3).
the line
x
y
is its
radius ?
Where
does the line x
=
2 cut
it
?
Also
0?
Find an equation giving the abscissae of the points of intersection of = 0, and the #-axis. if + 2gx + 2fy -f c Hence find the general form of the equation of a circle which passes
6.
the circle x*
-f-
through the points (a, Oj, ( -a, 0). 7. Find the condition that a circle represented by the general equation should meet the axis oft/ in coincident points. Find the equation of a circle which passes through the origin and cuts lengths a and b from the axes. 9. Find the equation of the circle of which the join of (a, 0), ( a, 0) is
8.
off
a diameter. 10.
and
11.
lie
Show
that the four points
find its centre
and
(2, 2),
(5, 3),
(6, 0),
(3,
-1)
lie
on a
circle,
radius.
Prove that points whose coordinates are of the form {h -f a cos 0, k -f a sin 6}
on a
circle for all values of
0,
and i
find its equation, centre,
2
and
radiua.
THE CIRCLE
182
What
12.
is
the locus of the points of intersection of the lines *
tfcos0+ysin 6
where
is
\/2,
sin
0-ycos6
3y-f 5
=
+ 2/ -4#-8y+15
=
At what points does the
IB.
a;
=
A/2,
a variable constant ? 2
o?
line
a?
2
cut the circle
0?
Find the length of the intercept made on the line by the 14. Find the locus of a point whose distance from (a, distance from (~a, 0).
Where does the
locus cut the axis of
The join of the points
15.
(2, 8),
circle.
0) is double its
#?
(-1,
2)
subtends a right angle at P.
Find the equation of the locus of P. 16. Find the equation to the circle whose centre is (-6, -8) and whose radius is 5 determine whether the origin lies inside or outside the circle. Find the coordinates of the extremities of that diameter which passes ;
through the origin. 17. Find the equation of the circle circumscribing the triangle whose sides are
5
= 0.
Find the condition that the point dividing the join of the points 1 1 should lie on the circle x^ + y = 9. 5), (2, (3, -4) in the ratio 1 Deduce the ratios into which this line is divided by the circle. 19. A point P moves in a plane so that its distance from A, the point of intersection of #-2y 4 = 0, 7o?+ Hi/ 3 = 0, is a mean proportional between OA and PN, being the origin, PJV the perpendicular distance of P from the first line. Prove that Plies on one or other of two fixed circles. 20. A circle passes through two points on the axis of x whose distances 2 c /a, and through two points on the f/-axis whose from the origin are 2 distances from the origin are &, c /&. Find its equation. 21. Find the equation of the line joining the centres of the circles which pass respectively through the points (2, 1), (3, -2), (4, -3), and (4, 6), 18.
:
,
-4),
(4,
(1, 5).
Find the coordinates of the centre and the radius of
22.
Find also the coordinates of the point on the circle furthest from the origin.
Determine the length intercepted by the circle on the line 2#-//-f 3 23. Write down the equations of circles which satisfy the conditions
=
0.
:
fc), and passing through (h, k), and touching the a>axis at (a, 0), and touching the coordinate axes.
(i)
centre
(a,
(ii)
radius
r,
(iii)
radius
r,
Prove analytically that if any circle cuts the axes of coordinates at and PQ' respectively, then OP. OQ OF.OQ'.
24.
PQ
25.
Find the equation of the
(0,<X), (0,0).
circle
through the points
(a, 0),
(/3,
0),
THE CIRCLE Find the locus of the centres of
26.
circles
133
which pass through the points
(3,2), (5,4).
an equilateral triangle ABC such that Find the locus of P. 28. Show that a circle can be drawn to pass through the origin and the points (1, -3), (2, -4), (5, 5), and find its equation. 29. Find the condition that the circle circumscribing the triangle whose
P is a point = PB* + PC
in the plane of
27.
PA*
2
.
1 should pass through the origin. x = a, y = b Ix + my A BCD are four concyclic points and P is any other point prove PA 2 A B CD -f PC A A B I) = PB* A A CD + P/) 2 A A B C.
sides are 30.
t
;
that
2
.
4.
Every straight
.
.
.
line
meets a
circle
in
two points whose
coordinates are obtained by solving simultaneously the equations of these two points may be (i) real and distinct,
the line and circle coincident,
(ii)
:
imaginary.
(iii)
Definitions. (i)
When
a straight line meets a curve in
points, the join of these points called a chord.
two
real
and
distinct
is
the particular case when (ii) In the points of intersection are coincident, the line joining them is called
a tangent to the curve at the point. is the 'point of contact'.
This point
A
straight line through the of contact perpendicular to
(iii)
point the tangent
called the normal to
is
the curve at this point.
Such a
straight line cuts the curve at right
angles or orthogonally.
PQ
is
drawn at two points P and Q meeting at T, then chord of contact of the tangents from T to the curve.
If tangents are
(iv)
called the
Let
'
'
Q (#2*2/2) be two
JP(#i, #1),
points on a circle.
Draw PMj Ox and
Then and
if
QL
QN
perpendicular to
perpendicular to
PL =
ih
PQ makes
the axis of x
PM.
y9f
an angle
with
THE CIRCLE
184 Let the equation of the
be
circle
= Then,
two points P, Q
since the
lie
1
we
Subtracting,
Hence
on
0.
it,
+ 21 + =
0,
get
the direction of the chord can be otherwise expressed, thus (i)
Now
approaches P, i. e. as x2 and #2 approach the values xl and y l respectively, the fraction (y\ y^/(Xi x ^ approaches a limit. The chord PQ then becomes ultimately, by our definition, as the point
Q
t
The value
the tangent at P.
#2
= X}
^ y\
and y2 tangent at P(x v y
The equation
{)
i*1
is
(0
>
of this limit can be found
thus
we
by putting
see *h ftt the direction of the
given by
of the chord
PQ
can be written
down
once as the
at
equation of the straight line joining the points (# ly #,), (# 2 y 2 ) ; since, however, the tangent is defined as a special case of a chord, it is a ,
demand
that the equation of the chord joining two points should be found in such a form that the equation of a tangent can be deduced by making the points coincident algebraically.
logical
Equation of
the chord joining two points (xly y^, (x 2 ,y 2 )
on
the circle
We
can now obtain the equation of the chord in a form symmetrical with respect to the two points, for the mid-point of the chord is {}(xl + x 2 ) (#1+2/2)}* an<* consequently the chord is 9
a straight line passing through this point whose direction by equation (i) above : thus the equation is
is
given
{^-H^i^2)}K + ^2 + 2 5f)+{y--i(y + y2)}(y 4-y2 + 2/)} = 0. Now let us call the coordinates of the middle point A of 1
1
chord
PQ
(f,
TJ),
then
= \(x + x
the equation of the chord point
;
l
PQ
2 ),
TJ
= \ (y + y l
;
can be expressed in terms of
for substituting in equation
(iii)
we
the
see that its
mid-
obtain
(*-)( +0)+(y-'0(n+/) as the equation of the chord
2)
we then
(iii)
=o
whose mid-point
is
(,
(iv) n).
THE CIRCLE
We
135
can discover at once from this equation several properties
of a chord.
(i)
The
centre of the circle
is (7(
/)
g,
line joining the centre to the mid-point
hence the equation of the TJ) of the chord is
;
A (f,
f+9
Now
this line
is
perpendicular to the chord
Hence the straight line joining the centre of a a chord is perpendicular to the chord.
some
straight line
y
The
circle to the
mid-point of
Suppose the chord to be one of a system of chords parallel
(ii)
to
(iv).
=
mx.
(i)
straight line
(*-)(+) + (y-l)fo+/)= to
is parallel
(i),
so that
Hence the mid-point
of
any such chord
which passes through the centre the line y mx.
(
0,
=
Hence
the
middle points of
all parallel
lies
on the
f] and
is
line
perpendicular to
chords of a circle
lie
on a
straight line through the centre of the circle perpendicular to the chords.
(iii)
Again,
if
the chord
(*-)(^) + (y-l)Ol+/) = passes through a fixed point
(A,
A:),
we have
P + ip-(h-g)-i\(k-fl-l*g-f1* =
or
Hence the mid-point (A, k) lies on the locus This
is
a circle
of
whose centre
is
point midway between the given
the point ^(h fixed point
*
the locus
0\ \
(ft,
the given circle ( g, /) : that the two points this circle is evident from the form
Hence
0.
any chord which passes through the point
it)
(A,
(&
/)t
*
e the
and the centre of (3, /) lie on
fc),
of the middle points of all chords of a
circle
which pass
THE CIRCLE
186 through a fixed point
and
the centre
(iv)
a
is
circle
of the given
on the straight line joining the fixed point as diameter \
circle
The constant term
the equation of the circle does not
c of
occur in the equation of a chord whose mid-point (f, TJ) is given we conclude from this that 'all concentric circles have the same chord :
corresponding
to
a given mid-point
\
Note i. The reader should prove these and when the origin is taken as centre of the
case
circle is
2 a? -f
then
y
=
2
similar results in the simple The equation of the
circle.
i* 9
and the equation of the chord whose mid-point
is
(,
ij)
is
(*-) + to-?) if^O, # + */?
or
Note
ii.
=
2
If the equation of the circle
4Y.
is
given in the form
the equation of the chord whose mid-point
The equation of the tangent at
is (f ,
the circle
is
the point (r v y^} to the circle c
The equation
rj)
= 0.
of the chord joining the points
(a^, y^\,
(x,
-
=x
Hence, putting #2 becomes
=
\
an d #2
^
V\
>
l
)(x l + g)
l)
l
l
l
.
on the
hence the equation can be written c
Note
ii.
For the
circle
a?
2
-f t/
2
=
r2 this becomes
=
In connexion with the general equation
in
Chapter
III the notation
X'=ax' + hy'+g,
= gx +fy +
c,
0.
xxl -f yy l
ax* + 2hxy + by* + 2gx + 2fy + c*=
we adopted
0.
*he equation of the tangent
+ (y-y (y^f) = 0, x (x^ -f g) + y (/i +/) = ^ 2 + y ? -f gx +fy or But ^ 2 + y^ + 2gXi + 2fy + c = since the point lies (x-x
i.
on
is
(y
Note
3/2)
Z'~ gx' +fy' + c.
=
r2
.
circle
;
THE CIRCLE For the
=
#*+ y* + 2gx + 2fy +c
circle
187
t
we have
=b=
a
so that the equation of the
and h
1
=
; f
tangent at the point (x
y')
,
can be written
xX' + yY' + #=0,
x'X+ y'Y+ Z=
or
This form should be remembered
it
;
0.
shown
will be
later on to be true for
the general equation.
Note
2
where
<;
Thus
The equation of the tangent at (xl y t can also be written 2 c, (x +g) (x l +g) + (y + /) (y, +/) = / +/
iii.
,
)
a c is the square of the radius. the equation of a circle is given in the form
-f/
if
(*-a the tangent at (x l9 y a )
J7^
Definition. point where
it
is
antpfe betiveen
a straight
line
and
the tangent at the
cuts a circle is called the angle at which the straight line
cuts the circle.
Ex.
To find
the angle at which the straight line
x* +
the circle
Let (x lt
i/i)
is
=
at (a^
,
y
t)
cuts
0.
be a point of intersection of the line and
The tangent
hence, if 6
f + 2gx + 2fy +
c
h + my + n = circle.
is
the angle required,
Now
because Also
xf + y^ -f 2gx
m (y l -f /) +
since (#j, t/J lies
Thus
1
(x l
-f
=
=
Jo?! -f
-f
2fyl -f
+
^
1
c
myj -f ty +
/ar-f wy-f n =
on
tanfl
g)
l
=
0.
wi/
=
lg
-f
m/- M,
0.
+ wt K^ +/ - g )"(^ + w/E5a >
>
.
^* ( x\ yi) can be either of the points of intersection, it follows that the straight line cuts the circle at the same angle at both points of intersection, or tangents are equally inclined to the chord of contact*. *
THE CIRCLE
188 If
P
the point of intersection and to the centre
ia
ON line,
the perpendicular from the OP is perpendicular to the
tangent and the result can be more from the fact that
ON where
The equation of point
(
the
OP is
easily obtained
ON
the radius.
chord of contact of tangents which meet at a
TJ).
Let the points of contact of the tangents be P(x^ y^ and Q(x2 y%\ and let these tangents meet at T(, TJ). ,
If the equation of the circle is
the equations of the tangents are
and
and since these pass through
But
T we
have the conditions
these are also the conditions that (x ly yj,
(rr 2
,
y 2 ) should be on
the straight line
hence this must be the equation of the straight line PQ.
The equation at the point (,
of the chord T})
of contact
of tangents
which meet
is therefore
It should be noted that this is exactly the same form as that of the tangent at a point (f, TJ) on the circle, viz. j?X / + yJ / + / 0. The symmetry of the result leads to the following proposition
^ = :
If the chord of contact of tangents from a point T passes through a point T', then the chord oj contact of tangent from T' will pass through T.
For
let
T be
the point (&
of contact of tangents from
TJ)
and T' be the point
(f,
TJ')
;
the chord
T is *,
(i)
THE CIRCLE
139
and the chord of contact of tangents from T'
is c.
The on
(i)
T
conditions that
should
lie
on
(ii)
lie
are identical, viz.
We shall discuss these equations more To find
the coordinates
Ix+my+n = (f, tj)
of tangents
must be
of
the point
# 2 + y* + 2gx + 2fy + c
the circle
(f,
with
lx
+ my + n =
of
to
with
The coordinates
;
then the chord of contact
0.
m Ig
n
+ mf
n'
of the point are therefore
of c
^j)
intersection
to the circle
>y)
Z
(j'i,
of intersection of tangents
at its points
be the point of intersection
from
identical
To ^fd
=
fully later in the chapter.
0.
Thus
Let
(ii)
and that T' should
=
the tangent
from any point (,
chord of contact of tangents from (,
Now (length
0.
be the point of contact of the tangent
hence
TJ)
j),
;
this lies
on the
on the
circle]
viz.
x(t+9)+y(-n+f)+9+fv+c *i(+0)+y
=
0;
of tangent) 2
[since
(#1, t^) lies
f)-c [since (x v , y a ) lies
on the chord of
contact].
THE CIRCLE
140
Note
OP is
i.
If
TP
is
the tangent and
PT a
perpendicular to PT,
Note
ii.
the centre of the circle, then, since
0T3 - OP
2
The square of the length of the tangent from J>
\^
T to
a circle
obtained by substituting the coordinates of
is
Tin
^T
Hence,
if
T
is
outside the circle the result of
this substitution is positive, if on the circle zero, if inside the circle negative, the tangent being
then imaginary. This agrees with the condition
TO2
>, =, or
TJie equation oftJie
normal at
The tangent
yj
at (x^
and the normal
is
<
2
(radius)
the point (x^
.
y^
to the
cmle
is
the line through (x lt
yj perpendicular
(#~#i)(#i+/M#-#i)(#i + 0)
=
to this, viz.
0,
which is the equation of the straight line joining the points (x^ y^ and (-0, -/). Hence the normal at any point of a circle passes through the and conversely all radii cut the circle centre, i. e. is a radius ;
Hence, also, the join of the centre to the point of orthogonally. contact is perpendicular to the tangent or, in other words, the per;
pendicular from the centre on a tangent
is
equal to the radius.
We
have endeavoured so to discuss these preliminary equations as to discover naturally by analytical methods the well-known the object of this properties of tangents and chords of a circle is to create in the mind of the reader, if possible, a feeling of :
confidence in the analytical method. The work can obviously be simplified by assuming these wellknown properties (see notes appended above), but if this is done
no opportunity occurs of exhibiting in a simple manner the certainty and directness of analysis. When solving problems the student should take the simplest form of the equation of the circle which is allowable under the given conditions.
THE CIRCLE Illustrative I.
the
To find the condition that circle x 2 + y* + Zgx + %fy + c
HI
Examples.
lx+my + n
the line
The perpendicular from the centre (-0, -/) on the radius
To find
II.
yj
n)
the equations
which are parallel If (x l9
a
mf
(Ig -f
is
=
2
(I
+ m2 )
+ my
=
equal to the
2 (
H-/
2
c).
1.
the point of contact, then, since the tangent
through this point which by hypothesis is
is
to the circle
of tcmgents
to the line Ix
line
Ct
^g>-+f*
Thus
Q should touch
= 0.
is
parallel to Ix
ffc-rej + wfy-y,) = 0. But the perpendicular on this from the centre
-f
my =1,
its
is
a line
equation (i)
(g, -/)
is
equal to the
radius
Hence or
xi
and substituting
+ my l
in
(i)
=
we obtain
as the required equation.
There are then two tangents to a
Note.
If
the line
circle parallel to
any given
given in the form tfcosft-f ysin
is
equation of the parallel tangents follows more simply, (x + g) cos
To find
III.
(X
+ (y +/)
a =
%-p = 0,
viz.
V
the equation oj the circle touching the line
4:X
and
sin
line.
+ 2y+3a \/5 =
,
cutting off equal chords, each of
from
length a,
the portions
of the
coordinate axes between this line
and
the origin.
The
straight line
cuts the axes at points
whose coordi-
nates are negative hence the centre of the circle lying inside the triangle ;
OAB has
negative coordinates.
If the circle is
B the values of g and / are consequently positive.
the
THE CIRCLE
142
This circle cuts the o?-axis at points whose abscissae are given by
xl and
If these are
a?
2
,
then by hypothesis
= a. = a 4^ 8 4/ ~4c = a o?
1
~o?2
2
Hence
2
4c
.
2
Similarly for the ^-axis
Hence
.
$F*=f*>
and since both are
g
positive
= /.
Expressing the fact that the perpendicular from the centre (-#, equal to the radius, we have 2 = 20 (f -f-/ 3 - c). (40 + 2/- 3 y^)
is
2
2
Since
/)
,
we have
,
i.e.
or '
^
The
centre
for the point
therefore
is
l-/5a
2a y^5)
2a\/5,
(
=
Hence
or
- \ \/5 a, - \ \/5 ( c
=
and the required equation is x* + y* -f
lies outside
$ta
| -y/5
2
2 -}a =
(
a),
the triangle
t^ a?
+ y) +
^
>
2
0,
or
IV. Investigate ^(a;
a)+2/(y
bisected
/3)
tlie
condition that from the point
=
it
may
/
to
P (a,
0) ow
draw two chords each
by the axis ofx, and stow that the angle between tJiem
tan" 1 Let
le possible
A(,
0)
sp
be the mid-point of a chord
^-
is
PQ
of the circle
-v.
The equation of this chord
is
(*-*)-!) +*(y-o. This passes through the point
(Of, ft)
which equation gives the values of These values are real if
i.e. if Ot
2
>8/3
2 ,
which
is
the required condition.
.
if
THE CIRCLE If (j,
,
H8
are the roots of the equation,
The equations of the two chords PQ, PQ' are then (i-*)tfi
(&-*)(fi thus the angle between them
snce
and
Examples
V c.
1. Find the equation of the chord of the circle c'-fy'-f 6# + 8t/ + 9 0, whose mid-point is (2, -3). 2. Write down the equations of the tangents at the point (-2, 4) to the = 0; circles (i) gf + y>4 4a?-10y-f 28 (ii) (iii)
(*-l) + (y-5)
(^45)
2
2
-fy
=
10;
-25.
Find the equation of the tangent to the circle x 1 + y8 perpendicular to 3o?-f4# = 5. 4. Show that the line 3# + touches the circle 4^ + 10 = 3.
=4
which
is
Find the point of contact and the equation of the other tangent parallel to this. 5.
Is
the point
(5,
6) inside or outside the circle
Show that the point (1, 2) lies inside the circle, and find the equation of the chord of which it is the middle point. 6.
Find the coordinates of the point of intersection of tangents at the 7 5. a?-f 4y *= 14 meets the circle a? + y*-3#-f2y Prove that the point (3, 4) lies outside the circle
points where 7.
and
find the lengths of the tangents from it to the circle. that their points of contact lie on the circle
Show
and on the straight
line 24a? +
ly ~2
*= 0.
THE CIECLE
144
Find from the definition of a tangent as the limit of a chord the 2 equation of the tangent at (2, 5) to x* + y = 29. a touches # + ya -3# ~3y + 2 = 0. 9. Showthat# + 3y-l = Find the equations of the tangents to the circle which are perpendicular 8.
to the given one. 10. The point (2, 3) is the inid-point of the chord of a circle and the 16. Find the locus of the centres of such equation of the chord is 5a? + 2y
circles.
Find the equation of the circle inscribed in the triangle 3# + 4y
11.
y =
x = 0,
Also of the circle escribed to the
0,
=
12,
first side.
Prove that the line (a?~l)cos0-f (y 2) sin0 2 a may have. (aj-l) + (y-2) = 9 whatever value 12.
=
3 touches the circle
Find the coordinates of the point of contact. 13. Show that the lengths of the tangents from any point on the straight line
x~y + l =0
to the circles
3 2 + y2 -f 7#-9y + 6
=
0,
# 2 4-f/2 -5;r + 3t/-6
=
are equal. 14.
A
circles
o?
point moves so that the lengths of the tangents from 2
2
+y + 6#-4y
equation of
=
12, x*
=0
+ y2 -4o?-4?/-f 5
are equal.
show that the locus
cuts
Find the locus of a point P which moves so that PT and PT' are tangents from it to two given
Also
when PT*
PT"* I
17.
points
A
point
= .
constant,
PT
2
and
+ m PT'
2
.
and
in general
=
(#-l)
PT
2
2 -f
(y-2)
2
=
4,
+ PT'-~ constant,
circles.
when
constant.
XOY
P moves
(5, 1), (3,
circles
it
orthogonally. 15. Find the equations of tangents to the circle which make an angle tan" 1 f with the axis of x. 16.
to the
Find the
its locus.
Find also the equation of the line joining the centres of the
where
it
in the plane so that its distances from the in 2 1. are the of constant ratio 2) :
Find the locus of P and show that 2# t/ + l = is a tangent to the 18. Find the equation of the circle passing through the points A (
locus. 4, 3),
B(-8,-4),
C(4, -3). Prove that the tangents at
to the tangent at 19. 2
a? -f
or
A
and C are
parallel
and each perpendicular
J9.
that the straight line #cos0 + ysintf p = meets the circle in real, coincident, or imaginary points according as p is <, =,
Show
y*r*
> r. 20. Find the equations of tangents to the circle #*-f
x cos a -f y sin O 21.
t/*
== r2 parallel to
= jp.
Find the locus of the centres of
circles
which pass through the
points (!, ^),(a a &*) 22. Find the locus of the mid-point of a chord of a circle which
constant length. 23. Find the locus of the mid-points of a fixed distance from a given point.
all
is
of
chords of a circle which are at
THE CIRCLE
145
/
Show analytically that two circles can be drawn through a given point each of two given straight lines. touch to Find the equation to the circle when the straight lines are axes of coordinates and the coordinates of the fixed point are (3, -6). 24.
25.
from
A it
point P moves so that the squares on the tangents PZ\, PTZ to a number of given circles aie connected by the relation
,
PT
3 ...
r^PTV-f 2 PTa l a s PTJ + ... = 0. on a circle whose centre is the mean point of the centres 2
Show
P lies
that
of the given circles for the constants a ly 2 26. A point moves so that its distance
3 ....
,
from a fixed point
length of the tangent from it to a given circle. a circle which lies entirely outside the given circle.
Show Find
twice the
is
that its locus
is
also the distance
of the fixed point from the centre of the given circle if the centre of the locus lies on the given circle. 27. Find the locus of the foot of the perpendicular from the origin to
tangents to
# 2 -f
We
2 ?/
-f
2 gx + 2fy
+
=
c
0.
forms into which the equation of the circle can be put, and convenient methods of and also to illustrate the types of manipulating these equations to which form is each problems specially suitable. 5.
now propose
to discuss various
;
The centre
at the origin.
x z +y-
=
r2
(I)
.
In the majority of problems dealing with one circle the work is simplified by using this, the simplest, form of the equation of the Since the equation is the same for any pair of rectangular axes through the centre, we can further choose the coordinate axes so as to make some other detail of the problem simple, e.g. the axis
circle.
of x
may be taken through some special point. The various equations already found become The chord joining (x v yj, (x 2 y 2 )
in this case
:
,
x (x l + x2 ) + y(y The chord whose mid-point
l
+ y.2 = )
is (f,
xl
x.2
+ y y + r2 l
2
.
i?)
The tangent at (x ly y^\ xx l 1 The chord of contact of tangents meeting .
at (,
We
have hitherto denoted any point on the circle by (xv yj, coupled with the conditional equation xf + yf r2 It is often possible to find a particular form for the coordinates of a point on a given curve containing only one variable, and such that the conditional
=
equation
.
is identically satisfied by the coordinates. Such coordinates are called parametric, and the single variable is called the parameter. 1267
THE CIRCLE
146
Now 2
r cos
2
=
in the case of the conditional equation xf + yf r2 , since 2 2 2 + r sin fl=r is identically true, the point (rcos0, rsin0)
We
the equation of the circle for all values of 0. can to therefore use (rcos0, rsin0) signify in general a point on the circle provided that every point on the circle can be so denoted. satisfies
P
is any point on Geometrically, if the circle, the position of is known when the angle xOP(S) is known, and
P
provided that the method explained for polar coordinates is adopted to measure the angle 0, one definite point on the circle corresponds to a given value of
and, on the other hand, a definite value of corresponds to every point on the circle if is taken between 0,
and
2-jr.
we see from the maximum and minimum real values Algebraically,
cos 6 and sin
=r
that the
x and y are r and
r ; since
equation of
o?
2
2
2 -}-t/
between 1 and 1, the coordinates (r cos 0, r sin 0) lie between r and r, and can have any value between these limits. In future we shall refer to the point (r cos 0, r sin 0) as the point
on the '
point
lie
circle
x2 + # 2
on the
circle.
= r2
whose parameter
is
i
or briefly
0,
the
The equations of the, chord joining two points lt 2 and of the can be found from the above equations by tangent at the point substitution we leave this as an exercise for tlie student and give an ;
independent investigation. (i)
To find
The
the equation of the chord joining the points
line joining (rcos0j, rsindj), (rcos0 2
# r (cos t
'
or,
rcos0! 2
_
"""
cos #1)
,
1
rsin0 2 )
y rainfli r (sin 0% sin
and
2.
is
'
X)
~
sin* (0!-0 2 ) sin i^+ey sin^-^) cos e the sin denominators by dividing J(0i 2 )> we 8 * = x cos \ (01 + 2) + y sin J (0l + 2) r cos \ (0 l -
2).
Note. The removal of the factor sin J(^i~*^a) should be cdmpared with the corresponding work in 4, p- 134. (ii)
The equation of the tangent
at the point
0.
This follows from that of the chord by putting
0i
= 0% =
;
THE CIRCLE (iii)
To
The
intersection
of the
line
Ix
147
+ my + nzzQ and
find the points of intersection of
two
loci
the circle
we have
to solve
simultaneously the equations representing those loci we may obtain an equation giving the ^-coordinates of the common points and a second giving the ^-coordinates. The following illustrates the advantage of the use of parametric coordinates in such investigations. :
If
(he parameter of
is
any point common to the
circle
and the line Ix -h my + n = 0, the coordinates (r cos 9, r Thus point must satisfy the equation of the line. Ir
cos
-f
mr sin 6 -f n
,x
2
+ y* =
r2
sin 6) of the
= 0.
any common point of the circle and line, hence the values of 6 given by this equation are the parameters of the points of This
is
true for
intersection.
The equation can be expressed
in terms of a single variable tan \
by substituting
in
which case
it
becomes
2 (n- Ir) tan
J
+ 2mr tan \0 + (n + Ir) =
0.
Each value of tan | given by this equation corresponds point only on the circle for suppose that
(i)
to
one
;
= &, \Q = WTT + tan" k, = 2n7i-h2tan-
tan \
1
then or
1 fc,
which gives the same point whatever value n has. The equation (i) is quadratic in tan i 0, and we can now apply our knowledge of quadratic equations to investigate the intersections of the line and circle. (a)
A
meets a (b)
quadratic equation has two roots circle in
The
two
;
hence a straight line
points.
roots of the equation are real and distinct, coincident,
or imaginary according as
m2 i.e.
as
Hence the
Z
2
r2
2
2
+tn r
line
>, =, or
is
/-
2
tt
2
is positive, zero,
te-fwy + w
= K 2
cuts
the
,
or negative. circle
#2 +y2
= ra
if
THE CIRCLE
148 2
meets
r2
2
ri
is positive,
touches
the expression
it if
is zero,
and
only in imaginary points if negative. condition corresponds to the perpendicular from the centre (This less to the line being than, equal to, or greater than the radius.) it
If
(c)
0],
the
are the values of 6
2
sum
which
= tan % O
of the roots
l
above equation,
satisfy the
+ tan
2
= ---~~
the product of the roots == tan i O l tan ^
2
=
Ir
-
=
cos |
(#!
+
2)
:
sin | (B l
+
*""""
VI
mr n
:
:
}0 l + tan
tan
:
2
IT
IV
Hence
,
n
\
-~(l
:
2
- cos \ (6 -
2) :
+ tan|
l
tan
2)
2 ).
l
m
The equation
terms of the parameters of its of the straight line therefore of the circle is intersection with points
x cos \ (0 1 -f 2 ) 4- # sin | (0 l + 2 ) - r cos \ (6 l - 6^ = 0, which agrees with the equation of the chord joining the points found otherwise. (iv) To find the points of contact of the tangents which can from a given point to the circle.
Let the given point be (A, k) x cos If this passes
through
(ft,
is
now
+ y sin k sin
=
2
be dratvn
the tangent at the point
=
,
is
r.
k)
h cos
This equation
;
t
-f
r.
true for the parameter of any point the tangent at It can be written in the form (A, k).
which passes through (A
+ r)
tan 2 \
6-2k
tan
\6 + (r-h)
=
0.
Then (a)
Since the equation
is
quadratic
has two roots;
it
tangents can be drawn from any point to a (b)
The
roots are real,- coincident, or imaginary as
k2 i.e.
z
as
h'
+ k*
is 2
r
>, =, or
<
-^ 2
r2
is positive, zero,
,
or negative.
Thus the two tangents which can be drawn from circle are real, coincident, or 7*
i.e.
as
(h,
thus two
circle.
2
+ #2
r2
imaginary according as
is positive, zero,
k) is outside, on, or inside
or negative,
the circumference.
(A, k)
to the
THE CIRCLE (c)
The square
149
of the length of a tangent from
(7i>
ft),
(if
given by the equation are
1?
2t
the
is
point of contact)
= (7- r cos 0) + (ft- r sin 0) 2 2
the two values of
(d) If
the
sum
of the roots
= tan i21 -f tan i2
2
=
= tan f 0!
J
2
1
the product of the roots
Hence 1161166
h "
-
i-**" r '
*
fl
i
*
*"*** _ ~
.
'
tan
cos j cos
2k r
#j
r
=
(
^-e
'" +
tan i
>
r-fft
sin \ (0
tan ^
then
}
fl
2)
cc
2
This result
evidently gives the coordinates of the points intersection of the tangents at the points O l and 2 It can be verified by solving the equations
of
.
+ y sin 0j = x cos d.2 + y sin 2 = x cos
No new
facts
0]
r, r.
have been found in the above analysis
;
the work
is
intended to illustrate the application of algebra to. geometry, and also to familiarize the student with methods which will be used later on in the book.
Illustrative (i)
To find
the point
Examples.
the angle between the tangents tvhich
(h, k) to the circle
=r
x2 + y2
can be drawn from
2 .
Suppose that the points of contact of the tangents are
0!
and
2
;
then
equating the coordinates of the point of intersection of the tangents at these points to the given values cos
(0,4-00)
cos \ (0X - 0,)
_
h r
But the equation of the tangent OJCOS0J
sin '
4-
(0,
cos | (0!
+ 0o)
-
at the point 6 l -f
ysh^ =
2)
is
r,
which makes an angle (\n + Q^ with the axis of x. Hence the angle between the tangents is (0i-0 9 ).
_k r
THE CIRCLE
150 But from equation
(i)
.*.
1
+ cos^-tfj
i
Cor. The locus of the intersection of tangents to the circle which include a constant angle 23 is a^-f-y 2 = f2 sec 2 ft, i. e. a concentric circle. The student should also work the problem geometrically.
(ii)
AA'
is the
circumference in the triangle
;
to
diameter of a semicircle and
P
any point on
is
find the locus of the centre of the
circle
the
inscribed
A PA'.
The geometrical
solution of this question is evidently easy; for if I be the point IA and IA' bisect the angles PAA'
and PA'A moreover, the angle P is a right Hence the angle AIA', which is the angle. supplement of %(A+A ), is 135. The locus ;
r
therefore of the
consists
arc
of
a
circle
passing through AA'. We have selected this problem, however, because the result obtained by purely algebraical analysis results in an
which
equation
needs
discussion
and
is
typical of a variety of problems in which the results are difficult for the student to interpret.
Now on the
let
-4',
P,
A
and
be the points
0, 0,
it
circle 2
-f
6 being
Then the
lines A'P,
PA, and
x cos
y*
< n.
AM are - r cos
0,
-arsin 0.
any point on the required locus, be (a?, y),. since I is the centre of a circle touching A'P, PA, and AA'> the perpendiculars from I on the three straight lines are equal. Hence (x, y) satisfies If J,
(#-r)cos|0-fysini04-y= y cos \Q- (x + r) sin \8 4- y =
0,
(i)
0,
(ii)
1 *
Hence
y
a
(y
+ a? + r) 3 -f y2 (y - # + r) 2
equation of the locus required.
,...v ^
- $ 2 2- 2 r (y + a? -
a
2 )
,
which
is
the
THE CIRCLE This can be simplified thus
the same as
it is
:
151
2 2 2 2 2y (y + #* + 2ry -f r ) = (y + a: - r ) 2 2 2 2 2y*(y -f * + 2^-r = (y + a-t*)-4*V> 2 2 2 2 a 2 + 2y (y a: -f2;y-^) = (y -f^-f2^-r )(y 4-^~2ry~^), 1 8 J J = 0, (y' + a * 2ry-t )(y*-* + 2fy + ) j (y* + a? + 2ry-f )(y + r+a:)(y + r-a?):=0. 3
2
1
,
i.e.
)
i.e. i.e.
i.e.
The (a)
locus then consists of three parts
radius r-s/2 viz.
=
the circle #* + y* + 2ry~i* (b)
;
the line #-ft/ +
:
0,
r=0,
and
a circle centre B(0, -r)
i.e.
AE\
viz.
(c)
# y-r =
the line
0,
A'B.
Solving equations
and
(i)
(ii)
for
?/,
we
find
=
y fl-f sin^ +0085) since B
than
is less
TT,
y
r sin 6
;
always positive and therefore the locus which is above AA'.
is
is
confined
to that part of the circle (a)
Now we
see from the equation
(i)
that
when #-fy + r=0,
6
is
and
TT,
#-y-r = 0, 6 is 0. When P approaches A', AP
when
approaches coincidence with AA'\ and A'P the -4'Pis where approaches position perpendicular to AA'. In the limit when P coincides with A', then the sides of the triangle are
Now
AA!, AA\ and TA'. BA' bisects the angle AA'T\ hence the perpendiculars from
the line
any point on
this line to
A A'
and TA' are equal,
i.
the condition that the perpendiculars from triangle APA' are equal when P coincides with A'. fulfil
So also
BA
Note.
Had we been
corresponds to the position
APA
when
e.
any points on the line on the sides of the
it
P coincides
with A.
finding the locus of the centre of the circle escribed would be changed the sign of perpendicular on y =
to the triangle ', in our original equations but in the process of elimination we have squared the y arising from this perpendicular, hence the locus of the escribed centre ;
is
contained in the equation found.
Examples Vd, 1.
Find the area of the triangle formed by tangents at
+ y2
= 325.
does the circle
#2 + y2
(10, 15) to the circle x* 2.
Where
=
(1, 18), (6, 17),
125 meet the line z + Sy
=
25 ?
Find the point of intersection of tangents at these points. 3. Find the coordinates of the points of contact of tangents from
= AB
to **-fy 4.
If
2
then OiV. 5.
the chord of contact of tangents from P to a circle centre 0, r and if these lines intersect at A to AB ;
.
Find the tangents to the
the axis of 6.
is
OP is perpendicular OP = (radius) 2
prove that
circle x*
+ y2
=a
x and the coordinates of the points
Show that the length of the chord + f r2 is 2rsin i(^~^2 )'
circle x*
(7, 1)
25.
2
inclined at an angle
(X
to
of contact.
joining the points
lt
a
on the
THE CIRCLE
152
7. Tangents at the ends of chords of fixed length in a circle intersect on a concentric circle. 8. Find the tangent of the angle between the two tangents from the
point (8J, 3J) to the circle ** + if = 29. 9. Prove analytically that all angles in the same segment of a circle are equal.
D AD.DB
10. A straight line cuts two concentric circles in the points A, B and C, respectively: show that the area of the rectangles AC.CB or are independent of the position of the straight line. 11. is a given diameter of a circle and PQ a chord of constant length.
AB
If
BQ meet
AP, 12.
B
At
at
E
t
show that the locus of
R
is
a
circle.
AC.BD
are perpendicular chords of a circle through fixed points on the circle: prove that CD always touches a circle concentric with
the given circle. 1 the circle x 1 + y 2 = a meet the axes of coordinates in four concyclic points. Show that P lies on one or other of two right lines.
13.
Tangents from the point
AA
P to
is a diameter of a circle and P is any point on its circumference. Find the equation of the locus of the foot of the perpendicular from A on the bisector of the angle APA'. Describe the locus.
14.
6. its
f
A circle can
centre
is
be drawn to circumscribe any triangle, and when taken as origin of coordinates many of the properties of
a triangle can be conveniently investigated.
In this paragraph
The
we
shall use the following notation 2 circumcircle is oP + y* its centre is 0. ;
=R
:
THE CIRCLE The
vertices of the triangle are .R sin y). ;
153
A (It cos a,
JK sin
OK)
;
B (R cos
/3,
E sin ft) C (It cos y,
The perpendiculars from the vertices on the sides are AD, BE, CF. The orthocentre is H. The mid-points of sides are A', B f C'. The centre of gravity, centroid, or centre of mean position of ABC ,
is
G.
The nine-point 2(7==
2S
It (cos
centre is N.
a + cos fi + cos y),
= E (sin a + sin + sin y), /3
The student should work straight through the following examples the easier questions are left as exercises, and the results obtained are used in the illustrative problems.
:
Examples
V e.
6.
The coordinates of the centre of gravity (G) are {$ C, 36'}. The orthocentre (//) is the point {20, 26'}. The mid-point (U) of HA is {CM Sftcosa, tf-f iflsinCX}. The mid-point (^') of#0is {0-^7* cos 3, S-Jflsinaj. The mid-point of 4' 17 is (o, Sj. Show that G trisects the line joining OIL
7.
The Nine-point
1.
2. 3.
4. 5.
Circle.
The equation of the chord Z?0(Fig.,
p. 152) is
^cos(#-i y)4f/smi03-fy) The perpendicular i.e.
to this through A, viz.
= R cos J(#-y)4D,
(i)
is
(^-^cosa)sini(/3 + y)~(y-7esin a) cos (# + -y) = = .Rsin {i(34-y)~a}. o?sini(/3 + 'y)-i/cosi|(/3 + y)
The coordinates of the point (D) of intersection of the are given by solving these equations thus 2 x = R (cos 3 + cos 7 + cos a - cos |8 + 7 - a)
0,
lines
(ii) (i)
and
(ii)
;
= 2S- R sin (|8 -f y - a). #- C - \R cos 4- y - a),
Thus
=
(j3
Hence the coordinates of the point
- 2 (x C) Le.
(a
-f-
- S)* (y
-Cf + fo -)
The symmetry of the
E and F will also satisfy
D satisfy the
= J-K* (cos8 = }#.
result
-f
y
equation
- a -f sin 2
-f
7
- a), (iii)
shows that the coordinates of the points
this equation.
THE OIKCLE
154
which represents a circle, is the circle DEF. by the coordinates of A', viz.
Thus equation
(iii),
This equation
is satisfied
and from symmetry by those of B' arid C'. lies on this circle, and So also the point U {<7+#cosa, from symmetry V and W. lie on a circle whose Hence the nine points E, F, A, B\ C', U, F, centre is (C, S) and radius |#. The nine-point radius is half the circumradius. 8. The orthocentre, nine-point centre, and circumcentre are collinear, the nine-point centre lying midway between the other two.
W
A
The angles^, B, C of the triangle are |(y-0), *r-i(y- a ) 103-). Show analytically that HA 2-Rcos A. 11. Show that AB.AC=2R.AD.
9.
10.
12. If
ABCD are
any four points on a
circle
AB.DC+AD.BC=-AC.BD. The
feet of the perpendiculars from to the sides of an insciibed triangle lie 13.
*
'
Simson line of the point Let the circumcircle be
A(Rco*&, let
P, or the a;
JRsina),
2
2
-ft/
=
any point P on the circumcircle on a straight line called the '
*
pedal line of P. 2 and the vertices of the triangle be 1Z ,
(cos|3,
.Rsin/3),
C (K cosy,
.Rainy):
P(jRcos0, JRsin0) be any point on the circumcircle.
Now BC is
the line (i)
and the equation of the perpendicular from
P to BC is
(x-R cos 0)8in$0+y)-(y-# sin B) cos or
#8in3+ y)-ycos J(/3+y)
.Bsin
(ii)
THE CIRCLE If
then x and y are the coordinates of L, the point of intersection of
and
(i)
155
(ii),
= Jf?8in$(/3 + y).sin{$(/3 + y)-0}-f tfcos $(/3 + y). cos(/3-y) 2x = .R{cos0-f cos # + cosy- cos (0 + y-0)j
x or
;
2y = 7?{sin0-f Binj3 + siny-sm(/34-y-0)}. Hence, using our former notation, we get
so also
therefore the coordinates (x, y) of the point
L
satisfy
The result is symmetrical with respect to a, 0, y, and consequently the also satisfy to CA, coordinates of the feet of the perpendiculars from
AB
P
this equation.
Hence
LMN is a straight line whose equation is =
(x-C-lKcoB0)8m(
Q,
(iii)
the pedal line of the point Q. 14. If the pedal lines of P, (), E all pass through the nine-point centre this
is
ABC, then PQE
of
Let
P be
is
the point
an equilateral triangle. then the pedal line is
0,
(x-C-lEco80)*m(
=
0.
nine-point centre is the point (C, S) ; hence if the pedal line of the passes through the nine-point centre we have
cos0. 8in((r-40)-sin0. cos(
=
or
sin(f0
o-)
= 0.
E
The
whose pedal lines possible values for the parameters of points P, Q, obtained are therefore tr-f n"f by giving pass through (<7, S)
n
same points P, Q E. Hence the angles which PQ, QE, EP subtend at the centre of the are each equal to TT, and the triangle PQE is equilateral. gives one or other of the
Note.
If
OK f
t
ft y, d are four points
2 & ==
jR (sin (X
on the
+ sin
-f
circle
sin
#2 4-y*
y + sin
I?
2 ,
circle
and
8) ,
has then the equation (x - 0') sin (a' - 0) - (y - S') cos (c/ - 0) = 0, when of line the correthe or one of the values #, /3, y, pedal d, represents any sponding point with respect to the remaining three points. The symmetry of this result 15.
of
is
helpful in problems relating to the pedal lines of four points.
The pedal
P to
line of
P
with respect to the triangle
the orthocentre of the triangle.
ABC bisects the join
THE CIRCLE
156
16. If the pedal lines of P, P' are perpendicular, then PP' of the circle.
17.
The pedal
20.
The pedal
is
a diameter
lines of PP' are inclined at an angle equal to half the subtended angle by PP' at the centre. 18. If PL meets the circumcircle at R, AR is parallel to the pedal line of P. 19. If the pedal line of P is parallel to EC, find P.
ABC with 21.
The
PQR on the circumcircle of a triangle a triangle similar to PQR. pedal lines of the extremities of any diameter intersect on the lines of three points
respect to
ABC form
nine-point circle.
PP'
22. If
is
parallel to
BC, then P'A
is
perpendicular to the pedal line
of P.
D
23. If A, B, C, are four concyclic points, and the pedal lines of each with respect to the other three be drawn, these four pedal lines meet in a point prove also that the centre of mean position of A bisects the join of the centre to the point of concurrence of the lines. :
BCD
pedal
24.
There are three pedal
which pass through any given point. 25. If the pedal lines of the points 0, ^ on the circle x* + if = R* with respect to the points ft, /3, y are concurrent, then Q + fy + ty^OL + fi + y + Znn. 26. A,B,C,D are fixed points on a circle on which moves a variable point P, and the pedal lines of C and D with respect to ABP meet at Q. Show that lines
,
the locus of 7.
Q
is
a circle.
The equation of
and radius
the circle whose centre (o,
r arc
given, is
W^r*.
(II)
This form has occurred previously, and the equations of a chord, tangent, and chord of contact were found. As in the first form, we can use a parametric system of coordinates for a point on this circle. The coordinates of the point {
+ rcostf,
/3
+ rsin0}
satisfy the equation of the circle for all values of If G is the centre
and
P
6. (a, /3)
of the circle,
any point 6 on the circle, then 6 is the angle which the radius CP makes with the axis of x. of
is
Thus Pis
in the figure the ^-coordinate
and the y-coordinate
MP= LC+NP When the ^-coordinate of
P
is less
than
OK
is in the second quadrant and the ^-coordinate greater
THE CIECLE than
/3,
which corresponds
when
157
to the fact that cos0 is negative
and
the second quadrant. The student should consider the coordinates of P when it lies in the third and
sin
positive
lies in
fourth quadrants.
To find the equation of the chord joining two points 2 2 (x-oc* + (y-p = r
circle
and
<J>
on the
.
The following
a variation of the former method of finding the
is
equation of a chord. Let the equation of the chord be
Ax + By+C=0. Since the given points
A
lie
on
(i)
this
B
+ r sin 0) + C + ) + B(i3 + rsin (oc -f
r cos 0) +
(/?
)
From
(ii)
and
0,
(ii)
0.
(iii)
by subtraction
(iii)
A (cos$
= J9(sin0 B
cos0)
A
sin
(/>)
;
sn
Now from on the chord
(i)
and
(ii)
by subtraction the coordinates
of
any point
satisfy
A (x-ot) + B (f/~/3) = Ar cos 0+ Br sin 0. Substituting for the ratio A B we get cos i (0 + + (y-/3) sin } (0 + = r cos 1(0-^). (a?:
Of)
The equation
is
then
/3)sin0 = r.
(x
Conversely,
<J>)
of the tangent at the point
all lines
a)cos0 + (# whose equations are of
form touch the
this
circle.
Since is the same angle in both forms I and II, these equations follow from those found for I by transferring the origin to the It is important to notice that when the single /3). point (a, variable
used to denote a point on a circle, the equation of the any line depending on a single point of the circle) takes
(0) is
tangent a definite form and (or
i.
To find
is
completely
known when
is
the equations of tangents to the circle (x
known. a)
parallel to the given straight line y = x tan If (xot) cos + (y{i) sin = r be such a tangent, = cot tan cot = = + or
2
+ (#
'.
7r
<
then
2 /3)
=
THE CIRCLE
158
The tangents
ii.
are therefore
Two e#waZ
(X
OL)
(x
&)sin<J>
(*/-/?) cos
sin
circles pass
are drawn from points on
(y
+r =
j8)cos<>
= 0.
r
each through the centre of the other: tangents
the first to the second.
Show
perpendiculars from the centre of the second circle
on
lie
0.
that the feet of the
to the
chords of contact
the straight line perpendicular to the line of centres
and midway
between them.
Take the- line of centres and the perpendicular line midway between them as axes. Then the centres of the circles are (a, 0), (a, 0) and the radii are each 2 a. Their equations are therefore (i)
(ii)
Any
point on
of tangents
(i)
from
is (a
+ 2a
cos
2 a sin0), and the chord of contact
0,
to the second circle is
it
+ a)(2a + 2{rcos0)+y2asin0= 4a 2 (x+ a) (1 + cos 0) +y sin = 2a 2 (x + a) cos \Q + y sin \Q cos \Q = a, x cos* \0 + y sin $0 cos \Q = a sin 2 0, (x
i
e,
or i. e.
,
which can be written x + y tan J '= a tan 2 \0. The perpendicular to this from (a, 0) is
(iii)
y-(x + a)iB,n\e.
(iv)
we
Eliminating tan J0 between equations (iii) and (iv) as the equation satisfied by the coordinates of the point may have. (iii) and (iv) whatever value
This
iii.
is
circle are
the locus
of the points of
In what
case is the locus
Let the (:c
a)
origin
=
common
to
then the locus required.
Chords of one
2
get x
circle
+y = JS 2
and the
The equation
2 ,
intersection
a
be x 2 i.
e.
drawn which are tangents of
tJie
to
another
:
find
tangents at their extremities.
circle ?
+
2 j/
=
we have
r 2 and
suppose the chords touch taken the centre of one circle as
line joining the centres as axis of x.
of a tangent to the latter circle can be written
(#-) cos 0+y sin = It.
(i)
THE CIRCLE Suppose that the
159
the point of intersection of the tangents to the ends of this chord the equation of the chord
(f, TJ) is
first circle at
:
must then be
=
#+y*? and therefore
(i)
and
are the
(ii)
2 **
same
(ii)
;
straight line.
f
o
Hence
"
sin 6
cos TT Hence
cos
Eliminating
we
0,
is
2
-c
R
get
Hence the point (,
which
fl
=r
iJ rs sin
TJ)
a circle only
always
when a
lies
=
0,
on the locus
i.
e.
when
the given circles are
concentric.
V
Examples
f.
Find the equation of the tangents to the circle #2 + y*-6a:-4y + 5*=:0 which make an angle of 45 with tho axis of x. Verify by a figure. 1.
Find the equations of the tangents which make an angle (X with the a?-axis. 2.
to the circle
8 2 (#-a) + (y ~&)
r1
3. Find the equations of the circles whose centres are at the origin and which touch (x - 3) 2 + (y - 4) 2 - 4.
A tangent to - a) a + (y &) 2 =
4.
the circle
x*+y***IP
is
perpendicular to a tangent to
Find the locus of their intersection. 5. Tangents are drawn to a circle from points on a straight line which does not cut the circle; show that the chords of contact "are concurrent. 6. Find the equation of a line inclined at 45 to the axis of a?, such that f the two circles a?Hy2 =*4, a?Hy -10a?-14y-f65 =* intercept equal
(x
r*.
chords on that line. 7.
Chords of a
circle (a?-a)
2
whose centre is the origin, are tangents to the =r*; find the equation of the locus of their mid-
circle,
-K{^-&)
2
points.
A
system of circles of radius r have their centres on a fixed .circle of radius r. Find the locus of points on these circles the tangents at which 8.
are in a given direction. 9. Tangents are drawn one to each of two concentric circles and include
an angle of 60.
Find the locus of their intersections.
THE CIRCLE
160
8. The equation of a circle referred to a tangent x 2 + y2 2nr coordinate axes is
and normal as
=
Let the tangent be the axis of y
the normal passes through the
:
centre.
Geometrically it is clear that, if the is r, the centre is (r, 0), and
radius
consequently the equation of the circle
#2 + #2
or
=
2rx.
is
(i)
This equation can be obtained algefor, if the equation of the x2 + y 2 + 2gx + 2fy + c = 0, then, since the line ,r = touches the circle
braically
;
circle is
at the point
(0,
0),
must both be zero The coordinates (i)
whatever value
;
the two values of y given by
hence
may
those of any point on the
Note. If two written,
circles
c
+
= 0.
(1
cos
6},
The angle
circle.
is
POx and
POx =
\6.
touch one another their equations can be
f=
axes,
2 roc,
x*+y
2
~
2Iijc.
P
a diameter of a circle and any point on the tangent at B: taken on the straight line are drawn ; and TE, touch the circle in and F. Show that Intersect the and
a point to
AB
and
+ 2fy + c =
r sin 0} satisfy equation have, and these coordinates may be used as
by a proper choice of x2 +
Ex,
/= 0,
of the point {r
2 ?/
T
is
AP
is
E
tangent at
B
TF
AE
In points equidistant
AF
from P.
Take the tangent at A as axis of y and the normal The equation of the circle is
as axis of x.
(i)
the tangent at
B
is
x
2r.
THE CIRCLE T be
Let
the point (,
161
then the chord of contact
17),
EF is (ii)
Now
=
* +f
the equation
"
2r*
by the coordinates of
is satisfied
therefore by those of E, F. Further, the equation is
(iii)
.
points which lie on
all
(i)
and
(ii),
and
homogeneous and therefore represents two
straight lines through the origin, i.e. represents AE, AF. The values of BE\BF' are the values of y which satisfy
(iii)
when x = 2
r.
BE', BF' are given by
/.
BE'+BF = the sum f
.'.
AT is the
Bufc
BE' + BF'
.'.
of the roots
=
^~; hence J9P
line y
= 2BP
or
=
Show
-^
Pis the mid-point of E*F'.
Examples 1.
-r-
V g.
that the equation of the chord joining two points (r(l-f cos0), rsin0}, {r(l
on the circle
x*
and the tangent
at the point 6
2 -f-
1/
-f
=
2rx
=
r(1
cos), rsin(/>}
is
x cos Q + y sin
-f
cos
2.
The
3.
The middle point of the chord joining the
line joining the origin to
0).
any point 6 on the circle
a?
2
2
4-
y
origin to the point 6
= is
4. The locus of the middle points of chords of the circle x*+y* = 2rx which are drawn through the origin is the circle x* + y 1 rx. 5. Two circles touch one another, and any straight line through the point of contact cuts the circles in P and Q. Show that the locus of the middle points of PQ is a circle which touches the given circles and whose
radius
is
the arithmetic
mean
of the radii of the given circles.
u circles touch one another at the same point through the origin meets them in the points Plt P2 of a point Q such that
A
6.
n.AQ = AP + AP^+ l
7.
Two
8.
+y
1287
=
.
.
one another, tangents to one of them are chords Find the locus of the mid-point of these chords.
that the straight line 3 (y cos 0-1) = # cos 20 cuts the circles 2 7 4# in four points which form a harmonic range. 2Xj # -f y
Show 1
+APn
and any straight line Pn Find the locus
...
circles touch
of the other. x^
...
,
=
L
THE CIRCLE
162
=
2nc can be represented that any point on the circle a?a -f^ r sin by (2rcos'0, 20). 10. Find the coordinates of the intersection of tangents at the points 6 and in Question 9. 9.
Show
Show
11.
that the pedal of the origin with regard to a triangle whose 2 2 2rx is y (see Question 9) on the circle # -f y
vertices are
=
(X, |3,
a?
cos (0
+ #4 y)-f ysin (OH
$-f y)
=
2 r cos
OK
cos ]3 cos -y.
12. If a, 18, y, 8 be four points on this circle, there are four pedal lines of the origin with respect to these four points taken three at a time. Show that the feet of the perpendiculars from the origin on these four pedal
Find the equation of the line. of which P is any point. AP meets at Q the straight line drawn through B such that l?Pand BQ are equally inclined to the tangent at B. Find the equation of the locus of Q. 14. is a fixed point on a circle and tangents to the circle from a point P touch it at QQ'. Find the locus of P if OQ OQ' are harmonic conjugates of a given pair of lines through 0. 1 2 15. Show that there are four chords of the circle x + y 4r# r2 = which subtend right angles at the origin and also touch the circle x~ + y* = 2 rx and show that they form a square.
lines are collinear. 13.
AB is a diameter of a circle
y
;
The equation of a system of
9.
passing through two given
circles
points.
Take If #
2
A -f y
A'
(a, 0) for the given points. + 2gx + 2fy + c = is the equation of
(a, 0), 2
y= and A'.
axis
A
the
meets
then the
circle,
it
at the points
Thus the values of x which are a and c
=-
a2
a
;
therefore g
=
and
.
The required equation which
satisfy
for different values
is
then
of/ repre-
sents a system of circles passing and A': the centres of through all
these circles
Ex.
i.
If one
its
centre
Let
AA
of
lie
on the
side
of gravity
of a is
axis of y, since
A g = 0.
triangle inscribed in
a
circle is fixed, the locus
another circk.
be the fixed side and
The coordinates of the centre
P
(x it y A ) any position of the vertex. of gravity (G) are given by
THE CIRCLE x* +
But if
the equation of the circle
Hence the locus of which
To find
ii.
APA'
is
j/j
is
is
a circle.
is
Ex.
G
A
constant, where
Let P(x,
y'}
AP&nd.
lines
of a point
the locus
The
=
which moves so that the angle
be any position of the point.
AP'
are
#- a Z APA'
P
and A' are fixed points.
x If
163
x -H* x +a
^y
a
y''
^
//
y
(X, r
y ,
tan
0(
=
is
a
x'
x'
+a
K 1
Hence the locus
y
_ _L
&
(i)
(ii) (iii)
2.
A,
passes through the point
touches the line x + y is
cP
V
5
* i*
-ry
_
Examples circle
%(ty -,.**
the two circles x^+if
1. Find the equation of a which also
=
= +2 ay cot 3. h.
through the two points
(h, k]
;
3a;
such that the tangents at the given points make an angle 30 with the join of the points.
B
are fixed points on the circumference of a circle AP, Prove that the locus of the intersection of AQ, ;
parallel chords. circle
through A,
A
a, 0), (a, 0)
(
B and
BQ are BP is the
the centre of the given circle.
through two fixed points find the equation of the locus of the points of contact of tangents to these circles parallel to a fixed straight line. 3.
series of circles pass
;
find the locus of 4. Circles are drawn through the points ( -a, 0), (a, 0) the point of contact of tangents to these circles which pass through the ;
point (2 a, 5.
X is
0).
2 are two circles of the system # 2 -f // 2 2Xf/ = a where a variable constant, which touch the straight line #cos3 -f ysintt -p 0,
Show that there
,
=
= or > X = (a
a cos
If Xj X 2 are the values of 2 2 2 When the value of p X for these circles, show that X t 2 j> )sec !X is such that the two circles are coincident, show that their equation is
and that they are real only
6.
Two
if;; is
fixed circles intersect in A, J5;
L 2
P
is
Of.
,
a variable point on one of
THE OIEOLE
164
that
X
PA
them, and
and meets the other circle in a fixed circle.
PB
meets
it
Prove
in F.
BX and AY intersect on
A and B. A line through A meets one circle a parallel line through B meets the other circle again in Q. Prove that the locus of the middle point of PQ is a circle. 8. The base of a triangle is fixed and its vertical angle is given; find the 7.
Two
circles intersect in
again in
P and
locus of
(i)
its
centroid,
(ii)
the inscribed centre,
(iii)
the orthocentre.
The equation of the circle on the straight line joining (xl y t ), as diatneter is y*)
10. (#2>
,
-y 2 =
0.
)
The
centre
Hence
its
is
the mid-point of the line joining (x i
equation a;
But
(a?|,
yj
lies
of the
is
^
y^), (# 2
>
j/ 2),
form
+ y a -(ar + 1
on
this
;
hence
xS+yS-xS-x^-yS-ytft + c = 0, or
=#
c
the equation
is 2
1
a;
2 -}-y 1 y2;
then
+y 2 -i(a? + 1
or
a
)-y(yi + ya) +
i
a
+y
1
y2
(^-i)(^-2)+(y-yi)(y-ya)
e.g.
the circle of which the line joining
0, ;
3) is a
5, 4), (2,
(
= =
diameter
is (
or
Examples Find the
real points where the circle as diameter cuts the # axis. 1.
Find the equation of the
2.
circle
V
i.
on the
line joining (13, 5),
on the
line
(
- 7,
15)
joining the points
a
(a\\ 2a\),(0/X -2a/X) as diameter. Where does it meet the line x = -a ? a 3. Find the equation of the circle on the chord of the circle # + y* = 125 as diameter whose equation is x + 3y = 35. 4. Find the equation of the circle on the line joining the points (1, 3), (5, 1) as diameter, and the coordinates of the extremities of the perpendicular diameter. 5.
A
Show 6.
circle is described
that
it
touches
on the line joining
# + y-4
(3, 7), (9, 1)
Find the equation of the circle whose diameter o?* + y* = 160 whose mid-point is (8, 4).
circle
as diameter.
0, is
the chord of the
THE CIRCLE
To find an equation giving the lengths of the segments of a chord drawn through any given point in a given direction.
11.
of a
165
circle
If the
given point
are both of the
is
same sign
:
outside the circle the segments OP, OQ if the given point is inside the circle the
segments OP, OQ are of different sign. Let the equation of the circle be x*
+ y* + 2yx + 2fy + c
and let the given point be (a, an angle with the axis of x. x-ot
ft)
=
r is the distance of
0,
make
the chord through equation of the chord is
The
cos
where
and
=
let
y-ff
_ ~" *
sin
any point
(x,
y)
on the line from the fixed
point 0.
Now,
if
r
is
equal to OP, the point
#
and
it lies
Hence,
= + r cos 0,
y
at
(#, y) is
=
/J
P
;
+ r sin
its
coordinates are
0,
on the if
r
is
circle. Similarly for Q. equal to either OP or OQ,
+ rcos0) 2 + (/3 + rsin0) 2 + 2#(a + rcos0) + 2/(/3 + rsin i.e. r* + 2r{(a + g)coB0 + (/3+f)8md Now, if we write for convenience, = x* + f(x, y) = a 2 + /3 2 + 2#<* + 2//3 + c. then /(a, ft) (cx
0)
+c =
0,
Hence the equation may be written ra
+ 2r{(a+0)cos0 + (/3+/) sin0}+/(a,
j8)
=
0,
which is a quadratic in r whose roots are equal to OP and- OQ. Hence (i) The product of the roots = OP. OQ =/(, /3). This
is
constant for a given position of 0, whatever the value of
hence the well-known proposition 4
The
which
;
:
rectangle contained by the segments of a chord of a circle passes through a fixed point is constant/
THE CIRCLE
166
we
Incidentally,
Also if/ the circle
;
the rectangle of OP.
Q
(a, /3) is positive,
but
if
meets the circle in two has a real value. lies outside positive, and
see that if the chord
imaginary points P,
/
(ff,
/3) is
OP. OQ
is
negative, CfP.
OQ
OQ
is
negative,
and
lies
inside the circle.
shown directly for the centre C OC* = (a + 0) + (/3 +/)
[This can be (
-3, -/)
;
;
hence
;
OC 2 -(radius) 2 =
.-.
Note that when
(ii)
2
+ /3 2 + 2
is at
the centre,
r*
e.
i.
e. if
OP,
OQ
=
-/(<*, /3), are equal and opposite for all values of
(a
now we
constant,
must
/,
0.
(Cf. p. 126.)
If the values of r given by the equation are equal and opposite, is the middle point of the chord PQ, then we must have
(iii)
If
$.]
i.e.
=
g and /3 = the coefficient of r vanishes and we have
a
i.
of the circle is
lie
i.
+ g)
cos
+ (p +/)
sin
0=0.
consider the mid-point (a, /3) of P() unknown and the chord to be drawn in a fixed direction 0, then (a, ft)
e.
on the
line
= 0, (x + g) cos + (y+f) sin which is therefore the locus of the mid -points of all chords of the circle drawn in a fixed direction 6. (The question of the chord meeting the circle in real poihts has not arisen ; hence the mid-point of the line joining a, pair of imaginary points of intersection is real.) Evidently then the locus of the mid-points of parallel chords of a circle is a straight line through the centre perpendicular to the '
chords (iv)
'.
To deduce
the point
P
(a,
the equation
of
the
pair of tangents
r 2 + 2r{(a + 0) cos0
OP,
if
+ (/3+/)
this chord touches the circle
OQ
from
).
If the chord through 0, drawn in the direction and Q, the lengths of OP, OQ are given by
but
to the circle
are equal.
should be equal,
The
sin 0}
0,
+/(.
cuts the circle in
)
= 0;
P and Q coincide and
(i)
the lengths
condition that the roots of equation
(i)
viz.
{(* + g) cos e
+ (p+f)
sin 0}*
= /(ot,
/3),
gives us an equation from which to determine the directions which to draw tangents to the circle.
(ii)
in-
THE CIRCLE If line
we
choose either of these values of
(which
is
now
a tangent)
sin 9
and sin d in
Hence, substituting for cos
then any point on the
0,
satisfies
cos 6
is
167
(ii),
we
see that
an equation
satisfied by any point on a tangent from the equation of the tangents from 0. This can be at once reduced to the form
to the
circle, i.e. is
Now we
have shown that the equation
represents the chord of contact of tangents from
(or,
/3)
to the
circle.
Thus,
=
if f(x. y)
is
a
circle,
and u =
of contact of tangents from a point
of these tangents is if
is
we
:
0JP (v)
If a chord
find the locus
in other words, the point
P and
2
R
=
.
(i) gives the as see, previously proved, that
= /(,$. cuts a circle in the points
R on PQ OR
respect to
=
through the point
of a point
P
or,
(a, /3)
to the circle, the equation
u /(#, y) ./(a, /3) the direction of a tangent, equation
Incidentally, length OP of this tangent
to
the equation of the chord
is
which
is sucli
P and
Q,
that
1
1
OP
which
* OQ* is the
harmonic conjugate of
with
Q.
(0,0)
Let the equation of any such chord be ...
cos 6
sin
r'
(i)
THE CIRCLE
168
where
the point
is
r>
(<x,
ft)
;
then the lengths of OP,
OQ are given by
+ 2r {(ex + g) cos Q + (p +/) sin 6} +/(<*, ft)
= 0.
OP+OQ=
Hence
-2{(a+0)cos0 + (/3+/)sin0}, OP. OQ =/(,/?).
X
_
J j.
+ (/?+/) sin fl}
2 {(a + g) cos
OP+OQ
OR cos0 (a+0) + OB sin 6 (ft +/)+/(, 0)
Thus But
since .R is
on the
#- = i.
e.
E lies
line
PQ
(i),
OJRcosfl,
if its
= 0.
coordinates are
(x,
y\
#-/?=OBsin0,
on the locus
This is a straight line, and, when lies outside the circle, it represents the chord of contact of tangents from 0. The locus is called in all cases the polar of 0, and is discussed later.
The above
T P and
Let at
results can be obtained in the following manner be a point (x, y) and let the straight line OT cut the circle :
Since
Q.
the points
(a,
ft)
P
and Q are points on the straight line joining and (x, #), their coordinates are of the form
TIT/' w ^ ere
9
(l
T
that
P and
Q should
lie
*h e va l ues of
on the
in the equation of the circle,
where /(#,
If
OT
y)
is
values of
I
=
is
circle.
we
I
are given
by the condition
Substituting these coordinates
obtain
the equation of the
circle,
and
a tangent to the circle, P and Q coincide, so that the are equal, hence /(#, #)./(<x, /3) = w2 .
by the coordinates of any point T such that OT is a tangent to the circle it is therefore the equation of the tangents from to the circle. This
is
an equation
satisfied
;
If
T is the harmonic
is
conjugate of
with respect to P and This gives us u =
are equal and opposite. the equation of the locus found in (v).
the valuea of
I
then
<2, ;
this
THE CIRCLE
169
Examples Vj. Find the equation of tangents from the point 2 tf' + y -/*. 1.
2.
Show
3.
with the notation of Chapter IV, the equation of a pair of = is uu' = (xX' -f yY' + Z'}*. (#', y') to the circle u
Find the angle between the tangents from the point
=
+ y*-2x-4y 4.
k) to the circle
that,
tangents from a?
(h,
(4, 3) to
the circle
0.
Find the locus of points the tangents from which to the
circle
are at right angles. 5. Find the locus of points the tangents from which to the circle #2 + y2 -f 4# 6y + 12 = include an angle of 120. 6. If PT, PT' are two tangents to a circle centre, C, Q any point on PT, and
QN the
perpendicular from this point on TT', show that PTiCP^QN: in the form found above.
QT
and deduce the equation of a pair of tangents 7.
Find the locus of the middle points of chords of the
which make an angle cos~ l 12.
9
with the axis of
circle
oc.
Poles and Polars.
If tangents te drawn to a circle from any point on a given straight line, the chords of contact will all pass through a fixed point. (2) If chords of a circle are draicn through a given point, tJie tangents (1)
at their extremities will meet on a fixed straight line.
To prove
these propositions take the equation of the circle in
simplest form
:
Let the equation given straight line be (1)
u If
=
Ix
P (x', y'}
straight line, tion lx
f
-f
+ my + n =
of
the
0.
point on this the condi-
is ariy
A'
we have my' + n
The equation
= 0.
(i)
of the chord of
contact of tangents from
Using the condition
(i)
(#',
/)
is (p.
145)
this can be written
mxx'y (n + lx') = or
x'(mx
w
2 ,
= ny + mr*. ly)
its
THE
170
CIjRCLE
This contains one undetermined constant x' in the first degree only, and therefore represents a straight line passing through the fixed point given by mx = ly and ny = wr2 ,
/
i.
e.
the fixed point
(2)
Ul
Let the fixed point be
Suppose the tangents
/r
2
i.
.
>
n )
U
(
Any
TJ).
straight line through this
with the
at its point of intersection
meet
at
(#',
circle
y'\ then since
x<xf+yy'~
r2
(ii)
the chord of contact of tangents from (of, y') to the circle, the lines (i)
is
and
(ii)
are identical. *'
_ "~ ~/_
Hence **
1
Jc
and therefore
Hence i.
e.
x'
the point
(x', y')
always
u
-f
t/'rj
lies
= r2
,
on the straight line
= #+
r2
rj
= 0.
Pole and Polar.
If tangents are drawn to a circle from any point on a straight line u, their chords of contact all pass through a point U. And if chords of a circle are drawn through
a point
the tangents at their extremities
C7,
intersect
on a straight
The point line u,
U
and the
line
is called
line
u
is
it.
the
'
*
pole
of the
called the 'polar'
of the poijit U.
Notes, (i) If the line u cuts the circle at A and B, real tangents cannot be drawn from points between A and B on the line it can be shown, :
however, that the chords of contact of pairs of imaginary tangents from these points pass through U.
When
the chosen point is A, the tangents from it coincide and their is the tangent at A so also at B hence U is the inter-
chord of contact
section of the tangents at
algebraical result.
;
A and
B.
:
This can also be seen from the above
THE CIRCLE For
if
u
=
7,r-f
wy-f n
/
=
7 is
0,
of contact of tangents from i.e.
the line u
(ii)
from
=
U
the point
is
x
--
\
--
**!*' \
7**^
(
,
n>
+y
-f
11
r3 ==
),
and the chord
/
or J# +
my -f w =
0,
0.
If the point J7 lies outside the circle,
U to
171
the circle, only lines through
UB
and UA,
V which
lie
are the tangents UA and (75
between
cut the circle in real points. The tangents at the imaginary points of intersection of the circle and other lines through may be shown alge-
U
braically to meet on u. The line UA meets the circle in coincident points hence the tangents at the points of intersection coincide and intersect at : thus A is on it. :
A
B
on
U
if
u.
outside the circle, the polar of
lies
Hence, Similarly is the chord of contact of tangents from U. (i\i)
is
Many
different definitions of the pole
and polar are used
U
in various
books on Geometry. We have adopted the above because it applies equally, as will be seen later, to all curves represented by the general equation of the second degree, and also, although in our notes we have introduced imaginary considerations, the pole and polar can be found from this definition
by a
real construction in every case.
be noted that the equation of the polar of any point (x, y) takes the same form as that of the tangent at a point (#', ?/') on the circle. A tangent to a circle is the polar of its point of contact. (iv)
It should
For the general
circle .T
the polar of the point (x,
2
=
+ y 2 -f 2gx + 2fy + c
f/') is,
in our usual notation,
xX' + yY' + Z'
=
0.
This should be remembered, as the equation
is
true for any curve of the
second degree. Propositions on the Pole (i)
and
Polar.
U passes through
If the polar of
V
V, the polar of
V passes
be the points (x^ y^), (o? 2 # 2) ; the polar of Let Uj a2 a 2 is xx\ + yy\ to the circle # 2 + y 2 ,
=
=
;
V
if x 2 x i + #2 #1 == a ** this passes through that U(xlt y } ) condition the Which is also
should
lie
on
## 2 + y# 2
=
a2
the polar
of V. (ii)
lines
If w u and
is the
v,
join of the poles of the
then
w
is the
polar of the
and v. u lies on the line line the the pole of (iii) If then the pok of the line v lies on the line u. These results are immediate consequences of
point of intersection of\\
v,
(i).
through U.
U with
respect
THE CIRCLE
172
If
U is the pole o/u
with respect to a circle whose centre is C, then (a)
(b)
CU is perpendicular to u. If CU meets u at L, then
= (radius) Let the circle be #*4-y = r (a) If U is the point (x^ y^, CU. CL
2
.
2
2
.
then u
is
the line
Also
CU is the
line (n)
and the (b)
lines
(i)
and
(ii)
are evidently perpendicular,
CU =
Again
CL = .e.
Definitions. (i)
(ii)
[7
and
i are
called inverse points with respect to the circle.
If the polar of
points.
If U, F,
(iii)
U passes through
F,
U and
Fare
called conjugate
W are three points such that the polar
of each
with
Respect to a circle is the line joining the other two, the triangle is said to be self-conjugate with respect to this circle.
UVW
(v)
If a
respect to
a
triangle
is
self-conjugate
circle, the centre
of
with
the circle is
the orthocentre of the triangle.
UVW
be a triangle self-conjugate with respect to the circle, then FTF is the have just shown that polar of U. If
We
CU is perpendicular to FTF, and similarly CV and CW are perpendicular to UW and
Z7F,
i.
e.
C
is
the orthocentre of the
triangle J7FTF.
The
circle in this case is called the
'
'
polar-circle
of the triangle.
In the figure
CM CT= .
CU. CL
= CN. CW = square on
UVW
the radius of the
circle.
must be obtuse, otherwise the Evidently, then, the triangle of the radius will be negative and the polar-circle imaginary. square
THE CIRCLE Ex.
To find
i.
(E cos a,
E sin a),
the polar-circle (It
cos /3,
The orthocentre of the
E sin
triangle
2C
the triangle
of
/3),
178
(JB
cos
y,
whose
vertices are
E sin y).
the point (2C, 2S), where
is
jR (cos CX
cos#-f cosy),
-f
2S = R (sin a + sin + sin y). Hence the equation of the
polar-circle is of the
# sin ft)
The polar of (R cos ft,
form
with respect to this circle
(#-2C)#(coB + cos y) + (/ - 2 S) -R (sin
i.e.
or
is
+ smy)-fp2
=
0,
2tfcos$(/3-fy)cosJ(/3-y)(;r-2C)
.e.
.e.
But the chord joining the points 0, y is by hypothesis the polar of the point CX hence the equation just found is identical with ;
#COsi(0 + y)+yBin(3-fy)-J?COs(3--y) =0. Thus
8
2
or
-
4/e 2
cos^(a-3)cos|(/3-y)cOB|(y-a), and the required equation is p
Ex.
To ,/wd
ii.
in a circle radius
Let the triangle. circle is
d
is
condition that a triangle
and
=
be ar* + y Then we have shown 2
<7)
1
+ (y -2S) 2 =
I?
9 ,
and
in Ex.
41? 2 cos )
i
=
1
[(cos
2
JK -f
let a, 0,
be
a
drawn
circle
(/3-y) cos $ (y-<*).
we have
+ cos a-0 + COB |3-y-f cos y-a],
and by elementary transformation
this
radius
p.
y be the vertices of the
a + COB /3 + cos y) a + (sin a + sin )3 -f sin y) 8 ]
2 JB1 [1
inscribed
that the equation of the polar-
(-0) cos
the distance between the centres,
R
may
self-conjugate with respect to
first circle
(a?-2 If
E
f/te
THE CIRCLE
174 2
But
p
= 4#
2
cos
(<X-P) cos
hence
which
=#
2 e*
is
2
(0-y) cos
+ 2p2
I (y-(X)
;
,
the required condition.
one such triangle .can be drawn, any number can. For take the circle (#-a) 2 + (y-6) 2 = p 2 where If
,
Then,
if
we choose
OK, ]3,
y so that
#(cosa + cosj3-f cosy) # (sin a + sin j3 -f sin y)
= a, = 6,
2 2 2 working backwards, that (#-a) -f (y-&) = p is the polarcircle of the triangle Of, /3, y. We can take an arbitrary value for oc and l = ^ 2 + 2p2 is satisfied, then find values for y; so that if the condition d
we
shall find,
,
not only one but any number of such triangles can be described.
Miscellaneous Examples for Revision. 1.
2.
Find the equation of the polar of the origin with respect to the
circle
Find the locus of the poles of the straight line x + y = 1 with respect which pass through the points (2, 0), ( 2, 0). Find the coordinates of the pole of the line 3# + 4y = 5 with respect
to a system of circles 3.
to the circle 4.
#2 + y2
If tangents be
all pass
=
25.
drawn from any point on one of three
circles (which through two fixed points) to the other two, prove that the ratio of
the lengths of these tangents is invariable. 5. Find the equation of tangents from the origin to a;
6.
A
chord of a fixed
tangents drawn from
2
+
2 t/
-6a? + 2t/ -f 5
= 0.
such that the sum of the squares of the extremities t6 another fixed circle is constant
circle is
its
:
prove that the locus of its middle point is a straight line. 7. Find the locus of a point the tangents from which to two fixed circles include equal angles. 8. Two fixed points are conjugate with regard to a circle of given radius. Find the locus of the centre of the circle. 9. From two points P, Q perpendiculars are drawn to the polars of Q and P with respect to a circle. Show that the ratio of their lengths is equal to the ratio of the distances of the points from the centre of the circle. 10. Tangents are drawn from a point P to a given circle and meet the tangent at a given point A in Q and R. If AQ + AR is equal to a constant
length, find the locus of P. 11. If
ABC
the vertices 12.
Show
lie
a triangle self-conjugate with respect to a circle, two of outside and one inside the circle.
is
that the polar of a point
(#', y')
with respect to the circle
2 2 (3* + 4y + 4) -f (4 x + 3y + 5)
is
24 (* + y) 2
THE CIRCLE A
13.
circle is
drawn
to
make the
175
touch one side of an equilateral triangle and it lie on the third side.
pole of another side with respect to Find the locus of its centre.
to
Show
14.
that the equation of the locus of the poles of tangents to the 2 2 2 2 2 -f t/ = & , taken with respect to the circle a^-f y = c is
circle (o?-a)
,
2
(a
The
15.
-62 )a?2 -&y -2c2 aa? + c4 =
sides of a triangle are
x/m + yip
1=0, y
Find
0.
=
0.
orthocentre and also the equation of the circum- and nine-point Verify that the centres of the two circles, the orthocentre and the centroid, lie on the same straight line. its
circles.
Find the condition that the straight line joining the points (x lt t/J, may touch the circle #2 -f t/2 = a2 Hence find the equation of the pairs of tangents that can be drawn from x i> ^i) to touch the circle. ( 16.
(#2
.
2/2)
17. From each of two points A, B pairs of tangents are drawn to a circle. Prove that the pole of AB is the intersection of two of the diagonals of the quadrilateral formed by the tangents. 18. A circle turns in its own plane about a point in its circumference. Find the locus of the point of contact of a tangent drawn parallel to a fixed
straight line.
L
19. If
and
Af are the feet of the perpendiculars
M
drawn from a point
P
'
are the feet of the perpendiculars to one fixed pair of lines, and L', from to another fixed pair of lines, prove that, if and L'M' are is a circle. inclined to one another at a constant angle, the locus of
LM
P
P
Show that the tangents from the origin to the circle whose equation x2 + y 5 k (x -f y] + 10 A; 2 = are the same, whatever value is assigned to &.
20.
1
is
For what values of k
will this circle touch the straight line 3o? + y-fl5-=0? that the equation of the tangents drawn from the point (h, k) 2 1 2 2 2 2 to the circle x* + if = a is (x + y - a ) (V + fc - a ) (hx + ky-a*)*.
21.
Show
=
Tangents are drawn to equidistant from the point is
2
cy
=a
this circle (c,
0).
from two points on the axis of
Show
Jc,
that the locus of their intersections
2
(c~#).
The vertices A, B, C of a triangle lie one on each and AB, AC are parallel to the tangents at circles 22.
;
Prove that
BC is
of three concentric C,
B
respectively.
parallel to the tangent at A.
Also prove that the circumcircle of the triangle formed by the tangents Cis concentric with the three given circles; and find a relation
at A, B,
connecting the radii of the four concentric circles. 23. From a fixed point P (#', y) is drawn a straight line to cut the circle 2 a?
to
+ y2
=a
P with 24. If
2
in
Q and
respect to
a straight
R
;
find the locus of the point harmonically conjugate
Q and R. line move
so that the lengths intercepted
upon
it
by
THE CIRCLE
176
two given circles are equal, the locus of its pole with regard to either circle will be a curve of the second degree. 25. Chords of a circle, radius a, subtend right angles at a point whose distance from the centre of the circle is c. Prove that the locus of their poles
is
a circle of radius
Properties of two circles.
13.
I. Definition. The points which divide the line joining the centres of two circles internally and externally in the ratio of the radii are called the Centres of Similitude of the two circles.
^
If the points Cx (c^, /3 a ), C2 (a 2 /3 2 ) are e centres of two circles, whose radii are r , r2 the centres of similitude, Si $2 are the points ,
,
}
,
ia-f2<*i ---
Of
The
S^
points
(Chapter
Show
8% evidently divide the line C^
harmonically
I, p. 15).
that
any
straight line
circles is divided similarly
Suppose two
through a centre of similitude of two
by the circles.
circles
= x*+y* + 2Gx+2I\j+C = Q x 2 + y* + 2gx + 2fy + c
have a centre of similitude at the origin
Rg-rG =
Then
G=
Hence
Now
C2
(0, 0).
F=~./
=A
.
= A.0r, F=A./
Then
(radius B),
Ef-rF = 0.
;
^.g;
put
(radius r\
JB
= Ar.
But
Hence the equation x*
of the second circle is
+ y* + 2\gx + 2\fy + \*c =
Let any straight line through the origin,
-= cc
of similitude, be
^
cos 6
it
-~L sin
=n
i.
0.
e.
through the centre
THE CIRCLE
177
x*+y* + 2\gx + 2\fy + A*c OQ are given by
If this cuts the circle
the lengths OP,
=
at
P
and Q
.e.
OP+ OQ = - 2 X (0 cos
Hence
~"
4(#cos0.f/sin0)
OP. OQ i.
+
e.
~*"
+/sin
6)
2
c
2=ia
quantity indsP 611 ^ 611 * of A and therefore the
same
for
both
circles.
Let the straight line cut the circles in the points P, Q and and let OP = ft OQ and OP'=ft'. OQ'; then it follows that
P', Q',
.
= + !/#, (ft-ft^ (ftft'-l) = 0. = I/ft' whence k = or = OP/OQ OPYOtf or OQ70P'. *+!/*
hence Therefore
The
ft'
*'
ft
;
straight line is therefore divided similarly
The
Definition.
II.
of similitude of two
as diameter
two
If the equations of the
nttg +
rgtt^
^l-H r2
\
/
circles.
the line joining the centres
is called the Circle
of Similitude.
circles are
the equation of the circle of similitude /
on
circle described
circles
by the two
is (v. p.
164)
tittg-rgtt rl~ r2
/\
This at once reduces to
r1 2{(^-a 2 ) 2 + (y-^) 2 }
or
which
is identical
-^{(af-a^-h (^-^i) 2 }
= 0,
(i)
= 0.
(ii)
with ft)*-/-!
2 }
Note that (a)
Equation
similitude of
then HIT
(i)
two
shows that if P (x, y) is any point on the whose centres are GI C2
circles
,
PC^
:
P02 =
r,
:
r2 .
circle of
THE CIKCLE
178 (b) If
S,
= (x - cOMy-ft) -/3)
are
two
circles,
we
see
from
-1 e= -| r 8
ft*
It
(c)
-^ = -r=
0,
.
a
can be shown that the circle of similitude
points at
which the two
Example. The circle
2
that the circle of similitude can be
(ii)
written
2
circles
centres of similitude oftJie circumcircle
of a triangle are
is
the locus of
subtend the same angle.
the orthocentre
and
the centre
and
nine-point
of gravity.
If the vertices of the triangle are (-Rcosft, 12 sin a), (7?cos/3, /?sin/3),
(J?cosy, J?siny),
we have seen
(p.
153) that the equation of the nine-point
(s-C)+ (y-S)* = Jtf 2
circle is
.
Hence the centres of the circumcircle and nine -point circle are (0, 0), and their radii are /? and %R< The coordinates of the centres of similitude are therefore ((7, |5) and (2 C, 2S), which are those of the centre of gravity and the orthocentre. The equation of the circle of similitude is (C, S),
or
III.
The Common Tangents
The equation this is the
of
any
circle
Two
for this problem.
circles
be 2
Pi) /3
we wish
Circles.
can be written in the form
most convenient form
Let the equations of two
of
=rS,
(i)
= r *; 2 )*
(ii)
2
to find the equations of those lines
which touch both of the
circles.
The
coordinates of any point on the circle
and the tangent at this point (x
(x and
(x
8)
a2)
cos 6
If either of these coincides
of the two circles.
ft^siaO
to the circle
cos
can be written
is
a^cosd + fy
There are two tangents
(i)
(ii)
+ (#- /3 2 + (y /3 2
with
t
.
(iii)
parallel to this, viz.
)
sin
)
sin
(iii) it
=r
=r =
(iv)
2
r2 .
will be a
common
(v)
tangent
THE CIRCLE Comparing the equations should be identical is
and
(iii)
179 the condition that they
(iv)
2,
(! - <X 2
e.
i.
)
cos 6 + (ft
-
/3 2)
+ fa - r2 =
sin
0.
)
(vi)
This equation then gives the values of which make the line a tangent to both circles. It can be written
(o^-ogq-tan 2
|0) + (fa
-
13 2 )
2 tan
4-
fa
- r,) (1 -f tan 2
J0)
(iii)
= 0,
or
a quadratic and gives two values for tan 6 with correfor costf and sintf to be substituted in (iii). values sponding The two tangents are real, coincident, or imaginary according as
This
is
2
(/3i-/3 2 )' i.
e.
as
(oc,
>, =, or
- oc,Y + (& -
/3 2 )
< K-r^-K-o^) 2 2 > = or < fa - r2 2
Since the centres of the circles are the points Ci this condition is the
same as C^Og
> =,or < ,
(oc }
,
ra
.
t'l
,
.
)
,
,
/^
C2
(ot 2
,
/
If one circle is outside or cuts the other, clearly real common tangents can be drawn.
and two
If
one
C = 2
circle r<
TI
2
lies
within
and
just
the
touches
and the common tangents
other,
then
coincide.
If one circle lies entirely within the other, C/iCa
r2
and the
tangents are imaginary.
Again,
if
equations
(! -
2)
(iii)
cos
and
+ (& -
Treating this in the same to
(v)
way
are identical, /3 2 )
sin
+ rt -f r2
as equation
two common tangents which are
M2
real,
we
get similarly that
= 0.
(vi),
(vii)
these correspond
coincident, or imaginary
THE CIRCLE
180 according as
(7a
C2
is
>, =, or < rx + r2 i.e. according as the one and touches, or cuts the other. ,
circle lies outside, lies outside
We see then that equation (vi) gives the values of common
tangents and equation
(vii)
for the
for the 'direct
transverse
'
common
tangents.
Now
let
PP' be the
equation
(vi)
with the
first circle.
points of contact of the tangents given
Since the coordinates of (a,-fr1 cos0,
where
is
1
one of the values given by
we have
x
<x l
by
P are
+ *iBin0),
(vi), if
+ t^cosd, y =
/3 1
Pis
called the point
(x,
y)
+ r1 sin0.
Hence an equation satisfied by the coordinates of P, and (by precisely the same argument) by the coordinates of P'. In other words (viii) is the is
equation of the chord of contact PP'. Now the chord of contact of tangents from the external centre of to the first cirde is
similitude^
i.
+ ,
e.
which
is identical
with equation
ff,
(y
(viii)
i
of the line PP'.
THE CIRCLE Hence the similitude
common
direct
2
181
tangents meet at the external centre of
.
In the same way it can be shown from equation (vii) that the chord of contact QQ' of the transverse common tangents is which
is
the chord of contact of tangents from the internal centre of
similitude
/Sx .
The following example
the methods of finding the equations of common tangents in numerical cases: the equation of each pair can be written down generally as the tangents from the centres of similitude to the circles.
Ex.
i.
Find
the
common
illustrates
tangents of
These equations can be written
(*-3)'+(y-2)' (*-)' + y2
The condition that and should be identical i.e.
or
is
= =
(*)*
+ (y 2) sin B = f (#-^)cos0 + ysin0 = + $ cos0 + 2 sin 6 + $ =* ^cos0-f S 9cos0-2ain0 + 2 = 7 tan'&0 + 4tan}d-ll =0. f ) cos 6
(a?
Hence
tan
0=1
or
A sin 0ei f
f
-V.
- *-$-<>
Then
($)',
or -If,
-
.
,
--4-~-*
=1
or
The corresponding common tangents are y-4 and if (*-V) + Again, the tangents (x are identical
if
1) cos
6 + (y - 2) sin 6
f cos 6 -f 2 sin 6 + f
=
^ cos 6 -
9co80-2sin^-7
i.e.
=
=
8tan a id-f2tan^-l
or i.e.
Then
tan$0=as above
cos sin
=
or
g,
0, 0,
4-
or \Z, == ^ or ^7 $
.
The corresponding tangents being or
3^-4y-9=0;
15a?
+ 8y-81
- 0.
THE CIRCLE
182
Ex.
Show
ii.
that the equations
- a) + (y - b) 2 = a _ y b 2
to the circles
(jc
a
Any tangent
to the
r,
+/
can be written
first circle
(#- a)
a',
pairs of common tangents 2 r' are tftoew ty
ttco
- a')- + (y - 7/)2 -
a,
#
6'
(oc
;
A'
,
y
of the
2
-
1
ff-a',
A
r
costf +
parallel tangent to the second circle
=
sintf
(y-6)
is
(i)
one of the straight lines
Points on the common tangents therefore and (ii). But from these costf
r.
satisfy
sintf
2 Hence, since cos 0-f sin'^
x
=
a,
.r
a
of
Let the equations of the two x~ -f- y'
x
r
7 ,
\
+ r
Two
circles
f/-TT
x-a, y b' a common tangent
!
any point on
1,
The Common Chord
IV.
a',r
b
y
a,
x
j
(i)
1
x-a,
_
both equations
*-*,
satisfies
2
' |
Circles.
be
+ 2 gjc -f 2fy
-I-
c
=
0,
(
i)
The equation 2
-f y
(jt-
2
-f
2 ^; -f 2/// -f- r) - (x* + y 1 + 2
=
(iii)
by the coordinates of any point whose coordinates satisfy both the equations (i) and (ii) ; hence it is a locus through the com-
is satisfied
mon
points of the two circles,
But the equation which
is
Note chord
is
is
equivalent to
a straight line and
i.
If
one
is
therefore the
common
chord.
circle bisects the circumference of another, their
a diameter of the latter
Thus, if
.r
2
common
circle.
4 yM 2gx
-f
2fy
-f
c
(i)
bisects the circumference of t>
x + y* + 2Gx-\ 2/>4 then (-
-F)
lies
(7
=
0,
on
2(F-/)y+C-c -
i.e.
(ii)
2((?-flf)a? + + 21? f
2<7(0-d
or the coefficients of the
first
0,
(F-/Hc-(70,
equation
(i)
satisfy the linear relation
THE CIRCLE
183
circle can be drawn to bisect the circumference of any three we then get three linear equations to find g,f, and c. The only which this fails is when
Hence a circles, for
case in
i.
when the three given
e.
Note
two
If
ii.
of them, for this coincide.
Thus,
if
two
Suppose
its
Hence,
if
1
,
F
I
t
equation
a
is
colliixear.
common chord
when the two
is
points of intersection
kT
;
a tangent to the circle C = 0, at the point of contact of T =
It follows
that
if
two
circles cut in
(72
where
i
=
a tangent to each
touch, the line C\
where they touch. then Cl - Ca ==
T
is
=0,
have their centres
= 0, C =
at the point
T=
iii.
1<\
circles
circles C\
a circle touching C^
Note
t
the limiting case
Ci^Ct-kT.
i.e.
*\
circles touch, their
is
common tangent
i
i
fc
=
is
their
a constant,
C = kT v
is
represents
0.
imaginary points, the
join of these points is a real straight line. The result of this paragraph can be written briefly thus are the equations of two circles, C^ If C t = 0, C2 = :
equation of their Conversely, if of the circle C
common
we
=
C-ku =
0,
where k
the
some constant.
is
This relation will be discussed more fully in i.
is
are told that a straight line u = is the common chord and some other circle, the equation of the second circle
must be of the form
Example
C2 =
chord.
14.
To find the equation of a circle which passes through the 2 ofx + y* 2u; + 3y + 3 = and 5x 2ylO = 0,
points of intersection
and
also through the point (8, 1).
The equation of the required circle is of the form * + ya -2# + 3y + 3-&(5#-2y-10) Since the point
(3, 1) lies
on
this,
10-3A?
Hence the required
=
0,
i.e.
fc
= y.
circle is
2y-10) 109
=
=
0,
0.
ii. To find the equation of the circle which is equal x 2 +y*~ a2 and touches it at the point (a cos a, a sin a).
Example the circle
0.
The tangent to the given
circle at this point is
x cos % 4 y sin a Hence the required
circle
is
of the form
=
a.
to
THE CIRCLE
184
The
radius of this circle
is
V\ (W COR
2 )
+
(k* sin
8
a) + a
2
- A**.
Hence, since the circles are equal,
k
i.e.
The equation required
is
a2
2
V.
To
^d ^ aw#te a
es
4a.
4a(a7C08(X-f ysinCX
which two
to the circles at their
tangents
=*=
therefore
circles cut,
i.
a).
the angle between the
e.
common points.
P
If the circles cut at P, the tangents to them at are perpendicular to the radii respectively CjP, (72 ; hence
P
the angle between the tangents is equal oV supplementary to the angle CiPC2 Now, if we denote the angle CiPC*' by \l/, .
2C P.C,P 1
If the circles are
since their centres are
(
/),
g, 2
-c
(
G,
F) and
their radii
and
we have cos
*
=
2J2rcos^ = 2gG + 2fF-C-c, where jR, r are the radii. Note i. If two circles touch internally or externally, + is or rr 2Rr. This is equivalent to the sum respectively, and 2gG + 2fF-C-c =
i.
e.
or difference of the radii being equal to the distance between the centres.
Note
iL
When
the circles cut at right angles (i. e. orthogonally) ^ is is zero : hence the condition that the two circles should cut orthogonally is 2Gg + 2 jp/ C - c = 0.
a right angle and cos ^
@i
Note (~0
lii.
The square of the length of the tangent from the centre
/) of the
first circle
to the second circle is equal to
Hence, if the circles cut orthogonally, the length of this tangent is equal to the radius of the first circle. Similarly, the length of the tangent from the
THE CIRCLE centre of the second circle to the
second
The converse of
circle.
first circle is
equal to the radius of the
this proposition is clearly true.
known and we
are finding the equation of the orthogonally, this condition is linear in the three
If the first circle is
second, cutting
185
it
unknown quantities G, F, and C. The circle can therefore be made
two other conditions. which cuts three
to fulfil
Thus, for example, one definite circle can be found
given circles orthogonally. If the tangents to two circles which cut orthogonally at one of their common points be taken as coordinate axes, the equations of the pair of circles are
Ex.
If two
i.
circles cut orthogonally,
tlte
tangents at one point of
intersection meet the circles again in points whose join passes through the
other point of intersection.
Taking the tangents at a common point points in which the axes meet the circles 2 f/
are (2r,,
-2r # = l
as coordinate axes, the other
0,
(0;2ra ).
0),
T r
2
~T~S (2
2r1 2
/* '
2
'a
Ex. ii. Show that a line cut liarmonicully by must be a diameter of one oftJiem.
tivo
This
lies
on
^-
-f
^-
=
.
1.
ortliogonal circles
Let the circles be
and the Then,
if
PQRS, the OP,
OQ
lx+ my
line
=
1.
the points of intersection are lines
are x* + y*-2rl x(lx + my)
=
0,
and OR, OS are ** + y'-2rs y(/# + my) i.e.
a
0,
(l~2rl l)-2r1 may-l-y0,
These form a harmonic pencil
if
l (1 -2^0(1 -2rl m) +
[Chap.
Ill, p. 92.]
THE CIRCLE
186
= 0, (1-rjJ) (!-,*)= 0. 1-^ = or 1 - 2 w = 0, 1
i.e. i.e.
Thus
>
either (rlf 0) or
i.e.
Ex.
ra m-f rjfj/m
i^/
A
iii.
(0,
circle cuts the
and
at angles O l
2
gonally if rA cos
=r
2
2
it
cuts
The
2
(a-a
tMr
circle
of similitude ortho-
cos O l .
Let the centre of this circle be (a, the conditions
1.
two circles
Prove that
.
Ix+my =
r2 ) lies on the line
*^-
2
and
)
=
2
2
*
its
radius be
circle of similitude is ~~
r M2
*- 2 r i
V {(*-i) + (y-W-^f^-a^ + fy-W f
or
and
2
a
since the coefficients of
tangents from
(ft,
and
y* are
2
2
= Hence the
Ex.
_^r
l
2
p
2
-
2
) ;
of the
ilic
circle
Ix+my+n = 0.
=
rj
.
equation of the circle having for a diameter that
%2 +y 2
+2
gx + 2fy + c
common
# + t/*-f20a?-f2/y+c=i 2
equation
is
=
whose
equation
points of
and Ix 4- my + n
x* + y* + 2gx + 2fy + c + 2k(lx +
=
0,
my + n
-(f+km)}
=
lies
on
my + n) =
a diameter of this
is
0.
circle,
hence
its
this line.
= l(g + M) + m(f+ km)-n 2 =B k (I* -f m ) (^/ -f w/- n).
or
Thus the required equation 2
(/
-f
is
of the form
Moreover, fc + ),
)
1
cos #2
Since the circle passes through the
its
-2r2 pcos0 2
2
circle cuts the circle of similitude orthogonally.
Find
iv.
0,
r2 ^( ; g COS 6 l ^ ~~_ -^jtjCOS 2 ) ;: 2~ ^ 2
since r2 cos 6 l
,
}
(rf-rf), the square of the
2
p
-
8
#) to this circle
< ?
chord
then we have
p,
-2
m)
2
(a;
is
4^4- 2^o? + 2/iy -f c)
2(^ -f MI/- 11)
(Ix
-f
iwy
-f
n).
centre
THE CIRCLE
V
Examples 1.
Find the equations of the
-V-8a? + 2 =
2
2.
187
k.
2 tangents of the circles x
common
2 -i
y
^
1,
0.
Find the coordinates of the centres of similitude of the 1
11
=
circles
0,
=0,
and the equation of their real common tangents. 3. Find the equation of the circle cutting orthogonally the three
circles
=0, 0,
or + y 2 -4rr 4.
Find the equation of a
circle
+ 6t/ + 9
which
=
0.
bisects the circumferences of
= 3. -f if + 2 y Find the equations to the two circles which have their centres at the 2 origin and touch the circle ^-f y -6;r-8y + 9 = 0. 2 6. Find all the common tangents of the circles .r -f if 3a?-54 = 0, tfHt/
2
^!,
+ y* + 2a:
3*
#2
3,
5.
;c
2
-ff/'-21oM 90
=
0.
Find the equation to the
circle which passes through the point (1, 1), and the points of intersection of 3#-f 7 = 5i/, 3#-2a:2 ~ 5y-f 2y2 -17. 8. Find the radius of the circle of similitude of two circles in terms of their radii and the distance between their centres. 9. Write down the equation to the circle which passes through the point 2 (a, b) and the points common to Ix + my = 1, ar-t f/ -f 2gx-t 2fy + c = 0. 10. A and B are the centres of two orthogonal circles intersecting in 0, and Pis any point on the circle AOB. Show that P is equidistant from the two points in which PO cuts the circles. 11. Find the equations of the two circles which cut orthogonally the circles aH t/ 2 4 2#-9 = 0, #2 -f ^-8^-9 = 0, and touch the line y-# = 4. Show that the distance between their centres is 10 */2. 12. Show that the circle on the line joining the centres of similitude of 7.
as diameter 13.
is
x
1
-f
tf
- 2 x (W + 6 2 )/(fc + V) + 6 2 =
0.
Find the condition that the circles (x -*c)
2
- 2 Or c)
-f
X (x -f
-I-
-f
2
c) 4- (1 42
ji
(x
4-
c)
(1
X)y
4- fi)
2
y
= 0,
2
should cut at right angles. 14. Show that the equation of the circle which cuts each of the three
= a2 a^-f y*-bx-cy + a2 circles
2 a? -f
2
t/
,
(x-b)*+ y
=
a2
#2 -f (y - c) 2
,
=
a 2 at right angles
is
= 0.
Show that the circle x + y*-2bx + &2 cos 2 a * O'is 15.
2
of similitude of
2 a?
4
2 f/
-2aa? + a 2 cos2 a
'0,
= 2abx. (a 4 &) (a; -f y ) 16. If two circles intersect, their circle of similitude passes through their points of intersection. 2
2
2
THE CIRCLE
188
17. If two circles cut orthogonally, the extremities of a diameter of either are conjugate points with respect to the other. 2x = 0, 18. Find the equation to a circle which touches #2 + y2
a^-f y* + 3x =s
at their point of contact, and has internal or external contact with the circle (x - 3)2 + (y - 1)2 = 1 19. If a circle of fixed radius p cuts a circle C = whose radius is r at an .
8 angle #, the locus of its centre is <7 = r - 2rp cos Of. z Find 20. the locus of a point at which the two circles y? + y 1 - 2 a x x + b
=
0,
= 2a,iX + b*
a^-f y8
subtend the same angle. 21. Find the locus of the centre of a circle which cuts the circles (x
at angles
and a
j3
- atf + (y - fcj)8
c
2
cos 2 a,
respectively.
Find the locus of the centre of a circle which cuts each of two given circles at a given angle. 23. Find the equation of a circle which cuts orthogonally 22.
-f
c
=
0, 0,
and
also the line Ix +
my =
1.
24. Find the equation of the circle which has for its diameter the chord 8 8 = 8 8 cut off on the straight line ax + by + c = by the circle (a 4- fc ) (x* -f y ) 2 c 25. Circles are drawn through the point (c, 0) touching the circle .
#2 -f y* = a2 Show that the locus of the pole of the axis of x with respect to them is 4a2 (x-c) 4 = ( 2 -c2 ) (a2 -(c-2a?) 2 }y 2 is the 26. If C=x* + y* + 2gx2fy + c, and U = XCOB a + ysiircx-j>, and value of w at the centre of the circle C = 0, then the equation of the circle on the chord which (7 = cuts off from u = is C 2ww = 0. .
.
If
27.
28. If ft
it
/LI
of the chord
may be
= =
=
R
C == a?8 -f y8 + 20a? -f 2/y -f c =
P
=
Cx 0, C2 0, C3 orthogonally, cuts all circles of the system X Ct -f C2 -f i/ Cs orthogonally. are the points of contact of the tangents drawn to the circle
a circle cuts three circles
prove that
from an external point (h,
QR and show that the
written in the form 2r*P
=
&), find
circle described
8 8 {(k+ $r) + (&+/) } 0,
on /-
the equation as diameter
QR
being the radius
of the circle.
=
C8 =
=
are three intersecting circles and and c are constants, then all where a, three circles pass through the same two points and if this condition is satisfied the circle IC^mC^nC^ for any values of I, m, and n also two these passes through points. 29.
aC}
-f
If
Cj
&C3 -f cC8
is
0,
0,
<78
identically zero,
fc,
;
14.
Systems of
Circles.
= 0, = being the equations of two
circles,
we have shown
that
THE CIRCLE the equation of their
is
meet in
common
=
chord
;
189
when
the circles do not
represents a straight line satisfied the of all coordinates to GI common it is the 0, <72 by ; points of their of intersection. join imaginary points real points, (?t
(72
still
=
Now (a?',
=
the square of the lengths of the tangents from any point
= is found by substituting x' and y' for (7, Another geometrical interpretation can now be
to the circle
if)
x and y in
Ci.
given to the equation
d-C^O, which enables us to define the
line in real terms in all cases.
The equation represents the locus of a point, the tangents from which to the two circles are equal ; hence the definition :
The equal,
locus of is
a point, tangents from which
a straight
line
which
is
two given circles are called the Radical Axis of the two to
circles.
Evidently, when the circles cut in real points, the radical axis if the circles touch each other, it is the the common chord common tangent at the common point. In every case the radical
is
:
axis bisects the
The equation
common
tangents to the
circles.
of the radical axis reduces to
2*(pi-02) + 2y C/i-y^-fCi-Ca^O.
(i)
Now and
the centres of the circles Clf (72 a?e ("-0i> A)> /i)> (~ #2* therefore the equation of the line joining the centres (the line of
centres) is
(x+ffi)(A-A)-(y+fi)(ffi-ff*)
This
perpendicular to the line
is
circles is perpendicular to their line
Now It
;
hence
(ii)
the radical axis
of two
of centres.
consider the equation
x*+y*+2gx + 2fy + c + A(te+wy + n) = represents a circle whatever valus A may have.
But the
is
(i)
= 0.
radical axis of this circle
the straight line Ix + my + n
Hence
and the
(iii)
circle
= 0.
for different values of A, equation
of circles each of
0.
which has the same
(iii)
represents a system
radical axis
with the circle x*+y* + 2gx+2fy+c = 0.
Ix+my + n
Q
THE CIRCLE
190 Further,
be any two
if
system, their radical axis
circles .of this
is
.e.
=
represents when A varies a system of circles, such that every pair has the same radical 0. Such a system of circles is called a Coacal axis lx + vny + n
Thus x2 -f y* + 2 ## 4- 2/y -f c + A (Z# -f m?/ -f n)
System.
In abridged notation we can write
briefly* if
C=
is
a circle and
=
=
a straight line, C-h \u represents a coaxal system of circles is the radical axis, of which u
u
=
Since the radical axis of two circles is perpendicular to their line of centres, the centres of all circles in a coaxal system lie on a straight line.
The equation of a system of coaxal circles is simplified by taking the line of centres as the axis of x and the radical axis, which is perpendicular to it, as the axis of y. In
this case, since the centre is 2
equation is of the form x 4- y Now the radical axis is x
2
+ 2 gx + c
= 0,
circles is
x* + y*
for different values of
or,
Jc,
on the axis of x
=
f=
0,
and the
0.
hence the equation of the system of
+ 2gx + c + kx=
e.
i.
writing 2 A for the variable coefficient of
The constant
t
c is fixed
;
by varying the
x,
we have
coefficient
A the equation
of all the circles of the system can be obtained, and further, to every value of A there corresponds a circle of the system.
Properties of a system of coaxal circles. (i)
The general equation
can be written
and
if
+ \) 2 '+y* = A 2 -c, values + Vc, this reduces
(x
A has either of the (a;-v
of a circle of the system
/ c)
2
+y = 2
These equations represent
or circles
(x+
to
v^+y = 0. 2
whose centres are
(
and
THE CIRCLE and whose radii are zero
\/c, 0)
(
which can
:
191
the only real values of # and y of these centres.
them are the coordinates
satisfy
These points (\/c, 0), ( Vc, 0) are called the Limiting Points of the coaxal system they lie on the line of centres at equal distances from the radical axis on either side. They are often referred to as :
the Point-circles of the system.
The limiting points when c is negative.
Now
the circle # 2 -f y'2 +
in points
when
are real
c
c is positive,
+ 2 \x =
and imaginary
meets the radical axis
whose coordinates are given by
y* + c =
0,
i.
e.
oc
=
the radical
axis meets the circles of the system in real or imaginary points according as c is negative or positive.
Thus a system
of coaxal circles
points has real limiting points
:
which
intersect only in imaginary a system of coaxal circles which
intersect in real points has imaginary limiting points. Hence #'2 -f7/ 2 -f 8 2 -f 2 A x represents in general a
=
coaxal circles which do limiting points are
(8, 0),
system of
not intersect in real points, and whose
(-8,
0).
The
limiting points are conjugate with respect to every circle of the system and have the same polars with respect to all circles of the system. For the polar of (8, 0) with regard to (ii)
s. i.e. i.
(a?
+ 8)(8 + A)= x+8
e.
0,
= 0,
a line through the other limiting point parallel to the radical axis. Thus the polars of the limiting points are the same for all circles of the coaxal system and are conjugate with respect to every circle of the system.
The equation
(iii)
hence
if
the circle
is
of
real
circle of the system has (iv)
Any
circle
any
circle of the
A 2 >8 2
;
its centre
system can be written
is (A, 0), hence no between the limiting points.
the centre
through the limiting points cuts
all
the circles of
the system orthogonally.
Any
circle
through
which cuts any
(8, 0),
circle of the
(-8,
0) is [p.
162]
system
+ y* + 2\ (The condition 2 Gg + 2 Ff~ C+ c x*
orthogonally.
real
is satisfied. )
THE CIRCLE
192
Ex.
All
i.
circles
which
bisect the circumferences
of two given
circles
form a coqxal system. Let the equations of the two given circles be
Any diameter
of
Hence the
-/,
(
(i)
0.
(ii)
of the form
(i) is
x+f+ky = since its centre is
0,
0,
0).
circle 1
ff*+y + c + 2/r + X (x+f+ky) bisects the circumference of (i). The common chord of the circles (ii) and (iii) is but this ference.
The
(iii)
2 (f-9) x + \(x+f+ky) = 0; a diameter of the circle (ii) if the circle (iii) bisects Hence the centre of (ii), viz. ( - g, 0), lies on this line.
is
circle bisecting the circumferences of the circles
(i)
and
its
circum-
Hence
(ii) is
accord-
2fx + 2g(x+f+ky) = a? + y* + 2(f+g)x + c + 2f0 + 2gky=*Q, z? + i? + c +
ingly i.e.
which
=
for different values of the
system of coaxal
circles,
t
undetermined constant k represents a
the radical axis being y
0.
are two circles of a coaxal system, show that the point-circles are given by
theeqwtionC^f+g')*-2CC'(f+g')(f'+g)+C'^f+g) = <>. Any circle of the system is C+\C' ~ 0, or written in full a? + y* + 2gx + 2fy-2fg + \(x* + y* + 2g'x + 2f'y-2f'
This
is
a point-circle of the system
TTx (9
i.e.
\~T+T
i.
e. if
<
TTx
+ *g'Y + (f+ X/')' + 2 (1 + X) f
(fg + X/y)
X(^+/') + 2X(^+jr+/y+>
or
is
if its radius is zero,
- 0,
+ fo+/)-
0,
Hence, if X has either of the values given by this equation, a point-circle of the system. Thus the coordinates of the point-circles satisfy /,
x-.
where X
satisfies
the above equation.
Therefore the equation of the point-circles
is
C+\C'
=
THE CIRCLE Three
II.
Circles.
The radical axes of three
taken in pairs are concurrent.
circles
=
are three circles, their radical axes 0, C3 C\ == 0, C2 taken in pairs are the lines if
For,
when
193
(--C.^O, Ca -C/3 =0, Ca-C^O, which evidently are concurrent. Their point of intersection is = = it is the Radical Centre of the three (7 called given by t C% C^ ;
circles.
The lengths
of the tangents
from the Eadical Centre
=
=
are equal, since its coordinates satisfy C\ 6'2 The circle whose centre is the Radical Centre
one of these tangents
equal to all
is called
to the circles
3.
and whose radius
is
the Radical Circle, and cuts
three circles orthogonally.
= 0,
If Ci
similitude
lie
the
=
are three circles,
six centres
their
of
external centres of similitude of
C^C^
C
x
C3 C ^1 C6J and
the
and
the
O.^C3) <72 ,
.
A
f
(72
tivo
external of (d)
Cj
two internal centres of similitude of
external of (c)
0,
in sets of three on four straight lines, viz.
(a) the three
(b) the
=
<7>
6a
.
internal centres
of similitude of
C^Cl
,
of similitude of C^Cl
,
C^C/j,
C^.
the two internal
centres
C^Cj, and the
external of C^C^.
These four lines are called
Axes of
Similitude.
Let the centres and radii of the circles be
-
(o^,
/tfj, )\
;
(<* 2
,
/^ 2 ),
r2
;
ta 3 , /^), '3-
The
external centres of similitude are ('2<X3~*a<X2 ra ^2 (
s
Let
-
,
fa^-^a^al> -
'a-^j
,
)
\Wi-W3 ---'j-''i
t
+ my + n = Q
--
r*i*\-rif**l
-
-f
}
^-^i
>
;
3
be the straight line joining the first two; then, of the first in this equation, we obtain the coordinates substituting to
^ a -f n
l
and, substituting the coordinates of the second,
+n 127
It
THE CIECLE
194
Hence
lot
+n
l
+*
and the third centre of similitude
To
also lies
Thus
line.
+ wy +n =0, = n + 0, J! #/*!
/#
~Ajr2 -fcr3 If
on the
obtain the equation of this line, let
we
eliminate
w,
Z,
and
n,
we
A;,
= 0, = 0.
obtain the equation in the form
1
y
If we change the sign of /^ we obtain the equation of the axis of similitude which passes through the external centre of C2 C3 and the internal centres of Q, C2 and Clt Ca and similarly for the ,
,
other two.
Ex.
i.
Prove that the locus of
the centre
of a
same angle is the perpendicular centre on an axis of similitude. circles at the
Let the circles be
and suppose (,
17)
the coordinates
,
17
(rj
(7 4
<72
Similarly they satisfy 2
^(^-j3,)
and
2
-^
fall
from
given
their radical
2
0,
p the radius of the cutting circle.
- ft) 2 = r* + p8 4- 2
r,
p cos
a
;
satisfy
f
Eliminating p
2
to be the centre
- a,) 8 -f (f
Then i.e.
~(^~^i)
C,
circle cutting three
let
and p cos
=p =p
2
-f
2
-f
2 ;*j p cos 0(. 2 r2 p cos Of,
from these three equations, we get the
ft
equation
.
,
This
which
is
a straight
line,
=
and
(72
=
r3 r,
=0,
it
evidently passes through the radical centre,
(73
.
given by Cj The equations of the axes of is
1 1
verified that the straight line
-^
similitude were found above; it can be perpendicular to the axis of similitude
(i) is
which contains the three external centres of similitude.
THE CIRCLE
195
The angle at which two circles cut can be taken as the acute or the obtuse angle between the tangents at the point of intersection. We might therefore have taken
giving as the locus of the centre
i.e.
,
I
r,
2
1
r,
*
1
r
=0,
four different straight lines,
n(Ct -Ci 1 (Ct -C l s (C 1 -C1 )-0. These will be the perpendiculars from the radical centre to the four axes
)r
)r
of similitude.
Ex.
//
ii.
d = 0,
C2
= 0,
CYa
=
are three circles, then
A 1 C 1 -fA 2 C 2 + A 3 Oa f
f
represents for different values of A 1
a fixed
,
=
A a , A 3 a system of
circles cutting
circle orthogonally.
Let (a, 0) be the centre of a circle and p its radius, and let C,', (?,', stand for the values of Clt C3 (73 when a, are substituted for a? and y.
C3
'
,
Now
this circle cuts the circle
*iCl + A t Ct + A t CI
:0
(i)
orthogonally, provided that the square of the tangent from (a, circle (i) is equal to p 2
j8)
to the
.
The coefficients of a:2 and / in equation (i) are (Xj-f X,-f X s ). Hence the square of the tangent from (a, 0) to the circle is A^CI 4-XjCj
"f-X 8
C8
Xil^TAi Thus,
if
the circle whose centre X1
i.e.
It will is
then cut
all circles
CV-p = This
ia
possible if
((X,
is
and
0,
and radius p cuts the
ft)
circle
2
+ X t + X 8 ), f = )-|.X i (Ci '-p ) p
(X t
(i)
0.
orthogonally
if this
X s , i.e. if it is possible to
condition
have
C9 '-p' = 0, C^-p^O = C8 = ps 0,' '
.
3) is the radical centre of
and p 2
,
= C,/
i.e.
(Of,
represented by
true for all values of X lf X a
simultaneously,
is
CV + X 8
orthogonally
- 0, C = 0, C9 = 0, 2
the square of the length of the tangent from the radical centre to
either circle.
Hence
for all values of X lf X 8f X 8 the circle
XjCj
+ XaCa + XjjCg^O
cuts orthogonally the circle whose centre
N 2
is
the radical centre of Cj
0,
THE CIRCLE
196
Ca SB
C9 =
0,
0,
and whose radius
is
equal to the tangent from this centre to
either circle.
In particular the circles orthogonally, corresponding ^i ^j ^s are zero.
C = 0, C2 =
0,
t
to
the cases
C8 =
each cut this circle
when two
of the coefficients
>
Examples VI. Find the radical centre of the
1.
circles
1=0, 5
=
0,
0,
and
find
Hence 2.
whether
it lies
inside or outside the circles.
find the equation of a circle cutting all three orthogonally.
Find the radical axis of
2#a + 2y2 -3#-i-5y42 x*+ yf + 8a? + 4y-5 and show that the circles cutting these two two fixed points on their line of centres. 3.
= 0, = 0,
circles orthogonally pass
through
Find the equations of the three radical axes of the circles 3 (*- a) +(y -&)> = &,
(*-&)' -My- a)
(>-a-&-c)
2 -f
y
9 3
= a', = afc-f c
2 ,
and prove that they are concurrent. Find also the equation of the circle which cuts them
all
three ortho-
gonally. 4.
Find the coordinates of the limiting points of
5.
Two
axis.
circles
whose centres are
(a, 6) (6, 0)
If the radius of the first circle
two
is r,
have the axis of y as radical
find that of the second.
circles cut a third circle orthogonally, the radical axis of the
two through the centre of the third circle. 7. Show that the locus of a point the tangents from which to two given circles are in a constant ratio is a coaxal circle. 8. The polars of a point P with respect to two given circles meet in Q 6.
If
circles passes
:
show that the radical axis of the circles bisects PQ. 8 9. In the equation a^ + y * 2gx + c = 0, if g is a variable parameter and a fixed the then (x, y) point, pol&rs of (#', y') with respect to the circles all a fixed point lying on a circle through (x\ y') and the limiting pass through points of the circles. 10. Show that the three circles of similitude of three given circles taken in pairs are coaxal. 11. If the equations of one circle and of the radical axis of this circle
and another are respectively l l a(# + y ) + 2#i: + 2jfy + c find the equation of the other circle
and
fc?
+ my +
0,
with the proper number of arbitrary
constants and the coordinates of the limiting points.
THE CIRCLE 12. Circles
which cut two
fixed circles of
197 a coaxal system at constant
angles will cut all circles of the system at constant angles. 13. Two systems of coaxal circles are such that the radical axis of either
Show that the product of the radii of any one of each system, which touch one another, is constant. 14. A certain point haa the same polar with respect to each of two circles: prove that a common tangent subtends a right angle at the point. 15. Prove that the locus of the middle points of chords of a fixed circle which subtend a right angle at a fixed point is a circle, and that the fixed pojnt is a limiting point of the two circles. 16. A common tangent is drawn to two circles so as to intersect the line
is
the line of centres of the other.
two
circles,
joining the centres when produced, and S is a limiting point external to one Prove that twice the perpendicular from S circle and internal to the other. to this tangent is a harmonic mean between the greatest distances of S from
each of the
B
circles.
D
A
are four circles the radical axis of C, also the radical axis of A and dicular to that of C and 17.
A,
y
:
D
B and Z); B and C.
to that of
that of
prove that the radical axis of
A and D
and
C is
;
is
B
is
perpen-
perpendicular
perpendicular to
=
8 a 2 and an equal that the limiting points of the circle o? -f y8 + n lie on the locus circle with centre on the line Ix + my 2 2 0. (a? + y ) ( lx + my + n) -f a (1x + my)
18.
Show
=
Find the limiting points of the system of circles defined by the 2 2 2 2/y -f c') = 0, and show that they subequation x* + y + 2gx -f c -f X (a? -f y tend a right angle at the origin if cgr* + c'f~* *= 2. 20. Show that the circle of similitude of any two of the circles described 19.
-I-
sides of a triangle as diameters cuts orthogonally the circle circum-
on the
scribing the triangle.
E
ABC
CFare
such that Fare points on the sides of a triangle concurrent. Prove that the radical axes of the circle
circles
on AD, BE,
21. D,
t
AD, BE,
ABC and the
CF as
diameters meet BC, CA,
AB
in three collinear
points.
15.
Other Forms.
u
I.
(HO
C
Let the four sides of a quadrilateral
ABCD
u EEtfcoso^+ysino^ p l = 0, v EE #cosa a +ysin<x a p 2 = 0, w = x cos 3 + y sin <X 3 p3 = 0, * = x cos 4 + y sin <X 4 p = 0.
be (i)
(ii) (iii)
(iv)
THE CIRCLE
198
(x cos
!
= \v&,
uw
Consider the equation
which, written in
+y sin tXj pj (# cosa 3 +# sino^ # 3
= A (x cos where X
2 4-
)
y sin <X 2 -j?2) (* cos
This
a constant.
is
full, is
4
+ y sin
4 -^4),
(v)
by the coordinates of each
is satisfied
=
B
of the points A, B, C, D: thus, is the point of intersection of u and v its therefore coordinates make u and v zero, i.e. satisfy ;
=
the above equation. Hence equation (v) represents a locus passing through the four points A, B, C, D. In order that the equation should represent a circle two conditions must be satisfied, but we have
only one constant, A, undetermined: thus the equation can only represent a circle when some definite relation exists between the coefficients of the equations of the lines u,
The
i.
w,
z.
conditions for a circle give us
A cos
cos (Xj cos <X 3
and
v,
sin
<X
cos a x
3
-f
and
4
sin c^ cos <X 8
cos (<*!
e.
cos
2
sin (o^
Hence the condition
+ <X 3) -f
3)
= sin sin a 3 A sin sin a = A (sin a cos a + sin 2 cos 2
!
4
= A cos = A sin
2
(
+
(<X 2 4-
that the lines
or
(Xj
where n
is
an
(<*!
+
8)
a2 + aa
4
,
4 ),
*
> 4 ).
(vi)
*
(i), (ii), (iii),
a quadrilateral which can be circumscribed by a tan
<*
2
4 ),
4
= tan = WTT,
(<* 2 4-
(iv)
should form
circle is
a 4 ),
integer.
This corresponds to the proposition 'the opposite angles of a cyclic quadrilateral are
Note.
It follows also
according as
supplementary '. from
we take the upper
(vi)
that X
=
or lower sign
and n
1, ;
is
even or odd
for example, taking X
1,
the equation
= (a? cos a + y
-f
y sin
represents a circle;
it
a -f y - a cos a - y }
{a?
cos
ft
+8
-I
y sin
can in fact be reduced to a^-fy*
=
=
=
-f
a
0, w 0, v suppose that the lines w z then A and the two such that coincide, points
II.
Now
D
locus represented
in
two coincident
by
wr
points, i.e. ^
r=
=
A
,
is
&
- a cos 3 - $}
9 .
= 0, z = = meets
vz
a tangent to the locus.
are
the
THE CIRCLE But
=
a line through the intersection of u hence [Chap. II, p. 52] #== Jw 4- mw, and are constants.
in this case z
and w = where I
is
=
:
w
Thus the equation of the locus (when A and D coincide) becomes
uw where
199
lu
+ mw
= v (lu + mw)
=
is
if u = 0, v =
the tangent at the
= 0, w = 0. = are the three sides 0, w luv + mvw + nuw =
point of intersection of
Thus,
A BCD
u
of a triangle,
represents a locus passing through the vertices, and the tangents at the vertices are mv + nu 0, nw + Iv 0, lu + wit; respectively.
=
Example, triangle
jfo
=
^nd Me the lines
formed by
equation of the circle circumscribing the ax 2 + 2 ftr// + fy/2 0. 0, fa? + m// + n
=
Suppose that the tangent to the
then, for
some values of A and fla? -f
2 /lory
=
=
circle at the origin is
Z?,
-f fey
2
represents the circle. The conditions for a circle give ug
Thus,
a-Al = b-Bm and 27* 4/-Bw + &-a = and
^
B
1
r
The equation of the circle is then t 2 (a? + 2/w^ + &// )(P + w*){^
{
which reduces to *
- 2^/m -f bP}
2 Au (1/10? + ty)
-f
w (a
- 6)
(te
- my).
THE CIRCLE
200
Again, in the quadrilateral
III.
opposite sides
(ii)
=
Xt>'
becomes
wt0
nnd
(iv)
uw
=
.
=
B
in coincident points at A, and in coincident points at (7, D, i.e. the locus touches
w=
the lines u ==
suppose that the pair of
2
This locus meets the line u the line
A BCD,
coincide; then the equation
0,
w=
at the points
where v
=
meets them.
Thus, (oc
cos
!
+ y sin a^-^) (x cos
3
+
?y
sin
or 3
represents a curve to which the lines line (ii) being the chord of contact.
As
in
I,
-pj)
= (i)
this curve can only be a circle
cos
(<*!
+ <X 3 = )
A cos 2 a 3
,
sin
A (x cos
and
(iii)
or,
+
y sin
<x
2
p 2 )-
are tangents, the
if
(^ +
3)
= A sin 2
2,
= ^7r+2a LEAC = LEDA, i.e. a 1 + cx 3
i.e.
2
,
the tangents must be equivalent to to the chord of contact. If this condition is inclined equally the above conditions the of either value of A. gives fulfilled,
which
IV.
is
Let
C=
x*
c
=
be any circle and
= 0, cos 2 -f y sin a 2 ~"^2 ^ ^ at A, D and B, C respectively. two straight lines cutting Now the equation C = huv represents some locus; tha coordinates and u = 0, and therefore of the points A and D satisfy both (7 = So also for B and C. Hence C = A uv represents lie on this locus. u
EE
v =
x cos Oi l
+ y sin a
l
pl
rr
it
a locus passing through the points A, B, C, D. This can never represent a circle unless the
locus
C
=
=
Awv
coincides with the original circle, i. e. A 0, for three points are It can, however, for certain values of A, sufficient to fix a circle.
represent the pairs of straight lines
AB and
CD, or
^4 (7
and BD.
THE CIRCLE
201
If a pair of straight lines is drawn through a fixed point fixed straight lines in four concyclic points, shoiv that the locus of the centre of the circle is a straight line.
Example.
to
meet
tivo
Let the two fixed straight lines be y mx =
0,
(i
t/4-wa? = 0.
is
(ii)
x2 + y 2 + 2gx + 2/y + c
Suppose that
any one of the
Then
circles.
A (y - w 1
)
=
(iii)
the equation
x 1 + y* + 2gv + 2/y + c
2
*-)
(iv)
by the coordinates of the four points common to (i), (ii), and (iii), and hence represents a locus passing through the four points. For some is satisfied
points,
A this equation represents two stiaight lines through the four and we are given that these straight lines intersect in a fixed point,
say (p,
q).
values of
The equation and (Chap. we have
(iv)
Ill,
can be written
11), if this represents (1
two straight
+ m*A)p+g
lines
through (p
t
q),
= 0,
(l
Hence, eliminating A, we have
But the centre of the
circle is
(-0, -/)
;
it
therefore lies on the straight
- = q (x -p) + m*p (y q)
Examples
0.
Vm.
1. Find the equation of the other two pairs of straight lines which pass a 4 = 0, and the circle through the intersections of x* 2xy + y
2.
A
4y = at the origin and cuts at the points Pand Q. -8j, find its equation.
circle touches the straight line 3a? +
the straight lines 7#* +
PQ passes through
ll#y + 3y
2
=
the point (1, 3. The common chord of a given circle and any other circle of given radius a passes through a fixed point. Find the locus of the centre of the If
circle of radius a. 4.
Show
analytically that if a parallelogram
must be a rectangle. 5. Show that the two pairs of straight
is
inscribed in a circle
lines
y*+10*-6y-24 =
0,
form a cyclic quadrilateral. Find the equation, centre, and radius of the circumscribing
circle.
it
THE CIRCLE
202 6.
through the intersections of
Circles are described
0,
a** + 2Jwpy + &y2
and
Show
(i)
0.
(ii)
that the other chord of intersection with
(ii) is
fixed in direction,
and
a form containing one arbitrary constant. Find the equation of a circle touching the o?-axis and passing through
find its equation in 7.
the points of intersection of the circles
0. 8.
The
^+y
circles
-2 for - d* =
2
x* + y*-2k'x-
A
intersect in
meeting the
and B.
circles in
0,
=0
Through
A
C and D
respectively.
circle circumscribing the triangle
a line
drawn perpendicular
is
to
AB
Find the equation of the
BCD.
Write down the equation to the circle which passes through the point 1) and the points common to the circles
9.
(2,
10.
Find the equation of the
line 8a? + 4y 11.
The
circle
whose diameter
=
is
<
the portion of the
12 intercepted by the lines bx *-lxy + 2y* points of intersection of the circles
=
= 0.
0,
subtend a right angle at the origin. Prove that #2 -/ 2 = 2c. 12. Find the area of the triangle formed by the three points where the circle a?
+ y*
=
2ax + 2by
is
cut by the pair of straight lines
Find the equation of the circle which passes through the intersection #* + y2 -3ar-2t/-6 = 0, #2 + y8 -5a? + A y + 2 = 0, and has its centre on the straight line x = y. 14. Show that the equation of the circle whose diameter is the portion 13.
of the circles
of the line
this passes
lx+my*=l
intercepted by the lines
through the origin
:
to
aa?
what geometrical
a
+ 2hxy + by*
=
is
fact does this corre-
spond ? 15. If
and are the equations of the pairs of opposite sides of a quadrilateral inscribed in a circle, show that H(b-a) h(B-A). 16. The straight line x cos Of + y sin a =*p being called the line (#,;>), find the equation of the circle circumscribing the triangle formed by the lines (Of, p) (0, q) (y, r), and show it passes through the origin if
grsin
(-y) -f rp sin (y-Ot) + l^sin (<X-^) -
0.
THE CIRCLE
208
Find the equation of the circumcircle of the triangle formed by the bx + cy + a = 0, ex + ay + b = 0, ax + 4- c = 0, and show that it passes 2 2 2 2 8 2 = if c a fc the + + ate (6 + c) (c + a) (a + b). + ) (c ) (a origin ) (fc through 17.
%
lines
Prove that the equations of the common tangents of the circle circle whose diameter is the chord x cos OK -f y sin OL = 15
18. 9 a?
+ ya = 289 and the
of the
are 3 (x cos a -f y sin a) + 4 (y cos OL - a? sin a) = 85. is a circle, where u = 0, ; are straight lines, 0, fv point on the circle is the point of intersection of two lines of
first circle
= uP
19. If wt?
show that any
=w
\u
=
=
=
\w. Indicating any point on the circle by the parameter X, show that the chord joining two points whose parameters are A t and A 2 is - (AJ + X a ) 10 = 0. XjXjj u + v
the form
v
t
20. Show that the equation of the tangent to the circle in Question 19 at the point whose parameter is X is X 2 w + t> 2\w = 0. Show also that the tangent at the points X and p. intersect at a point whose coordinates satisfy
the equation
w/2
tV2Xfi
=
M>/(X
+ fi).
Use the notation in Question 20 to solve the following OA, OB are tangents to a circle, P is any point on the circle, and the at C and D. Find the locus of the lines PA, PB meet any line through intersection of EC and A D. 22. OA and OB are tangents to a circle, a line through meets the circle at P, Q, and AB at R. If it = 0, v = are the tangents OA and OB, and the equation of the circle is uv = w*, find the equations of the lines AP, AQ, and thus show that 0, 7?, P, Q form a harmonic range. 21.
:
23.
Show
=
a circle uv
that the pole of the line lu + mv + nw = 0, with respect to 8 is the intersection of the lines nu + 2mf0 = 0, w> t? + 2to = 0, ,
=
=
uv = w*
is uv'
t>/2J w/ n. given by u/2m f 24. The coordinates of a point make u y v, and w equal to u', t>', and w Show that the polar of thia point with respect to a circle respectively.
i.e. is
whose equation is
is
+ uv
2 uw'.
that the triangle whose sides are u-lv 2 self-conjugate with respect to the circle uv = to
25.
Show
= 0,
u + Iv
0,
w=
.
= 0) are tangents to the circle uv = 10*, and any 26. 0-4 (u 0), OB (o f Show that the locus other tangent to the circle meets them in A' and B of the intersection of AB' and A'B is uv =
=
.
The
16.
If GI all
a
= 0,
values circle.
'
circular points at infinity
C2
of X
When
'.
=
are the equations of two circles, then for A C2 represents except unity the equation Cj the circles Clf C2 intersect in two real points,
=
=
A C2 pass through these two the circles represented by GI In order to obtain complete generality we introduced in points.
all
Chap.
IV
the ideas of imaginary points and coincident points
;
so
THE CIRCLE
204
we may
that
say that
the circles represented by the equation
all
= X C. pass through the two points of intersection of the circles Now any other type of locus might pass and C = 0. Cj = 2
C?!
2
through these two points of intersection
;
this
is
property
not
therefore a geometrical explanation of the algebraical result that (7X AOg alivays represents a circle.
=
if u = 0, v = = Auv passes Cj
Again, locus
are the equations of two straight lines, the through the four points of intersection of
the straight lines u, v with the circle Cr This locus is not a circle except in the special case when A is zero. On the other hand, the
=
Aw always represents a circle this circle passes equation C^ and the the two points of intersection of the line u through does this not but circle C 0, property evidently correspond to have seen in Chap. IV that the fact that the locus is a circle. ;
=
=
We
in order to obtain complete generality we had to adopt the ideas Also we saw that of points at infinity and the line at infinity '. '
*
*
the properties of a locus with respect to the line at infinity could only be satisfactorily examined by using homogeneous coordinates. Euclidean Geometry fails to explain the facts given above ; we
proceed to examine, by the use of homogeneous coordinates, whether projective geometry offers an explanation. The general equation of a circle in homogeneous coordinates is 0. The points of intersection of the circle #2 _j_ y* + cez 4. 2Jy& + 2gzx and and the line at infinity, z 0, are therefore given by x* + y*
=
=
=
z
= 0.
(1,
t,
The homogeneous (1,
0),
ij
0),
i.e.
coordinates of these points are therefore a pair of imaginary points. Now the co-
ordinates of these points are independent of the coefficients
and
c
hence,
;
all circles Intersect the line at infinity in the
These points are called and will be referred to as li Of.
of imaginary points. at infinity
',
'
the circular points
,
C2 =
Notei. G! This equation becomes in homogeneous coordinates - (x 2 -}- y 2 4- c a*2 + 2/^ + 2g^x) (x* + y* + w* + 2/^ + 2 gi zx) 0.
^{2tei-^)*+2(/ -/ )y + (c -c8 )^}
i.e.
g, f,
same pair
l
l
= 0,
= 0.
;
represents two straight lines, viz. the straight line Thus in projective geometry we at infinity and the radical axis. chord of every pair of circles is the one common that may say
The equation
straight line at infinity.
Note
ii.
Cx
Xu
= 0.
This equation becomes in homogeneous coordinates C\ it
A*i/
= 0;
therefore represents a locus passing through the points of inter-
section of the straight
the straight
205 the circle
line u =
A
cuts the circle in the points represents a locus circumscribing the
Q
GI~-'\u
THE CIRCLE lines u = 0, z = 0, and
C -\C 2
Noteiii.
l
=
G' t
= 0.
If
and B, then quadrilateral
O.
In the same way, since the circles G'1? C'2 intersect in two finite points (real and distinct, real and coincident, or imaginary and distinct),
G\
say
AC2 = The
A
and B, and also in the points
11, 11', the equation a locus circumscribing the quadrilateral represents
general equation of the second degree
+ by 2 + cz* -f 2fyz + 2gzx + 2 lury = independent constants if we know five 2
cut-
contains five locus we can therefore find
;
its
equation, and the locus
points on the is
completely
determined
We
have
seen
=
AC2
and the equations C^ A*/e = form that the loci they represent pass only one more point on the locus is
both
that
their
Cj imply by through four given points required then in order to determine ;
its
equation completely
A#u
follows, therefore, that the equations C^
=
should contain only one undetermined constant
Note
iv.
The centre
;
and C^
AC72
;
it
=
this constant is A.
of the oircle in relation to the line at
infinity.
The polar of a point P, with respect to a circle, was defined as the locus of the points of intersection of tangents to the circle at the pairs of points in which chords, passing through P, cut the circle. Now chords which pass through the centre of the circle cut the circle in pairs of points the tangents at which are parallel In Euclidean Geometry, therefore, the centre of the circle has no
In protective geometry we say polar with respect to the circle. ' that pairs of parallel straight lines meet in points at infinity ', and '
*
points at infinity is the straight line at 'the Thus infinity. polar of the centre of a circle with respect to the circle is the line at infinity ', and conversely the pole of the
that the locus of these
*
line at infinity with respect to a circle
is its
centre
'.
In homogeneous coordinates the equation of the polar of the point (#!, y lf tfj) with respect to the circle
is
x (x l
+ gtj + y (y +&) fr * (gx +fy + ctj l
The homogeneous coordinates of the
v
l
= 0.
centre of the circle are
(#,
/
1),
THE CIRCLE
206
and the polar of the centre is therefore z(g 2 +f* c) = 0, which is the line at infinity. is the same thing,
= 0,
or,
what
Note v. The equation of any circle, by a proper choice of axes, can be written xP+y* a2 or, in homogeneous coordinates,
=
,
This equation may be written (x + iy)(xiy) = a 2*2 which is in the form uv %w 2, and represents a locus touching the imaginary ,
=
straight lines
x + iy
= 0,
x
iy
= 0,
=
the line at infinity
being
the chord of contact.
We
have seen that every circle passes through the two fixed imaginary points 12, 12' on the line at infinity. Conversely, every locus of the second degree which passes through 12 and 12' is a circle. and The points 12, 2' are determined by the equations # 2 + #2 = = = i. e. are the and the z points of intersection of the line z ;
lines
x + iy
=
0,
#
Now we showed
i#
=
respectively.
in Chap. IV that, having adopted the ideas there explained, we could state that every straight line meets a locus of the second -
degree in two points. If, then, a locus of the second degree passes through 12 and 12', it meets each of the straight lines
= 0, # iy = A and B. Let
x + iy
is
in one other point,
=
ax + Iy + cz be say the equation of the straight line AB. Since the locus circumscribes the quadrilateral 1212', its equation of the form -f + + 0, cz\ (x z(ax+ty iy) (x iy)
AB =
xt
i.e.
which
is
a
+ if + czi + lyz + axz^Q,
circle.
We may therefore define a circle in projective geometry as
a locus
of the second degree passing through 12 and 12'. Thus, if the locus second the of the degree represented by general equation
=
ax* + If + cz* -f 2fyz + Zgzx + 2hxy 6 and h and 11', then a passes through the points this locus is called a circle.
=
&
Let the centre of a
i.
if
circle
be the point
P(r, 0) be any point on the circle
(c,
we have ,
is
R
a) and
P02 + OC 2 -20P. OCcosZ COP = PC 2 r2 + c 2 - 2ro cos (0 -a) = B 2
e.
which
;
and
Polar Coordinates.
17.
then
=
the general polar equation of a circle.
,
the radius
:
THE CIRCLE The equation special cases
(i)
The
(ii)
207
of the circle takes the following simple forms in
:
P(r,8)
origin at the centre of the circle
The
r= E. on the circumference origin r
The
=
2JRcos(0-a). on the and the circumference origin
initial line passing the centre r 2 0. A* cos through Polar coordinates can be used with advantage in certain types of problems ; the following examples illustrate the method. (iii)
=
Ex.
i.
A
triangle given in species lias one vertex fixed,
moves on a given
circle ;
find tlw locus
and a second
of the third.
Let the fixed vertex be at the origin and
let the initial line pass of the centre the circle. The through given equation of this circle is r2 + c 2 -2rccos0=JR2 (i) .
Suppose that
OPQ
is
one position of the triangle, and let the Let the coordinates y.
given angles of this triangle be a, /3, and of the point P be (r, 0) and those of Q (/, 0^ is
;
then, since the triangle k . r'.
given in species, the ratio r r' is fixed, let r 0' a. It is evident from the figure that 6 :
=
=
of P, r and 0, satisfy equation in this equation, we obtain
(i)
;
substituting r
Now the coordinates
= kr'
and
= 0'
a
Hence, the equation of the locus of Q is the circle JkV + c2 -2fcrccos(0-a) = B*. Note. In general, if Plies on the locus r=/(0), then the locus
THE CIRCLE
208
Ex, ii. Ttvo The origin leing
circles
of radii
the centre
centres, find the equation
fonmr and
cutting
tlte
be
their centres distant c apart.
b have
of the former and the
of
of centres
the locus
initial line the line
of
of
circles touching the
tatter orthogonally.
Let P, the centre of such a this circle
a,
circle,
be the point
(r,
6\ and
let
the radius of
p.
Since this circle touches the circle A, r = a + p. Also, if Q be a point of intersection with the circle, .e.
+ c2 - 2 re cos B 2recos0-2ar+a3 + &2 -c 2
Hence
r*
or is
2 (r- a) +
0,
the required equation.
.
is
= =
(a
Show
iii.
that the equation
Hence prove
+ b)r~2abcos0.
similitude
of
of the
circle
of similitude of
of the from any point on
the property
two circles, that the tangents
circle it
two circles are in the ratio of their radii.
The equation of the
circles
can be written -.
2
r
- 2 br cos 6 +
Hence their centres are respectively.
(a, 0)
(6,
2 ft
0)
3 a* sin
V sin8 a. and their
radii
a sin*,
of
to the
THE CIRCLE If
or
OK
= OL
thus 9
209
the corresponding value of r is given by (r - a cos a) = a meet the first, and similarly the second, circle in 2
OK,
and Q
;
=
coincident points, i.e. are tangents this is otherwise evident. The centres of similitude lie on the initial line one is at the origin and the radius vector of the other is :
;
a sin a. b -ft sin
a
OC.
2ab
__ ~"
a 4- b sin OK
a sin
a
-f
b
Thus, P(r, any point on the circle of similitude, since a semicircle and hence i OPS a right angle, if
6) is
OPS
is
0P = OS cos 6, r
i.e.
=
2ab
A
,
a-f 6
or
= 2 a&cosfl.
(a-f b) r
Let
FT be a tangent
from
i
But
this point to the first circle,
4
~ 2 ar cos
ar==
6
So
if
Pr
is
2 a cos
(>
/a
-f-
PZ"
a 2 cos 3 a.
6).
+ a3 cos 3 a
a tangent to the second 8
then
2abcos0,
(a-f fc)r
i.e.
coB0,
a
-(rt&co3 <X-r
-
1 (a& cos
0(
-
;
J ).
circle,
3
).
or
Ex.
Prowe Wta^
iv.
//t
polar equation
r*-krcos(d-oi) + kd
w/*m 4
<5
variable, represents
= 0,
a system of coaxal
circles,
and find
the
polar
coordinates of the limiting points.
Any two
circles of the
system are
/*-&!/ cos (0 a) + AV*
0,
The equation r a -fc l rcos(tf-a)+A? l rf-{i J -A- a rcos(tf-a) i.
(A-j
e.
i.e.
A-
2)
[r cos (0
+ X? a rf} =0, - a) - d\ = 0,
rcos(0-a)-d
=
represents a straight lino through the real or imaginary points of intersection of the circles.
This 1267
is
therefore the radical axis of the two circles and, since the equation
Q
THE CIRCLE
210
A: , every pair of circles in the aystem have the same a the system is coaxal. The equation of the circle can be written
does not contain k l or radical axis,
i.e.
thus the radius
The
circle is
is */\k*kd and the centre (\k, a). a point-circle when %k*-~kd = 0,
i.e.
=
fc
Hence the coordinates of the
or 4d.
point-circles are (0, a)
and
(2ef,
a), i.e. the
origin and (2d, a).
Examples Vn. Find the polar equation of the
1. (
ri
on the
circle
line joining the points
as diameter.
0i ) (*2> #9)
Find the equation of the tangent to the circle r = R at the point vectorial angle is ft. 3. The equation of the chord joining two points on the circle r= 2#cos0, whose vectorial angles are lf a i* rcos^-f-^-tf) = 2
whose
.
Deduce the equation of the tangent at the point 6 l
Show
4.
.
that the vectorial angles of the points of contact of tangents (rlt B^) to the circle r = 2R cos B are given by 2 - 2 rx sin 6 tan B -f 2 # - rt cos 0! = 0. *\ cos B v tan B
from the point
.
5.
The polar of
6.
Show that
(rl9 rt\
r2
.
QJ with respect to r cos (0
0j)
= 2# costf is
= JK (r cos = is the
-f t'j
~2a/'cos^~3a2
cos B^}.
polar equation of a circle
whose centre lies on the initial line. If OP is any radius vector of this 2 circle and a point Q is taken on OP so that OP.OQ = 6 a find the equation of the locus of Q and show it is a circle whose radius is double the radius of ,
the given
OA
7.
circle.
a diameter of a
is
PQ\ EL
mid-point of If
PQ
is
circle,
Q any
point on the polar of P,
perpendicular to
OA
M
and
that A, L 9 P, lie on a circle. a chord of r = 26 cos0 which touches r
Show
to OP. 8.
is
01? bisects the angle P00. 9. If P, Q are two points on the circle r
QM =
is
the
perpendicular
2 a cos
= 2JRcos0
E
at
<#,
then
whose vectorial
angles have (i) their sum, (ii) their difference constant, find the locus of the centroid of the triangle OP#. 10. PQ is a chord of the circle r=2fecos# which touches the circle r = 2 a cos B. Show that the locus of a point R on PQ, such that OP. OQ are
harmonic conjugates of 11. is
A
straight line
taken on 12.
OP such
Show
OR and
the tangent at the origin,
is
OP meets the circle r = 2#cos 6 at P OP OQ = k* find the locus of Q.
that
.
that the equation
and a point
Q
;
Y*-2(a-X)rco80 + 2\d
=
for different
values of X represents a system of coaxal circles, and find the radical axis
and limiting
points.
THE CIRCLE From a
13.
a line
fixed point
drawn
is
211 meet a
to
fixed circle at P.
A line PQ is drawn equal to and perpendicular to OP. Find the locus of Q. to a given circle, the 14. OP is the radius vector from a fixed point angle POQ is constant and equal to a, and the area of the triangle POQ is Show that Q lies on a circle. Show that 2/r = 4 sin 6 3 cos 6 is a common tangent of r and r - 12 rcos + 20 = 0. 16. The equation of the pair of tangents at points on the circle
fixed.
2 cos Q
15.
2
whose vectorial angles are (d'-fl
2 2
)
a
*
(d
-
,
is
2
2
)
= and show that when
{^co8 (d-
=
Oi
{2 cos
(0-a)
(<-0)-
cos
this reduces to 3
-.R a
=
0.
=
the line rcos(0 a, and circles are drawn about 6 ,. taken three at a time: triangles formed by the lines tij, t< 2 w 8 u 4l circles are then drawn through the centres of these circles taken four at
#w) cos#w
u n is
17.
,
a time
,
show that the five centres of these circles lie on the circle 4rco80 cos0a cos08 cos0 4 co806 = a cos (0 6^ 8 - s -04 -06 ). :
1
18.
We
conclude this
chapter with the
some
of
solution
important and typical problems.
Ex.
i.
Find
the condition that the
four points on a circle.
(MI*, Sw^),
(m a 2w 3), (w4 2m4 ) should lie Show that (0-25, 1), (2-25,3), (1-69, -2-6), (049, and find its equation. 2
2
,
,
The equation
-14) lie
(m 2
2 ,
on a
2m 2
),
circle,
of any circle is of the form
x*+y* + 2gz + 2Jy + c = 0. A point of the type (m j 2m) lies on this, provided that w satisfies w* + 4 m 2 4- 2#w 2 + 4/m + c = 0, the equation w4 + 2(0+2)tw2 + 4/m +c = 0. i.e. This is of the fourth degree, and for any given values of /, and c gives four values of m, the corresponding points to which lie on ,
the
circle.
On the other hand, four chosen points of the type will lie on a circle provided that we can find values of g, /, and c such that the four given values of m are roots of an equation of this type.
m 2m m a m 3 =
If the roots of the above equation are
2m = 0, 2m m2 = 1
20 + 4,
1
,
m<2
Hence, provided that
=
m3 m
4/,
1
o 2
,
0,
,
4
,
then
m^m^fn^m^
=
c.
THE CIRCLE
212 values of g,f, and
c
can be found from the remaining three conditions.
the necessary and sufficient condition that the given lie on a circle. should points In the numerical example This, then,
is
iiij
= 0-6 i
i.e.
Now 2g + 4
20 Again,
2
= 1-5 w 3 =
lie
on a
1-3
;
+ fn a + iM a +
1
the points
*
;
i>i
= 0-6 + l-5
4
w4 =
;
07
1-3
0-7.
= 0,
circle.
= 2 w2 = + wa (m^ + w4 - -4 + 0-75 + 0-91 = -2-34. = -6-34. )
(ffij
!
)
4/= SmiiHgW^ = (m -hm 2 )m = 2x0-91 -2x0- 76 = 032. 3
I
2/=0-16, c
= m m2 mj tn4 = 0-6825. 1
The equation
of the circle
is
then
= 0. Show
Ex. ii
Uiat all circles oj the family
s 2 +y2 + 2A# +
= 0, its
i>
common
0,
ortJwgonal
circle,
and find
equation.
Suppose that the cuts the circle
We
circle
x*+y* + 2gx + 2fy + c = Q x~+y~ + 2\z+2tJiy+v = Oat right angles; then
are given that
Eliminating v
20A + 2/m/-c-i> = A X + JB/x -f- Cv + D = 0.
from equations
In this equation A and (i)
have a
2^ + =
jz
(ii)
and
(iii),
0.
(ii)
(iii)
we
are independent
;
obtain
hence,
if
the circle
cuts every circle of the family at right angles, this equation
be true for
all
values of A and M
-4
:
(i)
so that
+ 2CJ/ = 0, J+2(7/=0, D-OJ =
Hence
^-J. v--. -f.
0.
must
THE CIRCLE and the
circle
(i)
becomes ,
or
This circle cuts
and
A
iii.
tiro
of
it*
2 1
c'+c^
'
C(* + y 2 )-Ax-By + D =
0.
all
the circles of the family at right angles.
triangle circumscribes the circle
on the
vertices lie
(.
show
that,
d
if
D
B
Ax ^a* -
Ex.
213
2
=
72*+
v
circle
-ap + y*-
72 2
;
on
the third vertex also lies
272r,
this circle.
n be the points of conthe sides with the circle tact of Let P, Q,
#2 + y2
C be
and B,
d)
(.r
Let P, $,
-R
=
r2
(i)
,
the two vertices on 2
+ #2
=
7t 2
.
be the points
(ii) Of,
y
fi,
; '
then, since 7? is the point of intersection of tangents at P and 72, its
coordinates are }(oi
jngos | cos
i
r cos
i.e. i.e.
4
> '
}.
this lies
(y.
+
)*
.V
(a
+ y)
)
y)
\
'
1
(a
on the
circle
y)-<1 cos i (a I (
+ r2 sin 2 i (^ + y) = 7? 2 cos 2 i (a- y), 2 2 cos 2 i (a /i x) = 0, y) + (<J (a
y)]
.1
i.e.
>
r
r/
(cos
cosa(7icosy Similarly, since cos
a
Hence, by
(72
/i
)
(OK
(oc
(sin
27?rcos 2
>)+ 27? cos
2
= y) =
i (a
y)
i (oc
+ cosy) + 72 (I + cos:x y) = (/cosy (/) + 72sin y sin&+ 72 + r
0, ;
;
on the
d)
circle
=
0.
(ii),
+ 72 sin /3 sin a + 72 +
r
r?
cos
/J
sin
a
=
0.
cross multiplication
cos 3 )
y)
i
OK
C lies
cos
(ii),
2
+ >)cos J ~2(/rcos \ (a + y) cos 2dcos(a + y)cos J
r--2drcos 2
sin
cos
y)
(a
and since ;
+ y)
y-sin^)- JVdsin(y
r- d
* j
(cos
^ - cos y)
1
-sin
y)
THE CIRCLE
214
=#* + 2Br,
Put d 2
and multiply each fraction by 2J?sini(y-/8);
then sin
cose*.
a
^(/^
Now
y)
the coordinates of -4 are
_ '
cosi(^->)
i(*-y)' hence cos (72
+
_ ~~
ex
sin
dr
r)x
a
1
__
dx
r#
llr'
Thus i.e. 2 2 Substituting for d in the coefficient of #
r2 # 2 +
dividing through by
i.
e.
A
also lies
,
on the
circle
2
2 2r7#-f d
?/
JZ
2
,
=
this becomes, after 0,
or
(ii).
P
Since we have chosen the point in any position, any of triangles can be drawn circumscribed to the one circle 2 JR + 27?r. inscribed in the other, if d 2
Note.
number and
Ex.
=
iv.
SJiow that the equations of any two non-intersecting circles can
be written in the
and find
forms
the equation
of
the
polar reciprocal of the first with respect
to the
second.
Note. The polar reciprocal is the envelope of the polars of points on the first circle with respect to the second. Since the circles are non-intersecting, their limiting points are real; the limiting points of the two circles be (a, 0) (a, 0), then the
let
equation of either circle
is
of the
form a2
Put g
=
-
A
=
A
;
This proves the
0.
then the equation becomes
(X-l) i.e.
=
(x*
+ y* + a*) + 2(\ + l)ax =
2
(rc-o) first
+^ = A {(x+a
part of the question.
0,
THE CIRCLE Let
(yf> y')
be any point on the (*'-a)
and the polar of
#')
(a?',
2
and from
(i)
Now
2
|(*'
+ a) + / 2 },
with respect to the second
=
((a?
/x
+
a) (#'
(i)
circle is
+ a) + y/}
.
(ii)
(ii)
(^-a)
Eliminating variable
then
first circle,
+ y'* = X
(#- a) (a?' - a) + yif Hence, from
215
2
--X(z' + a)
=
2
y' so as to get the
(A -I)?/'
2 .
equation of the polar with only one
#',
_ let the fraction
-,
=P
then
; '
or
The envelope
of this line for different values of
P
is (cf.
Chap.
II,
12)
which reduces
Ex,
v.
tangents
to
Show
to the
two
tliat
the locus
of
the point /;iom which
the pairs of
circles
t*+2Ay 4 c 2 = are harmonically conjugate to one another
is
If the pencil formed by the tangents from a point (a/, y') is harmonic, so will a pencil of lines through the origin parallel to the
tangents be harmonic. shall use the usual notation,
We
THE CIRCLE
216
The equation circle Cj
=
of the pair of tangents
from the point
(#', y')
to the
is
and
lines parallel to them through the origin (retaining only terms of the second degree in x and y) are
[
x*
i.e.
J*
2 [C/-(^+^) }-2('f 2)(/+/2 );r/4-^ {0/-(y'+/2 )2} = f?
These pairs of lines are harmonic conjugates
,
(see
TTT Chap. Ill,
If
we omit
locus of x
if
5).
the accents, the resulting equation therefore gives the
f
y'.
is
=
0,
2 +/ -c + {^(/a-/i)-ytoa-^)+/2i-/i^} 2 =
0,
2 l))
or
2C C 1
C,
SI
-C
(C2 +
1
a flr
a
{pf+tf-Ct} +
Definition. to
0.
K,
,
This equation
i.e.
= 0.
+/2 -e }-C,{C + 2
l
r
1
1
l
}
C.
If a radius vector
meet a given curve
fi
at 1\
OP
is
drawn from
a fixed point
then the locus of a point P' on
OP
such
that
OP. is
or =
constant
called tho inverse of the given curve with respect to the point 0. (a)
To find
the inverse
of a straight
line with
regard
to
any
point.
Let the origin of coordinates (0) le the centre of inversion, and let the equation of the ho line straight
The equation
of
any straight -
cos
The
distance
the given
from
line, is
line
=
through tho origin
y
=
is
r
sin
of the point P, in which this line intersects
given by the equation
THE CIRCLE
217
Suppose that P' is the point on the inverse corresponding and let OP' = /, then substituting for r it follows that
+ Bain 6) + r/ C =
* 2 (A cos
to
P
0.
Now, since the point P' is on the straight line OP, its coordinates are r' cos 0, r' sin ; hence tho equation of the inverse is This equation represents a circle passing through the centre of inversion, whose centre lies on the line Bx Ay = drawn through the centre of inversion perpendicular to the given (b)
To prove analytically
with respect
to
line.
of a system of coaral a system of concentric circles.
that the inverse
a limit ing point
is
circles
= be any one of the system of coaxal whose limiting points are ( + d, 0). Any line through ... x- + d = y
Let # 2 -f y* + 2gx + d 2 circles, i.
...
a limiting point r
is
-..
A
cos 6
This line meets the
r.
sin
whoso distances from the
circle at points
centre of inversion are given by the quadratic equation d ) 2 + r2 sin 2 6 + 2 g (r cos d) + d* = (r cos 6 r2
i.e.
+ 2rcosO(gd) + 2d(d + fl)
=
,
0.
The equation connecting the
distances of corresponding points on the inverse from the centre of inversion is accordingly
Hence the equation
of the inverse curve
is
-
-+,f}
which represents a
whose centre
circle
is
= 0,
the fixed point
Miscellaneous Harder Examples on the Circle. (For Revision.) chord of contact of tangents from subtends a right angle at (/*', //), prove that 1.
Tf the
'2
(;t
2.
Show
that
+ A/ 2 _ f j }
the equation
formed by the linos y o?
and
2
^/ 2 -(l
find the points
-f
2 (
7l
.
f
,
/v 2)
=
2 ,*
(h, *)
(M
4-
to the circle #*
^-
+ if
1
,*).
l
1
it
r
of the circle circumscribing the triangle
= am + x/m }J y = am^x/m^ y = aw s + x/ni^ 2 2m,m a ) a^-(m m a m 3 4-Sm)ay + a 2wi 3 =* 0,
where
=
cuts the .r-axis.
l
>i
is
THE CIRCLE
218
A
is drawn through the origin and is such that the a constant angle with the line joining the points makes at the tangent origin of intersection of the circle with the coordinates axes. Show that the circle
3.
variable circle
belongs to one or another of two coaxal systems. 4. Points P, Q are taken, one on each of the two circles 2 o? -f
=
2
2a(a?-f
t/
c)
0,
so that they subtend a right angle at the origin. and Q the point of intersection of ihe tangents at
P
c(x* +
y*)
+ a?(x + c)
Show
that the locus of
the circle
is
= 0.
Prove that the coordinates of the centre of the circle which passes
5.
through the points given by ax = ( ft by
=
(a cos
_ yj
b sin
ff,
COB \
- (a 2 - 52
)
(ft
ft)
+ y)
sin $ (3
4-
(a cos
COB I (y
/3,
b sin
(a cos y,
)
b sin y)
are
+ ot) cos J (a + 0), + a) sin i a + #)
>) sin | (y
(
Through the origin pass two circles which cut the rectangular axes of coordinates in the points (a, 0), (0, a) and (-, 0), (0, a) respectively. Prove 6.
that, if straight lines be drawn through the origin to cut both circles, the locus of the intersection of the tangents to the two circles at the correspond-
ing points where they are cut by these lines
is
2 2 tfx* -2ay(x* + if- a 2 ) - a 4 (x* + y )
Show
7.
=
0.
that the locus of a point, such that tangents from it to two equal two cui ves of the fourth degree placed
circles are at right angles, consists of
symmetrically with respect to the line of centres. 8. If JSj -f wi5a -f 7/53 = is the equation of a circle orthogonal to S2 = 0, then the tangents from its centre to ,, 5, are in the ratio
S
*=* l
0,
9. If the axes of a? and y are conjugate lines with respect to a circle, show that the general equation of such a circle is
#2 -f ya + cos 10.
ABC is a triangle,
[2^-2*17 -2y + A AB, Care axes of x and o>
*?]
y.
=
0.
A
point
P is taken on
the circle
4 y2 -f 2 xy cos A - 2 (c 4 b cos A) x - 2 (b -f c cos A) y 4 4 be cos A 0, and P7/, PJIf are drawn parallel to the axes to meet AB, A C in L, JW. On AB, AC points Z/, M' are taken such that BL' = BL, CM' CM. Prove that a?
2
L3/', Z/Af are perpendicular. 11.
Find the coordinates of the centre and the length of the radius of 2 9 is the inverse of (x - a) -f (y - b) =
the circle which
origin, k* being the constant of inversion. 12. Show that the envelope of chords of the circle
which subtend a right angle at the origin
is
i (2c-^-/ )(^-fy )-f^+/y4c = a
9
/
What
does this become
when
0.
the origin lies on the circle ?
THE CIRCLE
219
Show
that the system of circles C-frC'=*0 ckn be obtained by a do not intersect in real inverting system of concentric circles, jf C and 13.
'
points.
The axes
14.
AB
P and
AB AC are inclined at an angle of 60. A circle touches AC a chord whose length is equal to AP. Show 9
intercepts on that the locus of the centre at
is
when AP = c. If at any point
circle
a straight line and give the equation of
th.e
P of a circle chords PQ, PR are drawn making given 15. angles with the tangent at P, show that the locus of the intersection of QR with the radius at P is a circle. 16. If
each of three circles touches two parallel straight lines, prove that 2/3, 2y are connected by the equation
their three angles of intersection 2 Of, 0. sin (X sin 3 + sin y
=
Two
17.
PA meets the other circle in X,
them,
and
fixed circles intersect in A,
AY intersect on
a fixed
B; P
and
PB
is
a variable point on one of it in Y. Prove that BX
meets
circle.
when the given
circles are orthogonal. one another at 0; on their common diameter fixed cuts the points A and B are taken and a variable straight line through Prove that the locus of the intersection of ylPand BQ circles at Pand Q. is a circle which becomes a straight line if OA/OB = rj/r2
Discuss the case
Two
18.
circles touch
.
Find the equation of a circle through the origin cutting the axis of x 2 2 2 at right angles and the circle or -f i/ = a at an angle of 45. 19.
two given circles orthogonally, show that the locus of a straight line. 21. The three segments of the radical axis of the circles x*+y**=2c*
20. If a circle cuts its
centre
a^-f y2
is
- 16&r + 14
made by the
c*
and their common tangents
circles
are equal. is the point (;>, q) and Q, R are the feet of the perpendiculars 2 the straight lines n.r -f 2hxy + by* = 0, find the equation of the circle PQR and show that the length of QR is
22. If
from
P
Pto
{
\
)
Show that the
locus of a point, such that the square of the tangents drawn from it to three given circles arc in A. P., is a straight line which forms a harmonic pencil with the radical axes of the circles taken in pairs. 23.
24. If A, B,
C are
the respective centres of three circles
0,
and
the centre and
right angles,
p
the radius of the circle which cuts each of
show that the equation of the latter may be written OBC.S+ OCA S' + GAB S" - 2j>* ABC, .
where
OBC denotes
Show how
the area of the triangle
to find the coordinates of
.
.
OBC
t
&c.
and the value of p.
them at
THE CIRCLE
220
25. Given a circle and a point P in its plane, show that there is in the same plane a straight line L such that the square of the distance of any point on the circle from P is equal to the distance of the same point from L multiplied by a constant length, and find the position of L. 26. Find the equation of the tangent at any point of the circle
# -f y + 20#42/f/-f c = 0. c* of any point on Prove that the polar with respect to the circle x* -f y* the circle (# + <*)* + (y4&) a = fc a always touches the curve a
2
one of the external common tangents to two circles meet their R and a perpendicular drawn to it from the internal centre of similitude S in T, and if ST produced meet the radical axis in R', then V, the middle point of RR', will be such that VR is equal to the tangent from Vto either circle. 28. Tangents to a circle from a point P cut a fixed diameter of the circle at A and B. If the mid-point of the segment AB is fixed, find an equation 27. If
radical axis in
for the locus of P.
Show
29.
that four points of the type (a cos#, 6sin#), where a and b are any given circle and if four given points of this type lie on
constants, lie on
a circle, then 20 30.
then
;
=
6, Interpret when a 3 points of the type (X X) lie on a given circle ? 1, where Sr = sum of the X's taken
2n?r.
How many S^Sj-S^ + ^-S^ =
,
If four do,
rat a time.
point C is taken in the diameter AB of a circle on AC, CB as diameters circles are described PQ is a common tangent to these latter circles; show that AP> BQ and the common tangent at.C meet on the first
A
31.
:
;
circle.
32.
A
point
P moves
ABC so
in the plane of a triangle
PA* BC* .
PB* CA* + PC* AB* .
.
that
;
prove that the locus of P is the circle which passes through B, Cand cuts the circle orthogonally. 1 33. Show that the circle through the three points (aX, aX" ), (a^, a^T 1 ), 1 aiT ) passes through the point (tf/X/i>, a\^v}. (rti>,
ABC
34.
Prove that the difference of the squares of the tangents drawn from circles is proportional to the distance of the point from the
any point to two
radical axis of the circles.
Two
circles are
such that the sums of the squares of the tangents drawn
vertices of a triangle are the same for each circle prove thart the radical axis of the circles passes through the centroid of the to
them from the
;
triangle.
Show
that a homogeneous equation in x and y of degree n represents through the origin, and find the equation giving the six lines straight joining to the origin the points where the curve 35.
n straight
lines
ac
=
8
+ 3*ya +5i/ a + 6a;~ 7
cuts the circle x* + y* 1. 36. Given three non-intersecting coaxal circles, prove that the lengths of
THE CIEOLE
221
the tangents drawn from any point of one of them to the other two are in a fixed ratio.
Prove that the locus of the middle points of chords of a fixed circle
37.
Wu
which subtend a right angle at a point is a circle, and that the fixed is a of the two circles. point limiting point 88. Find the locus of a point the tangent from which to a fixed circle bears a constant ratio to its distance from a fixed point. If
Sla
denotes the circle coaxal with the circles
Slt
,
and having
its
centre at their external centre of similitude, and 581 529 arc the circles obtained in the same way from 3 S\ and S2 S8 prove that 1S S, 8 S3l are coaxal. ,
,
39. (
+
Form
,
,
,
,
the equation of a system of coaxal circles of which the points
a, 0) are the limiting points.
Prove that, if log {(x + iy + a)/(x + iy - a)} = u + t'c, the curves u = constant and v = constant are two families of coaxal circles.
B are
three collinear points circles are described with centres cut orthogonally a circle of variable radius r, whose centre is 0. on all common tangents Prove that the product of the perpendiculars from 40. A, 0,
A
and
to the first-named circles
41.
if
;
B to
Show
is
4 in a constant ratio to r .
that a circle can bo drawn to touch the four circles
(I/a- 1/6)
-(!/-
+ y'-2fcr
0,
**
0,
3? + y*-2dy
0,
-
0,
CHAPTER VI THE LOCUS EEPRESENTED BY THE GENERAL EQUATION OF THE SECOND DEGREE, S= Preliminary. In the following discussion it is assumed that neither a nor 6 is zero this involves no loss of generality, for if either or both of these coefficients is zero, by a simple change 1.
:
of axes the equation can be transformed into one in which a and 6 any value of x is substituted in the equation
are not zero.
S
= 0,
Now,|jf
we
obtain a quadratic equation giving two values of y ; these inky be real and distinct, equal, or imaginary. Thus, to every value of x there correspond two real,
two
two imaginary
coincident, or
points on the
locus.
Similarly, to
every value of y there correspond two points on the locus. As the value of # is increased continuously oc to +00, the two correfrom
sponding points move across the plane of the coordinate axes and trace out the complete locus for example,
;
P^, P2 &, P3 &
as, ...
in the figure.
We
propose to investigate by oo to +00, of x from the values as increases y elementary algebra and the values of x as y increases from -co to -f oo and to classify ,
the loci by their principal graphical properties so found. can be written The equation 6>
=
or
b
Hence
which
for
ax
= -\hy+g)y{(hy+g)*-a(by*
convenience of reference
we
will write
THE GENERAL EQUATION by = = - (luc + f)
Similarly,
223
(For the notation refer back to Chap. Ill, 10.) The values of x and y will be real only when and
D
E respectively
are positive.
Now
D
(vide
same sign
as
Hall and Knight's Higher Algebra, has the 120) C for all real values of y except when the equation
has real distinct roots (say y and 5), in which case D has the opposite C when y lies between y and 8, but otherwise has the same sign to sign as
C.
The condition
that the equation
= should have real and distinct roots is that F (y-2Jfy + U
i.
A
a should be negative. Thus -D has the same sign as
e.
that
and y has
(i)
BC should
2
be positive,
.
a value between y and
C
when A.
except
a is negative
8.
In the same way E has the same sign as C except when negative and x has a value between the roots of the equation
A
.
b is
GxP-ZGx + A =0, which we I.
C
will call
Since h
is
/i.
ab
>
i.
positive, 2
a and e.
h2
.
essentially positive, ab is positive,
a and b have the same sign. Hence A a and A b have also the same .
.
(a)
A a and A b .
.
Now
D
(b)
sign.
positive.
E
have the same sign as In this case the locus is wholly imaginary
and
and consequently
(7, ;
i.
e.
are both negative.
e. g.
A a and A b negative. .
.
Now
D and E
are negative, except
when y
lies
between y and
8,
between a and ft respectively. Hence x is only real when and > < 8, and y is only real when x > <x and < ft. Thus y y the real part of the locus lies inside the parallelogram whose sides and x
lies
are x
= a, x =
When
/3,
y
= y,
y
=
8.
D
is zero and the y has either of the valued y or. 8, then are the of of the two points values paths equal: corresponding of values each other (vide P4 &, which correspond to those two y meet
THE GENEKAL EQUATION = 8 touch the curve. Fig. p. 222), and the lines y = y, y Hence the locus in this case is a lines x = a, x =
224
/3.
So with the closed curve
inscribed in the above parallelogram.
7 The
locus
is
in this case called an Ellipse; the whole curve
at a finite distance
II. (a)
C
negative,
As
i.
e.
origin.
ab
A a and A b positive .
.
x and y are
(b)
from the
A particular case
is
is
the circle.
.
:
real for all values of
then
D and E are
positive, therefore
y and x respectively.
A. a positive, A.b negative. in (a) x is real for all values of y.
E has the same sign as
(7,
between a and /3. Hence y is real except for all values of x except those lying between a and /3. As before, when x = a or x = /3, the values of y become coincident, and the locus touches the lines x = oc and x ~ /3. The curve thus consists of two branches which extend in opposite
i.
e. positive,
directions
from x
when x
lies
= a and x =
respectively to infinity.
OP THE SECOND DEGKEE
225
A a negative and A b positive gives a similar result (c) A. a and A.b negative. In the same way it can be seen that x and y are always .
.
real
except when x > ot and < /3 and y > y and < 5 respectively. The curve touches the four lines x but a, x fi, y y y no part of it lies within the parallelogram formed by these lines.
=
/
=
=
j
=
,
*;/*
In each case the curve consists of two branches extending in the curve is called an Hyperbola, opposite directions to infinity a special case being a pair of straight lines, viz. when A = 0. Note. It must be borne in mind throughout that no straight line can meet the curve in more than two points. :
1267
THE GENERAL EQUATION = h2 e.
226 III.
C=
0,
flfc
D
In this case result.
i.
and
As an example
E will be positive for all values of
The values
of
x
for all values of x greater less
is
than
^-~
>
* n d negative
A
than
^or
= ^A
consists of a single branch extending
A = ^-~
only of the line x
A
E are both linear: all cases give a similar E = 2GxA. If G and A are both positive,
y are coincident when
Thus the curve
The
.
locus in this case
on one side
to infinity. is
called a Parabola, a special case
when
zero being a pair of parallel straight lines.
Summary.
5=
If
C positive, A ^ C positive, A = C negative, A ^
is
0,
an Ellipse which
0,
a pair of imaginary straight lines,
0,
an Hyperbola which has two branches
is
a finite closed curve,
ex-
tending to infinity.
C C
zero,
A= A^
zero,
A
negative,
0,
a pair of straight lines.
0,
a Parabola
which has one branch extending
to infinity,
C
= 0,
a pair of parallel straight lines.
OF THE SECOND DEGREE
227
Examples Via. Classify the following curves
1.
(i)
(ii)
;
=
f/
(iii)
;t
2
#?/ 4
-f
6 .r
?/
=
3
^2 ~4^-f4//2 -f3or-6y-f
(vi)
5 x1
2xy
Show
the lines y
2 ^/ 2
-f
-f
4 ?/
2 jc
;
;
(iv)
2.
:
9#2 -f 3.ry-f Jy2 -f 2r-f 5y -f 6 2#2 + #// 21 2 4 fix 5y + 4
= 0;
2
-f
2
=
0. 2
2
that the curve 4.c +
=
1
and
=
//
-f
12.r//
-f
10//
S.r-f
Sy +
=
7
lies
between
3.
Whore do
these lines tourh the curve ? Find the equation of two lines parallel
which touch the
to the y-uxia
curve. 3.
Show
that the line
points for all values of 4.
=
./'
cuts the curve
<
?
-f
4.n/-f 3f/
2
-l =
in real
c.
Prove that the curve 4 (#-2) (y-
the same side of the lines
x
2,
y
=
3)
=
(x-\
y
2 -i
lies
6j
altogether on
3 as the oiigin.
that the curve 4#--8.ry-f
.r
Show
=
touches the sides of the
3,
y
y
0,
=
4,
and that no
part of the curve lies within the parallelogram. Find the points of contact of the sides and the chords of contact.
= Find the condition that the axis of y should meet the locus coincident, points. imaginary (iii) (ii) 7. Prove that the tangents at the points where the straight line aie parallel to the a>axis. meets the curve X = ax + hy + <j = 8. Find the equation of the chord of contact of tangents to S which 6.
in (i) real,
are parallel to the y-axis.
Find an equation giving the ordinates of the points where the line x-y = cuts tf= 0. 9.
If these axe real, prove that 2 // -10. Prove that the line r-f //
A
{
II f
2.
2 II
=
> A + B. touches the curve
5=0,
provided that
0.
To find the locus of the middle points of chords of
S =
r/,/
2
+ 2/W/+ to/
2
-f 2//.r
x which arc parallel
to the straight line
yi
+
2 /}/+ c
PQ
;
if
the chord
= in-
is parallel
to
'i
m p 2
curve
?/
Suppose that the point M, whose coordinates are point of a chord
tlte
=
(^ 1 ,# 1 ), is the
=
,
)tl
its
middle
equation
is
THE GENERAL EQUATION
228
The coordinates of any point on this line -are point lies on the curve 5=0, we have
(x l
+ It, y l + mt)
if
;
this
=
= 0, (al + 2 him -f 6m + 2 1 (IX + mYJ + S 8 = ax * + 2hx y + ly 2 + 2gxl + 2fi/ + c. [Chap. Ill,
or
t
where
2
0,
2
2
)
l
l
L
l
1
(i)
l
l
10.]
l
This equation gives the two values of t which correspond to the P and Q in which the chord cuts the curve. Since the point
points (#i>
!/i)
form t
*s
(xl
^e
+U
J
mid-point of P$, the coordinates of P and Q are of the U yl mt); in other words, the values of y\ + vnt\ (x l J
given by equation
This
is
(i)
are equal
and opposite.
the condition that the point
(o? lf
Hence we have
y^ should be on the
line
all chords of the curve S= 0, which are on a straight line whose equation is IX+mY = 0. Such a straight line is called a diameter. Now if the coordinates x l9 yl satisfy both the equations
Thus, the middle points of
parallel to
w#
the coefficient of
Zy = 0,
t
lie
in equation
(i)
vanishes for
all
values of
I
and m,
and the equation gives equal aijd opposite values of t. 2 is not zero, there is a finite point whose coordinates If
abh
satisfy the equations
X = 0, Y = 0.
this point cuts the curve in
In this case any straight line two points which are equidistant
through from this point and on opposite sides of centre of the curve.
it.
This point
is called
the
OF THE SECOND DEGREE
= 0, there is no the equations X = 0, Y = A2
If a&
;
The
229
point whose coordinates satisfy in this case the curve has no centre. finite
=
loci,
according as
0, can be divided into two classes represented by S abh? is or is not zero. They are called central and
The
on referring back to the Figures in 1, will see that this is in accordance with the general shapes of these loci. It is convenient to examine the two classes separately.
non-central curves.
Central curves.
I.
Since the centre
is
reader,
/^^fcO.)
(a&
= 0,
= 0, Y given by the equations = 0, it is evident that IX +
X
mY
equation of a diameter is
all
and the
diameters
pass through the centre.
Conjugate Diameters.
We
have seen that
mx
parallel to
= ly
0,
are bisected
0, which are by the diameter IX + mY =
;
the equation of this diameter written in full
x (al + Am) + y(M + bm) + gl
This
is
parallel
m' shows that
1'^
is
is
+fm =
0.
to the straight line m'xl'y = all' + h (lm' + I'm) + bmm' = 0.
The complete symmetry
mxly = Q
S=
chords of the curve
all
if
of this result in the
the diameter which
parallel to
w'o;
fy
= 0,
if
numbers
Z,
pi
and
bisects chords parallel to
then the diameter which
=
mxly=z
is parallel to 0. bisects chords parallel to m'x-l'y Such a pair of diameters is called a pair of conjugate diameters.
The
pair of diameters
lX + mY=Q, Z'X + w'JT=0 are conjugate,
therefore, if all' + h(lm' + trn)
+ l>mm'
=
0.
are parallel to conjugate diameters lines a'a? + 2h'xy + b'if provided that they arc parallel to (tnx ly)(m'x l'y) = 0, where
Ex. The of
S
all'
= r-
0,
h (lm
Axes
f
-f
of
I'm)
+ bmm'
=
;
that
is, if
+ a'b
ab'
=
2hh'.
Symmetry.
been shown that the diameters IX + mY = 0, FX + m'Y = and are conjugate are parallel to m'xl'y = 0, mxly 0, It has
=
all'
+
ft
(lm
= 4- I'm) + bmm'
0.
If this condition is satisfied, these
diameters are therefore perpendicular
when
the lines
m'xl'y
mxly = //
if
=
0,
are perpendicular. the coordinate axes are rectangular, the condition for this is
Thus,
if
the ratios l/m, l'/m' are determined by the equations all'
+ h (lm' + I'm) + bmm' =
0,
tt'
+ mm'
= 0,
230 the diameters
THE GENERAL EQUATION are both ZX + wY = 0, rX + tn'Jf =
conjugate and
at right angles.
These equations give
A(?-m ) = (a-b)lm, these from which we obtain two directions for the line mx ly = directions are at right angles, and therefore correspond to the directions of the two diameters lX + mY = 0, l'X + m'Y = 0, which The equation of these diaare both perpendicular and conjugate. 2
;
meters
is
therefore
-T =
2 - 6) h (X 2 (a ) the equation of the
or
This, therefore,
is
both conjugate and perpendicular for
the condition
X F.
(iii)
two diameters which are
the diameters are always real, h(X*-Y*)-(a-b)XY should have real
that
;
2 2 that (a should be positive; this is always &) -f4/* Now, a straight line which bisects all chords perpendicular
factors
is
true.
to itself clearly divides the curve symmetrically. two straight lines, at right angles to each other,
There are then which divide the
curve symmetrically. These are called the axes of simply the axes of the curve.
Note.
Since the axes are parallel to Ix +
(p-w their inclinations
h
2
)A
- sin
2
(cos'
2
0)
toi20
or If
lt
0% are
of the axes are
the values of
X sin
}
= 0,
where
=
(a
=
by tan
6
- 1) sin G cos 0,
-
a
= -m
In this case both the lines hence every diameter IX +
l
(ab
and
X sin
2
=
3^ cos at
2
.
an angle
o>,
h 2 =0.)
X = 0, J = Oare parallel to ax + hy = 0;
mY =
The diameter aX + hY
thus
given by this equation, the equations
= Y cos
Non-central curves.
;
V' b
Ex. If the axes of coordinates are oblique and inclined show that the equation of the axes is
II.
or
= (a-&)Z/w,
to the #-axis are given
(Q)' v
my
symmetry,
is
is
parallel to this straight line.
therefore parallel to
ax+hy
=
and
chords parallel to bx liy = 0; hence, the diameter aX + 7iY=0 bisects all chords at right angles to itself. A noncentral curve has therefore one axis of symmetry.
bisects
Ex.
all
If the coordinate axes are oblique,
axis of a non-central locus,
8=
0, is (a
show that the equation of the Y = 0.
-h cos o>) X+ (h -a cos o>)
OF THE SECOND DEGREE The graph of 8
3.
231
= 0.
Since the equations of the axes of the locus are of the form IX + 0, the axes pass through the intersection of the lines = and Y this point of intersection is the centre.
mY ~
=
X
Let
:
CA, CJB be the axes of
symmetry, then if Pl is any point on the locus 5 = 0, the points
P P3 P4 2
,
are also
,
by symmetry -
on the locus.
P
19
P P4
I\,
form a rectangle
3,
whose sides are parallel to the axes and are bisected by the axes.
we
Incidentally,
see that
if
a
P
on the locus and the centre of symmetry C are known, point l a second point on the locus Pu can be constructed, since CPi = CP3 bisects all chords We have seen that the line ZX + .
wr=0
parallel to
Two
x
y
I
m
important special cases arise when
I
m
and
are respectively
zero.
Thus the
line
X=
bisects all chords parallel to
y
= 0,
i.e.
parallel
to the #-axis.
Y=
So also
bisects all chords parallel to the y-axis.
of practical importance when drawing the graph of a given equation the following general rules will be useful; the student will discover with practice their relative utility.
This
is
:
To draw the graph (a)
Draw
of a curve
5=0.
the graphs of the lines
X = 0, Y =
;
these intersect at
the centre C.
has simple factors such as axes + p.Y-0. the + \r=0, draw next (X + \Y)(X + pY), we know since A single point on each is sufficient for this purpose, the axes pass through C. Even if the above factors are inconvenient,
hX*-(a-b)XY-hY
(b) If
as
when
A and
ju
involve surds,
2
it is
X
X
often possible to find convenient
points on the axes, e.g. the points of intersection of
and the (c)
0-axis
Try
;
these points can be joined to C.
to discover a point
P on
venient to try the intersections of or
x = -.
the curve
S
=
:
it is
with x
= 0,
generally con-
y
= 0,
x
= y>
THE GENERAL EQUATION
232
Having found one point Construct a chord
(i)
by
X = 0.
(ii)
by
Construct a chord
F=
(iii)
(iv)
P, other points
PQ
l
PQ 2
parallel
to
can be found thus
:
the #-axis and bisected
parallel to the y-axis
and bisected
0.
=
Construct a chord PCQ 3 such that PC C%. Construct points which are symmetrical with
P with
respect
to the axes.
The
last
method involves drawing perpendiculars
but
is
useful
Qit
Qz
ma y
when
be
ff
(as in
to the axes,
the case of the hyperbola) the points
th e paper.
It will be helpful to the student at this stage to draw a few curves and thus to acquire a knowledge of their shapes and to become familiar with the notation we have employed. (See also Chap. VIII, 9.)
Illustrative (i)
Examples.
108^-124
=
0.
\
OF THE SECOND DEGREE Here
C=
9xl6-12 2
;
X = $x + 12^-22 = The
axis of
.'.
the curve
is
233
a parabola.
symmetry hX-tbY*=Q is and the axis are parallel.
X = 0, Y = 0,
The points (-2,
X= Y=
0), (2,
-3)
passes through (245, 0) passes through
The point P(
(
2, 0) is
4*5, 0)
on the axis, draw this line. and is parallel to the axis. and is parallel to the axis.
on the curve.
The other points shown (iv)
lie
in the figure are found
by
(ii),
(i),
above.
C=
= 50, i.e. is + ve, the curve is an ellipse. = 0. Points on this line are (1, -2), (0, 2-5). 10 = 0. Points on this line are (1, -2), (-2, -1). The axes are 2X+ Y = 0, X- 2 Y = 0, which give 2x + y = 0, #-2t/ = 5. Point on curve P (0, -3-06). or x = y in real points. The curve does not cut either y =
Since
54
-
4
5
or
THE GENERAL EQUATION
234
2# 2 +5#j/+2#2
(iii)
Since
C=
4
^,
i.e. is
X=2#4|y-3 = 0. r== #+ 2y-3 0. The axea are
ve,
6#
6y
the curve
Points on
is
X- Y= 0, X+ Y = 0, (0, 4), (0,
=
0.
an hyperbola.
this line are (1-5, 0),
Points on this line are
Points on the curve are
8
i.e.
-1),
x-y =
(4, 0),
(-1,
(0, 1-5), (2,
0,
(-1,
2).
-1).
and x + y 0).
NJ
\
\
\ \ \
\ N
y ^\
X
'
^.i
X fla.
i
\ \
OF THE SECOND DEGREE Examples VI 2 1. or -f-4^t/-f4t/ !
b.
and draw their graphs
Classify the following curves
=
-6^-3?/4-8
235
:
0.
2. 3.
4. >
5. 6.
2#"-f 5;n/-f2y o;
2
2
+
9*/
-6#-6t/-8 =
+ 6;ry-5a;-9t/-f
=
1
0.
0.
7.
8. 9.
10.
=
2
11.
or-;ry-f2f/ -2;r--6f/-f4
12.
Draw with
9,
=
9#a -24:rj/-f 16t/> + 21#-28 y + 6 2ar + ay-y 1 -aH-2y = 0.
0.
0.
the same axes of reference x^ + xy 4
=
2 t/
c,
when
c
=
1, 4,
16 respectively.
In this section we propose
4.
lines relative to the general curve
in the following chapters considered. (I)
To
when
to find the equations of certain
= 0;
this will prevent repetition
special
forms of the equation are
S
find the equation of the chord of the curye^ S ax 2 + 2hxy + by 2 + 2gx + 2fy + c 0,
=
=
whose middle point
is (x x
y^.
,
-y
T
1 Let the equation of the chord be -T1 (
=J ^ 77
}/
*
then the coordi-
in
nates of any point on this line are (x l + It, yi 4- wO- As in 2, the values of t corresponding to the points of intersection of the line and the curve are given by (al*
and since
(x lt
+ 2 him + bw 2
7/ t )
is
t
)
2
+2
the mid-point of the chord,
ZX + wr
l
1
Thus the equation """ '
of the chord
= 0.
is
(i-^^ + to-yjr^o, i.
ocX l
e.
-f
y
Y = l
or (II)
To
point (x
7 ,
find the equation of the tangent to S
y
=
at
the
7 ).
*""
the two points of intersection P, Q of a line and the curve become coincident, the mid-point of the chord evidently coincides with this point hence we can deduce the equation of the tangent at (/, y') from (A) by writing x l remembering that */' x', y l
When
:
=
=
;
THE GENERAL EQUATION curve and therefore S' = 0, the
236
on the
(#', y') is
Note.
If
becomes
(#', #')
s;r+yr+z' =
o.
(B)
the axes are rectangular the normal at
(x-x')Y'-(y-y')X'=
(a?',
is
y^)
therefore
(T^x^^yX' ^x'T-yX'.
Tojlnd the'equation of the chord of contact of tangents
(Ill)
from
equation of the
*"""
tangent at
y) to the curve S
(x',
==.0.
Let the points of contact of the tangents be
The tangents
(xl9 ^), (# 2 , y%).
at these points are l
=
xX 2 + yY2 + Z2 =
;
;
and since by Hypothesis each of these passes through
(#', y')
we have But these are the conditions lie on the line which i.e.
yj
therefore the equation of the line joining
is
y2 ) should
(#2 ,
(os lf
yj, (x 2
equivalent to
is
x(ayf + hy'+g)+y(hx' + by'+f)
to
+ gx'+fy' + c
= 0,
0.
(C)
To
find the equation of the polar of the point with respect to 8 0. (IV)
=
Suppose that the tangents at polar of (x' 9 j/). (#!,
,
of the chord of contact.
This equation
i.e.
that the points (x l1
yj
its
(x', y')
any chord passing through (#', tf} and that meet at (x l9 yj the locus of (a? lf y x ) is the Since PQ is the chord of contact of tangents from
PQ
is
P and Q
is (vide (C)
equation
;
above)
This line therefore passes through
(a/, y'},
hence
a'Xi-tyri which
is
algebraically equivalent to
*iX' + yi^' +
Now
this is the condition that (x ly
yj should
lie
on the line (D)
Q,
which
is
Note.
consequently the polar of If (#', y') is the centre,
centre has no polar the polar of the centre ;
we ia
(a/,
y'\
then X'
=
and Y'
= 0,
so that the
shall see later that, as in the case of the circle, '
the straight line at infinity'.
OF THE SECOND DEGREE
287
(V) To find the coordinates of the points where the straight line joining the points (x^ y^, (x 2 y 2 ) cuts the curve 8 0.
=
,
Let of
P be the
point (# 1? y^) and
any point on the straight
If this point lies
a (Z^
+ mxj
= 0,
,
2/ 2 )
the coordinates
we have
2
+2 which may
on 8
the point (#2 are
Q
PQ
line
-f
{0 (to!
mfy) +/(Zyi
+ *w2/ 2
) } (Z
-f
m) -f c .(I + m)
2
=
0,
be written
f8 + 2lm{x2 X + y^T + Z \ + nfS^ = l
l
l
l
0.
(E)
This equation gives the two values of I m corresponding to the two points L, in which PQ cuts the curve. Cor. i. If the points of intersection, L and Jf, are coincident, then PQ is In this case the two values of / m given by (E) are a tangent to the curve. 2 equal, hence 5 53 = {# 2 Xl + ytYl + Z ] :
H
:
t
Thus, of
Q
is
.
1
PQ is a tangent to the 5,5= {a^ + yri + Z,} 8 if
.
curve from the point P(a? 4 yj, the locus ,
(F) that of the pair of tangents from (x ly y^) to the .
This equation, therefore,
is
curve.
Cor. Cor.
y^ lies on the curve, then S = 0, which is the tangent -}-ijYl + Zl
If (x l
ii.
reduces to
xX
l
,
Again, suppose that Q line through P cuts the curve at L, P(XI,
iii.
yi) is
xX
l
=
-^yYl + Z
i
a JTl
(E)
the locus of
Q then
to the curve at (xlt yj. lies on the polar of P, so that a straight
M and the polar of P at Q.
0, so that, if
a?
The equation
;
{
Q(#a y 2 )
+ yi F1 + ZI =
,
I* 68
The polar of
on ^> we
0-
now becomes
so that the two values of
/
:
m are P
equal and opposite
:
hence the points L,
M
and Q. are harmonic conjugates of Thus any chord of the curve 5
= through P is divided harmonically by the polar of P. Cor. iv. In the equation (F) of the tangents from (#', y') to the curve 5 = 0, omit all except the terms of the second degree in x and y we find are parallel to the then that the tangents from (a? M y ) to the curve 5 = P and
;
t
pair of ljnes through the origin whose equation
is
which reduces to
x*(Cy'*-2Fy' + These are perpendicular (the axes of coordinates being rectangular) Ca:' 7
i.e.
if
(a?
,
y') lies
a
if
+ Cy -
on the
/2
circle
(0)
THE GENERAL EQUATION
238
Hence the locus of points the tangents from which is
a
this circle
.circle perpendicular centre is the centre of the curve
Note.
If
;
S=
a parabola,
is
equation is 2 Gx + This straight line
its
=
S
(7=0,
this locus
is
;
its
a straight line
0.
called the Directrix of the
is
are
0.
i.e. if
2Fy-A-B =
S=
to the curve
called the Director Circle
is
parabola
its
;
;
(H) other
properties will be discussed later.
Find the equations of the director
Ex.
directrix of a parabola
The student
a central curve and the
circle of
when the coordinate axes
are oblique.
advised to read the remainder of this chapter as a study carefully through so as to maintain a logical sequence it may be postponed until the end of Chap. VIII. is
:
5.
To
find the lengths of the axes of a central curve.
Definition.
symmetry meet the curve
If the axes of
AA
A, A', B, B' respectively, then
f
and
at the points
are called the axes of
J97?'
the curve.
x-x
this axis is
l
the centre of the curve
if
Using rectangular coordinates, and either axis makes an angle
with the
_yy
cos
Hence,
if r is
r (a cos
But
2
+ 2 h cos
since (xl9
v
[
r sin
6)
in the equation
by substituting
+ b sin 0) + 2r(X l cos0+r
X
the centre,
is
-f
r2 (a cos'2
S
l
=x
l
+ 2h cos Xi +y l
Thus
'
1
sin0)
and 1^
l
7/ t
+S = 1
= 0,
i.
e.
7=-
O
when
(x l
,
+ I sin 2 0) + S = l
l
l
l
=
0,
/MI+ fy/i+/=
0,
-f
hy l
+g
we have a
h
li
1}
9 Si =
sin
Y + Z = Z = gx ax l
Eliminating xl and
S=
0,
2
sin
yj
us,
y
equation becomes
Also,
} )
the length of the semi-axis, the point
on the curve. This gives
2
y
the equation of
#-axis,
sin
(x l -f r cos 0, lies
is (x ly
yj
f is
g
f c-
the centre.
=0,
l
+fy v
0.
-f c.
0.
(i)
so that this
OF THE SECOND DEGREE The length
of the semi-axis in the direction
r 2 (a cos 2
Now, we showed ft
.
+ 2ft
2
0-sin
(cos
+
a cos
.
,
so that
ft
2
= (a-b) cos 0)
sin
=
ft
cos
= =
cos
+ sin + b sin
sin
+ 1 sin 2 =
a cos ft
=
C
ft
by
0.
are given by
sin 0,
+ b sin sm0
=
:
cos0
therefore
+
230) that the values of
2, p.
(
therefore given
is
sin 2 0)
+6
cos
sin
239
Jc
cos
0,
Jc
sin
0.
A',
say J
;
Hence a cos 2
+ 2ft
cos
(cos
=
fc.
fc=-~.
Thus
We
+ sin 2 0)
2 fc
now have
and
A
+ ftsin
(a
+
(b
+ ^Jsin0 + ftcos0
--^jcos
= =
0;
This equation gives the squares of the semi- axes of the curve.
Note
If r}
i.
is
one of the semi-axes, we have f
But we showed the
rr-axis is 0, is
in
TT-;,
Jcos0-f
ft
sin
=
Hence the equation of the semi-axis of length
Note
S = l
G
ii.
,
if
?-
Since,
2
0.
2 that the equation of the axis, whose inclination to Fcostf.
A"sin0
and that of length
=
is
r, is
similarly
when
(x l9 y,)
is
the centre, we have
X
l
=0,
Y = l
and
the coordinate axes are changed to parallel axes through the
centre the equation
a (r-f x^-i-
2ft
.9=0 becomes
(# + a^)
(j
which at once reduces to
THE GENERAL EQUATION The Curve 8 = at infinity. Asymptotes.
240 6.
The
coordinates of the points of intersection of the straight line are x' + lt, y' + mt where t has the
=
(x-x')/l (y-y^/m with flf values given by the equation
2hlm+ Tm2
If aZ 2 +
=
= 0,
then in the Euclidean sense the straight line meets the curve in one point only as explained in Chap. V we can keep our results general by means of the conventional 'points at ;
'
infinity
we
straight line at infinity ', thus the curve 8 in two points.
meets
line
'
and al 2
= + 2Hlm + 1)m* =
say every straight
In the present
^
case,
the present ab A2 ), i.e. if the 2 straight line is parallel to one of the lines ax + 2hxy + by* 0, we say that the straight line meets the curve in one finite point if
then,
(for
=
and in one 'point at infinity'. Again, if al* + 2hlm + bm 2 = and ZX'-fwY' = 0, then in the Euclidean sense the straight line does not meet the curve at all in this case we say that the straight line meets the curve in two coincident 'points at infinity', or the ;
straight line touches the curve at infinity. If therefore the point (of, y') lies on a line at infinity, its coordinates satisfy the equation
=
The equation
0.
which touches 8=0 lX + 0, where
of the locus of
mY=
(a/, tf) is
therefore
represents two straight lines passing through the ; these lines are called the 'Asymptotes of the curve, be regarded as the tangents to the curve from the centre.
this equation
centre of
and
may
S=
The equation
of the asymptotes, written in full, is
(db-h*)(ax* + 2foy + ty* + 20a?+ 2fy) + fy
or
C/SA =
in the
2
-2/^ + a/ 2 =
'
0,
0, which only differs from the equation of the curve term independent of x and y.
The asymptotes
are therefore straight lines through the centre ax2 + 2hxy + ly* 0; if (xQ9 y ) is the centre,
=
parallel to the lines
their equation can also be written
a(x-x )*+2h(x-x,)(y-y Q + l(y-y oy )
^
a form which
is
sometimes
The asymptotes
= 0,
useful.
are real
or imaginary according as a&-ft 2
is
thus an ellipse has imaginary asymptotes, and an hyperbola has real asymptotes. When the asymptotes are real and at right angles, the curve is called a Rectangular Hyperbola ; negative or positive
;
the condition for this
is
a+6
2 h cos o>
In the case of the parabola, when ab
= 0. =A
2 ,
the equation
OF THE SECOND DEGREE
=
241
= 0, i. e. two coincident directions which a straight line can be drawn to meet the curve at infinity. There is no finite straight line which meets the parabola in two and M + bm = 0, points at infinity for in this case, since al+hm =
gives
(al
+ hm) 2
or
(AZ
+ 6m) 2
in
;
mY =
the equation IXf +
reduces to gl+fin
=
so that
0,
When it is possible to satisfy simultaneously the equations and IX' + mY'
= 0,
al
a/= gh. + hm =
the equation of the parabola becomes
which
is
lines.
The parabola meets the
points,
i.
not a proper parabola but a pair of coincident straight straight line at infinity in coincident
e.
touches the line at infinity.
Application of Homogeneous Coordinates.
The equation
of the curve
S
=
in
homogeneous coordinates
the points at infinity on it are given by have therefore for these points re 2
We
intersections with
its -f
2A?j-f
2
6rj
=
and
is
= 0. = 0.
The is
2 ab points are real, coincident, or imaginary according as h or the line an hence meets positive, zero, negative: hyperbola
at infinity in real distinct points, the ellipse meets it in imaginary points, the parabola meets it in coincident points or, in other words,
the parabola touches the line at infinity.
L When (1,
*?!>
the points
and
0)
(
The equations totes,
are
equation
l
is
2
,
rj 2
,
are not
0),
coincident, let their coordinates
of the tangents at these points,
X -f Y = rj x
be
then
and
%
X+ Y = ?? 2
i.
;
e.
of the
their
asympcombined
therefore
or as previously shown.
This equation can also be written CS the linear factors of S A(VC, then J7=
= 0, 7=
AC 2 0,
and
V
if U, are are the separate
The equation of the curve can then equations of the asymptotes. be written in the form Z7F+ AC2 /0= 0, which indicates that Z7= 0,
V=
If
U 1207
=
are tangents to the curve, f being the chord of contact. and are real, we can take the asymptotes as coordinate axes,
F
THE GENERAL EQUATION
242 in
which case the equation of the curve becomes
Cartesian coordinates, xy
Note
2
2
C
,
or, in
.
'
this case the origin,
Note
=
evident from the form of the equation that the centre, in is the pole of the line at infinity '.
It is
i.
=&
17
2
It is evident from this form of the equation of an hyperhola that, AOA', BOB' are the asymptotes, the curve lies altogether within either the angles AOB, A' OB' or the angles AOB', A' OB also, that the distances from the centre of the points of intersection of the hyperbola and a diameter ii.
if
;
increase without limit as the
diameter approaches the position of the
asymptotes.
The following important exercises for the reader
The
(i)
properties of central curves are left as
:
bisectors of the angles
between the asymptotes are the
axes of the curve.
The asymptotes
(ii)
are harmonic conjugates of every pair of
conjugate diameters.
An
(iii)
own
asyirptote, regarded as a diameter of the curve,
is
its
conjugate.
II.
When thepomts are coincident,
and
=
0,
i.e.
ab
=A
2 ,
we have af + A?j
hence the coordinates of this point are
The tangent which reduces
=
a, 0).
(/*,
at this point is
=
to
To obtain the when its equation
The equation
0,
the line at infinity.
separate equations of the asymptotes of is
of the asymptotes in the
may be used. When, however, proceed as follows
an hyperbola
given with numerical coefficients.
form
6X 2
ax 2 + 2hxy + ly 2 has rational
2hXY+ aY = 2
factors,
we may
:
Let the equations of the asymptotes of then and Vx + rn'y -f n' =
S=
be 1x + my + n
=
;
Vn + In' = 2g, = 2 (fl ~gm). n (Im'-l'm)
m'n + mn'
so that
hence
Thus the equation
=
2/,
of one asymptote
is
(Zm'-rro) (lx+mif) + 2(fl^gvn)
Hence the following in the
x
=m
form
and
(lx
y=
rule
:
'Express a#
2
+ my)(l'x + m'y) + 2gx + 2fy: /,
= 0.
+ 2hxy + + 2gx + 2fy in this expression put
except in the factor Ix + my.
The
resulting
OF THE SECOND DEGREE
243
expression equated to zero is the equation of the asymptote parallel 0. to Ix -J- my w', Similarly, for the other asymptote, put x
=
y
=
=
except in the factor
Z',
Ex. To find
l'x
+ m'y.
the asymptotes of the hyperbola
= 0. In the expression (2# + (2a? + 3y);
Again, put
3//)
this gives 7 (2a? a:
The equation
(x
4x + 5y put x =
2y)
=
+ 3y)-22
3,
y
=
=
2, y 1, except in (x 2y) this gives 7 (# of the curve can then be written ;
(7a?-14y-3)
(14a?
To determine in which quadrant
2,
except
0.
+ 21y-22)
=
the infinite part
3
2y)
=
0.
17.
of a given parabola
lies.
2
+ fiy) + 2gx -f 2/?/ -f c
be the equation of the parabola. to the the origin point of intersection of the lines Change c + 0, the equation then becomes 0, 2gx 2fy +
Let
(OLX
=
=
The
position of the infinite part of the parabola relative to the coordinate axes is clearly not altered by a change of origin.
= y/(*
&), where 6 is a very small number, be a straight the through origin nearly parallel to the axis of the parabola. This will meet the parabola again at a point which is at a very great
Let x/P
line
distance from the origin, and which lies in that quadrant in which
the infinite part of the parabola Substituting the point
y =
(e
a) x/fi,
J X =-2{g +/( Since
e
lies.
-
we
-
obtain for
the ^-coordinate of
-gfl/p - 2/c//*. have the same sign as
)//3}
2 (fa
very small, x will will havo the same sign as
is
and therefore y Hence,
if
the equation of the parabola
is
=
and gx-\-fy is the signs of x and y, determined so that Oix + fBy negative, indicate the quadrant in which the infinite part of the parabola
Ex.
lies.
5x + 2
+ 7x-Sy + 8 = 0. = and 7# 3y = a 2
Take 5#-f2/ negative quantity; then y and (7 -f *f)x is negative, therefore x is negative and y is positive.
The
infinite part of the curve This rule will be found useful
lies
=
\x
therefore in the second quadrant.
when drawing a parabola from
its
equation,
THE GENERAL EQUATION
244 7.
The
To find
foci
and
directrices.
of a point tvhich moves so that its distance from a fixed point is proportional to its distance from a fixed straight line. the locus
y') and the #cos0 + #sin0 p
Let the fixed point be S(x',
fixed line be
= 0,
using rectangular coordinates. If P(x, y) is any point on the required locus, and dicular from P to the fixed line, we have SP = e a constant
which
is
j
.
PM the perpenPM where e is
SP 2 = e 2 PM 2 or 2 (x-xf + ly-lf)* = e {#cos0 + ysintf-p}*,
hence
.
,
the equation of the locus.
Now
(i)
this equation contains five
arbitrary constants, #', #', 0, p, and e we may expect then that the can be put into this equation SEE ax* + 2hxy + by* + 2gx + 2fy + c :
=
form, and we proceed to show that this is always so. The locus represented by the general equation of the second degree can therefore be described as the locus of a point which moves so that its distance from a fixed point S is a constant (e)
times
its
distance from a fixed straight line. is a focus, the fixed line
The point S
directrix, and the constant
e is
is the corresponding the eccentricity of the curve.
Note. This is the fundamental property proved in geometrical conies for the various curves which are sections of a cone by a plane. The locus is therefore usually referred to as a conio section, or briefly, a conic.
5=0
The equation (i) above will bo referred to as the focus-directrix form of the equation of a conic. I.
To egress
the equation
Using Rectangular
S=
in the focus-directrix form.
coordinates, suppose that the equations
- x')* + (y -yj = e2 (x cos 6 + y sin ~j?) 2 (x S=ax* + 2hxy+ty* + 2gx + 2yy+c =
and
(i)
(ii)
are identical.
Change the origin of coordinates to the point (x', y') these equations then become 2 2 x*+y* = e (x co&0 + y sin -jp') (iii) = where p' p x' cos y' sin 0, ;
(iv) 2hxy+by* + 2X'x+2Y'y + S'= 0. Observe the way in which the terms containing x and y and the independent term occur in equation (iii) this suggests writing
and
ax* +
;
(iv)
in the form
(X'*-aS')x*+2(X'Y'-hS')xy+(Y'*-l>S')y* (v)
OF THE SECOND DEGREE
245
This equation will b of the same form as equation (iii) provided we can find a point (xf, y') whose coordinates satisfy the equations
that
X*-a8mY*-b8i XY=hS, v/
_
V
tJ
::
..
a
(vi)
h
b
For a central conic, the equation h (X* - Y*) = (a - b) Y indicates that the foci (if any) lie on the axes of the conic. Note ii. For a parabola, the equation h (X2 T 2 ) = (a -b) reduces to F(aX+ HY) = 0, which indicates that the foci of a parabola (if any) lie on
Note
X
i.
XY
its axis.
Now,
we substitute we obtain
if
tion (vi),
the usual values of X, Y, and
S
in equa-
Cxy -Fx-
and Thus,
(1)
If the conic is a parabola,
2 Gx
linear equations
2Fy
-A
+
i.
e.
J5
C=
we have
0,
the two
=
There is therefore a single solution hence the equation of a parabola can be put in the focus-directrix form in one way. We have now proved that a parabola has one focus, and that this focus lies on ;
the axis.
If the conic
(2)
and If
C=
is central, i.e.
write A for
CxG and
fx
equations
= FG-CH =
(Cx-G)(Cy-F) we
0,
for
Cy~F,
(vii)
AA.
these give
= AA 2 = 24A A (a 6) X A A 0, = A (a -6) A v^ -.&)* + 4 A
X 2 - M2
= A (a -6),
can be written
AM
;
2
hence
2X 2
or
The expression under the
is
;
two real values of A and one negative, hence A has two positive therefore
;
.
we get always positive evidently one of these values is
radical 2
2
real
and two imaginary
We
have notv proved that a central conic has four foci, two real and two imaginary, and that these foci lie on the axes. values.
II.
To prove
Now,
if
(,
that two foci 77)
lie
on each axis
oftlie curve.
are the coordinates of the centre of the conic, the
equations for the foci
may
be written
THE GENERAL EQUATION is the equation cos & = f) sin (y
246
then, (x the coordinates
If,
axes, A;
TJ
sin
0),
>j)
of
one of the
(
X;cos0,
the foci on that axis are
of
where O 2 /; 2 cos
=
sin
Aft.
(#--) cos + Q/ rj)sin0 = 0, on this axis are ( + #sin0, so that the coordinates of the 2 2 cos sin 6 = -Aft. Hence 7c 2 + fc' 2 =0; this ?? + //cos0) where C // shows that two foci lie on each axis, one pair being real and one
The equation
of the other axis is foci
pair imaginary.
To find
III.
the distance between the real foci.
Let 2rx be the length of the axis which )sin0
(x
(#
T])cos0 =
and 2r2 be the length of the other axis. We showed (Chap. VI, 5) that rx 2 r2 2 a cos 6 +
,.
,
so that
+
2
along 0,
/C
3 ,
and
that
A
CV
cos0 a cos 6
=A
sin 6 __ ~~
ft
lies
ft
C2r
sin
2
A
cos0
Since the axes are perpendicular, substituting 77/2 + a sin
_ ~~
cos
ft
2
O' r x
C2 2
Thus
fc
=r
2
r2 -;
1
2 ft
sin 6 cos
evidently
case of the ellipse the real foci lie
between them
2
is
get
we have
-r A 2
(r
we
A
sin
Subtracting these equations
for 0,
2
7b
is real
rl >r%, so that in the
if
on the major
axis,
and the distance
Vr^r^.
If the ellipse is real, it is evident from the equation giving the must be negative hence the foci lying lengths of the axes that
A
on
the axis
(x
)
sin 6
;
(y
77)
cos
=
are
which meets the curve in
lie
imaginary
on the axis
real points.
IV. To find the eccentricity of the conic
S
=
0.
coefficients in the equations
Comparing
+ Zlwy + fy/ 2 + 2#t! + 2/y + c = 2 = c2 (x cos 6 y sin x') + (yy'f (a-
and
or
real
according as ft cos 6 sin 6 is negative or positive. In the case of the hyperbola the real foci must
(x
2
-t
p)*,
we have
l-e 2 cos 2
= Aa,
l-e 2 sin 2
= A6,
c
2
cos
<9
sin
(9
=
-Aft,
OF THE SECOND DEGREE 2-e 2 = A(a+6)
so that
- e 2 cos 2 6) (1 - e 2 sin 2 0) - c 4 cos 2 (1
and
l~e
or
Note (ii)
(iii)
Since
i.
If the conic is
If the conic
is
X
(a&
22
2 ft
:
)
an ellipse, ab~h* an hyperbola, ab
If the conic is a parabola, ah
(iv) If the conic
Note
is
a
a
circle,
A 2 (6 - ft 2 ),
)
5" 4
2
=
2
2 2
S1-e = T a&-& 1 - 1? =
sin
2
(2-e
.,
thus
(i)
(9
247
is
+ve,
W
is
h* is zero,
&
and
/.
=
e
e>l.
.*.
e
and
0,
<
/. .'.
ce,
= e =
1.
1.
0.
This equation gives two values for e, one corresponding to the real and one to the imaginary foci. When the eccentricity of a conic is ii.
referred to the eccentricity corresponding to the real foci
the example given below,
We
is
meant.
(Cf.
p. 249.)
have also from the above relations e
2
2
(cos e
so that
2
0-sin2 <9)= -A(a-&), cos
sin
b) sin 6 cos 6
(a
= A//, = h (cos
2
2
sin 6)
;
the equation which gives the inclinations of the axes of the conic to the re-axis of coordinates. Hence the directrices of the
but this
is
conic are parallel to the axes. The equations 1 c 2 cos 2 2
=
A a,
c
2
cos
=
sin
=
A/j
give
A/cos (/* cos # a sin 0) if 2r is the but, length of the axis inclined at the angle a shown that we have #-axis, 2 a sin (9- ft cos C2 ^2 C2 2 ) c
;
_ ~~ _
hence
c
To find
V.
Beferring
the equations
X'x+ Y'y + /S' = + 2Jwy + ly~ + origin, (a)
=
2
2
r 2 )/r 1
(r 1
of the
sin
cos #
'
2 .
directrices
of a
conic.
to
hence the directrix corresponding to a focus
is its
polar.
Central Conies.
The coordinates are
2
ft_
A
equation (v) we see that the directrix is the polar of the origin with respect to the conic 2X'x+2 Y'y + S'= 0; we took the focus as the
back
ax*
^ -^ ~~_
A
sin
to the
fccosfl,
on the axis
of the foci lying
(#--)sin0 r)4;&sin0, where (7
The corresponding with respect to S =
2
2 A;
?))cos'0
(f/
(f,
TJ)
is
the centre and
sm0cos0:=
directrices,
=
A//.
which are the polars
0, are therefore
of these points
THE GENERAL EQUATION Now X+ Yr) + Z = A/6 therefore the equation
248
1
,
of the pair of
directrices is
Y sin 0) 2
h (X cos 6 +
or
A cos
=*
sin
0.
(b) TJie Parabola.
The coordinates
of the focus are given
and the equation of the this point.
by the equations be found as the polar of
may
directrix
The
theoretically,
case of the parabola is, however, practically and most simply dealt with by reducing the equation to
the focus-directrix form.
Let the equation of the parabola be
where we have written a 2
2
eccentricity of the parabola is
Comparing the directrix
1,
,
and
we
axis of the parabola. then in the form
Hence, comparing
coefficients, 8
a,
fc,
and
=
get tan
perpendicular to oix + fiy
We may
for
a/3
so that the focus-directrix
x 2 and xy,
coefficients of is
/3
,
= 0,
The
h.
form
is
a//3, so that
and therefore
to the
write the equation of the parabola
we have 2
+ /i )a?'+/ty= -#, (a* + /3*) /-*=-/,
(a
2
From
these
(ftp
or
2
2 2 2 2 (a + /3 )(*' 4y )-^ equations we find
+ 9)* + (*P -/) a = 2
(a H-/32)p
2 (
=
4-2(/3p-a/) p+/+/ = j
(iii)
= (a + 2
)
2
(a
(ii)
c.
+ /3 2 ) 2 (x' 2 -h y' 2 2
(i)
2
4- /3
)^
2
2
2
/J
+ (a
)
(p
2
-h
+ c),
^
2 )
c
;
But so that the equation of the directrix
Using
this value of p in equations
the coordinates of the focus (a
(ftxoiy+p = 0) may be
;
(i)
and
(ii),
we can
written
write
thus
+ &)' = b(A+B)/2G-g;
(a
+ %'= a(A + B)/2F-f.
down
OF THE SECOND DEGREE It can be readily verified from equations (i) coordinates of the focus also satisfy the equations
=
VI.
Methods of finding
and
the foci
tvhosc equation is given with
;
and
(ii)
that the
2
of a central
directrices
numerical
249
conic,
coefficients.
The
coordinates of the foci involve complicated surds unless the factors of h(X*-Y*)-(a-l)XY are rational. If these factors are
not rational the coordinates of the foci may be found as on p. 245, When the factors are rational the following method is useful.
ax2 + 2hxy + bif + 2gx + 2fi/ + c
If
= k{(x-x')
2
+ (y--y') 2 }+l(xcoBe + ysmO--p) 2
(i)
on considering the terms of the second degree, it is evident that ax 2 + 2 to/ + fy/ 2 - k (x 2 -f y 2 ) is a perfect This is the case square. when (k a)(k b) = ft 2 and this equation gives two values of k which are rational when the factors of h(X 2 - Y*)-(a b)Xr are identically,
,
Take
rational.
either of these values of k
from which we can determine
Comparing the in equation
we
(i),
k(c-lp
cos 0, and sin 0. x and y and the independent terms
+ lp COB 6 = _ g> ky' + lpsinO= f, 2 2 2 = c, M*' +2/' ) + ^
obtain
whence, 2
Z,
coefficients of
kx
>
= WV+y" = 8
)
and then we have
)
(<7
(ii) (iii)
(iv)
+ tycos0) 2 + (/+Zpsin0)2
.
This gives a quadratic equation from which we can find p, and the f corresponding values of x and y' are given by (ii) and (iii). This gives the directrices, foci, and also the eccentricity which is J~-l/k.
Ex.
If
*=
To find
and
#2 -f 4#*/ + 7/2 -A'(#a -f if)
-1
or
When
directrices
is
of the conic
a perfect square, we have (fc-1) 2
3.
Thus and (a)
the foci
x
1
-f
4xy + f -f (x + if) l
=
a?
*=
-1,
so that
-1
or
-
2 (x -f y) 2
,
= 4,
i.e.
THE GENEEAL EQUATION
250
p=
If
.'= 1, y'= 1, and the equation - 1)2 -f (// -f I) 2 = 2 (a? + y - 1) 2 (a?
then
1,
reduces to
If
p=
so that
then x
*
=
=
f
J, y'
3x'-2p =
1,
6j>
and the equation reduces
J,
3y'-f2p
=
=
3,
3
2
to
(2/>4-l) -f(22?-3)
have now the real and imaginary and 3^-f 3#-5 directrices are x + y -1 =
The foci
The equation (x
2
y'
)
=
2 p2
;
2
We
sponding to the real
-f
(a?'
2
If i> = 2 + t, then x = J(5 + 2?) and y'= can be written in either of the forms
may
conic
3,
2
VII.
the
,
-,
When
(b)
of
-jK^ 2
foci
=
;
*)*
and tbe equation
the equations of the real and the eccentricity corre-
;
0,
foci is 2.
in relation to the circular points at infinity.
of a conic in the focus-directrix form
x'f +
(yy'Y
=
e* [x
cos
-f
y sin b -p}
2
be written
this is of the
form
itv
= fav
2 ,
from which
it
appears that the conic
touches the imaginary lines
(s-aO + ifo-yO
=
0,
(s-sO-ifo-yO
=
>
the chord of contact being the directrix
#cos0 + 7/sin These imaginary lines intersect
^
p
= 0.
at the focus
(x',
y')
;
hence the
directrix is the chord of contact of the imaginary tangents
from the
focus to the conic.
These imaginary lines are parallel to
#-fi?/
=
and x
fy
= 0,
x
they pass through the circular points 12, ll at infinity. This argument applies to each of the ways in which the equation of the conic S can be put in the focus-directrix form.
i. e.
=
Hence the foci are the points of intersection of tangents to the conic from the circular points II, !2/. Since these points li and 12' are imaginary, we cannot properly
OF THE SECOND DEGREE them in a figure ; the following diagram properties of these points and lines.
represent
Si
,
S2 $3 S4 are
Two is
,
,
foci
;
d^
r?
of the foci are real
d
When
the conic S-
is
shows the
relative
are the corresponding directrices. and two are imaginary, and the conic
2
,
inscribed in the quadrilateral
one focus,
251
,,
rf 4
SlJ $2
,
a parabola, since.
3,
12,12,'
4
.
touches
it,
we have only
THE GENERAL EQUATION
252 Note.
Since the tangents from a focus to the curve are parallel to 2 (x iy) = 0, i.e. to a^ + y = 0, the equation of a pair of tangents to the curve from a focus satisfies the conditions for a circle. This gives us (a? + iy)
another method of finding the
foci.
The equations of the various conies in
8.
their simplest
forms.
When
the axes of
=
are taken as coordinate axes, the
symmetry
reduces to
simplest form. Two cases arise (i) C is not zero, being positive for an ellipse and negative for an hyperbola there are then two finite axes of
equation
its
:
;
symmetry.
(7=0, the curve being
(ii)
axis of
a parabola
;
there
is
only one
finite
symmetry.
has already been shown that when the coordinate axes are changed from one set of rectangular axes to another the quantity C It
is
unaltered
:
this is also self-evident so far as sign is concerned, since
a change of axes cannot affect the nature of the curve. (i)
When
the coordinate axes are the axes of symmetry, on the curve, so also do the points (x, ?/), (
the
if
y) lies
y\ of in hence coefficients and the the y ( -x, xy, x, y) equation of the curve must be zero. The equation is then of the form point
(x,
#,
;
For an ellipse ab is positive, i. e. a and b have the same sign. and b are positive, c must be negative, otherwise no real values of x and y could satisfy the equation. (a)
If a
We can of
thus write the equation of the ellipse referred to
symmetry
in the
a
* 1
_L
a
(b)
a, b
axes
2
_ + y_ where
its
form &2
2
'
are evidently the lengths of the semi-axes. is negative : hence the equation of an axes of symmetry can be written in one of
For the hyperbola ab
hyperbola referred to
its
the forms
~l !_3/!-l Consider the points (-fa, axis.
0),
first (
equation
a, 0),
or r
;
and a
^-^a 2
""
-1
'
the #-axis meets the curve at the is
the length of the corresponding
OF THE SECOND DEGEEE
258
The axis of y does not meet the curve in real points usually called the length of the other semi-axis. (ii)
(x,
;
Take the
b is,
however,
finite axis of symmetry as axis of x if the point on the curve, so does the point (x, value whatever y), the hence coefficients in and of the of the have, xy y equation ;
y) lies
x may
curve are zero.
The equation aa?
Now
We
ab
= 0,
the form
is of 2
+ fy + 20r + c
=
0.
since the curve is a parabola, hence a
cannot have 6
=
parallel to the axis of
The equation
2
is
unless the curve T/,
hence a
=
is
=
or 6
= 0.
a pair of straight lines
0.
then of the form
Now take the origin at ( c/2jr, 0), i.e. on the curve of the equation then becomes zero. The equation of the parabola now takes the form y 2 2
;
the constant
=
f
2g x
9
or, as
=
4 ax. usually written, y The axes of coordinates are the axis of the parabola and a tangent to the parabola at the point where its axis intersects it. it is
In the following chapters we propose to discover the properties of the parabola, the ellipse, and the hyperbola from their equations in these simple forms. 9.
Envelopes.
In Chapter II we showed
how
to
find
the equation of the
envelope of a straight line whose equation contains an arbitrary constant in the first or second degree. It will be convenient to extend this method to the equation of any curve which contains an arbitrary constant, or the constants of which are connected by given relations which leave one of them undetermined. are three functions of the coordinates x and y, of the If P, <J),
S
first
or second degree, then
A 2 P+AQ +
#=0
(i)
an equation of the first or second degree, and represents some straight line or curve whatever value X may have. is
Two
of these lines pass through any proposed point (x' 9 when are the values of the functions P, Q, P', Q', substituted for x and y, we have the condition
K
to determine A for loci of the type
B
which pass through
(x' t
y')>
for if
x', y'
are
THE GENERAL EQUATION
254
These two
lines will be tangents, straight or curved,
from the
to the curve
enveloped by the system. If (.a/, y') is on the envelope, then these tangents become coincident. The condition for this is 4 P'E' = Q' 2
point
(#', #')
.
Hence the equation 4
PR = #
of the envelope of loci represented
(i)
is
.
Examples VI 1.
by
2
c.
Write down the equation of
the tangent at (x' y'} (ii) the normal at (#', y') ; the of (iv) a pair of tangents from (#', y'), (iii) polar (x' y') for the curves whose equations are given in Ex. VI a, 1. 2. Find the coordinates of the centre of these curves, and explain your (i)
;
9
t
;
results. 3.
4. 5.
Find the eccentricity of the curves in Ex. VI a, 1 (i), (iii). Find the focus-directrix form of the parabola (x -f 2 ?/) 2 = 4# -f 2 y -f 1. 2 2 in the focusExpress the equation 8# + 15f/ -f 24xy + 2x + 4y-5 =
directrix form. 6. 7.
= 8.
Find the equations of the asymptotes of the hyperbolas in Exs. VI b. If the coordinate axes are oblique, show that the foci of the conic are given by the equations X*-aS = Y*-bS = sec o> (XY-hS). Find the lengths of the axes of those curves in Exs. VI b which are
ellipses or hyperbolas. 9. Find the equations of the ellipses in Exs. axes as axes of coordinates.
VI b
referred to their principal
10. If the coordinate axes are turned through an angle 0, what does the 2 2 become ? equation of the conic ax + 2 hxy -f by -f c = For what values of B does the xy term in the result vanish ? Explain this
result. 11. Show that the equation of the axes of symmetry of the curve and # + 3y-2 = 0. are 3x-y + 1 = ll^ 2 4-6^-f 19y2 -2#-26# + 3 =
Determine p, q, and r so that p(3x-y + l) 9 -f-g(^-f-3/~2) 2 4-r = maybe identical with the given curve, and hence show that the equation of the curve referred to its axes as coordinate axes is 5#2 -f 10/a = 3. (Note.
When
the separate equations of the axes take a simple form, as in
this example, this is the easiest method of finding the equation of the curve referred to its axes. Exs. VI a and VI b give other curves which can be
similarly treated.) 12. 13.
Where are the AA' is a given
moves so that
focus and directrix of a circle ? finite straight line
PN*:AN.A'N
is
and
PN is perpendicular to it.
a constant
ratio,
a curve of the second degree. If the given ratio cases of the circle, parabola, ellipse, and hyperbola. Find the eccentricity of the curve in terms of X.
show that is
X
:
1,
its
If
locus
P is
distinguish the
OF THE SECOND DEGREE Draw graphs
14. (iv)
a?-4^ =
15.
1
;
of
(v)
= 4x = -1 -4^ if
(i)
.r
2
if
(ii)
;
2
;
(vi)
a://
= -4rr = 4 (vii) ;
;
255 2
(iii)
xy =
.r
-f
=
2
4?/
1
;
-4.
Find axes, eccentricity, centre, foci, equations of axes, asymptotes, and x 1 4 4#y + y2 -2#-6# = 0. Draw the curve and find the foci and eccentricity of = 0. bxt + lxy + Find the condition that an asymptote of S = may pass through the
directrices of 16.
Sf-Wx-Wy
17.
origin.
Determine completely the position and nature of the curve 2 2x'* + 4xy-y' + 4#-6*/ + 2 = 0, its its Sketch the centre, finding eccentricity, and the equations to its axes. 18.
curve.
Find the condition that the line Ix + my = 1 should be (i) a tangent, a normal at a point on the curve aaP + Zhxy + ty* = 1. (ii) 20. Find the coordinates of the centre and of the foci, and the equations of the axes, asymptotes, and directrices of the conic whose equation is 7#2 -48;r*/-7y 2 + 60# + 80t/-50 - 0. 21. Find the position and magnitude of the axes of the conic whose 19.
equation 22.
is
ax*
+ 2hxi/ + by 2
1.
Find the asymptotes of the hyperbola
6#2 -7;ry-3i/ 2 -2tf-8y-6
=
0.
whose equation is (x-4y) 2 = 5 IT/. Find the eccentricity of the conic whose equation is x* + xy + if = 1. 24. Show that one focus of the conic x* 4- 1/ 2 -f 2 hxy -f 2g (x + y) +g*/h = is the origin, and that the other is x = y = (-20)/(l +/?). 25. Trace the curve 81#2 -f 90#*/-f25i/ 2 -f 59# + 21t/ + 9 = 0, and find the coordinates of its focus and the equation of its directrix. 26. Trace the curve 3#2 + 8#y-3f/ 2 -40a;-20t/ + 50 = 0, and find the 23. Trace the conic
equations of
its directrices.
Reduce
and draw the figure of the conic Ixy- 12y*-x + ly = 26. 28. Mark on a diagram the position of the focus and directrix of the 2 parabola whose equation is x -2xy + y* + #-3y-f 3 = 0. 27.
to its simplest equation 12o?
29.
Show
2
4-
that the equation (a-l/^ar-f hy ax 1 + 2 hxy -f by* 1, if r
axes of the conic
Show
that the equation of the directrix of the parabola a x*
is
represents one of the a root of the equation
is
= 0.
Trace the curve 3#a + 4#y + i/ 2 --3a?-2t/4-21 30.
=
2a/t/-f ca-p -/ = 2
2
-f
2gx -f 2fy -f
31. If the axes are so inclined that
x*-xy + y*
~
c
=
0.
x + xy -f if = 1
a2
is
a
circle, trace
2
and obtain the lengths and positions of the coordinates of its foci, and its eccentricity. 32. If the straight line y = rrtan is an axis of the conic conic
a
,
ax*
and the length of this axis
is
+ 2hxy + by*=
2r,
show that
its
the
axes,
1, 2
I//'
=
a
-f
h tan 6
=*
b
+ h cot A
THE GENERAL EQUATION
256 33.
Show
that the equations to the tangents to the ellipse
nx*-l2xi/ + 8y* + 50#-20?/-f 21
=
at the extremities of its axes are
34.
Prove that
where A 35.
8=
is
the discriminant, is the diameter parallel to the tangent at x y'. the coordinate axes are oblique, the eccentricity of the conic
When is
given by (ab
36. If (x', y') is
- *)
(2
- e*}* sina o>
a focus of the conic
corresponding directrix x cos 6 + y sin S
Y
f
cos 37.
= M,
Show
and
fl^cos
6 sin 6
=
(a
+ & - 2 h cos
a )
(1
- e*).
5=0, and d its distance from the p = 0, show that X'Bin0*=hd
hd*.
that a pair of intersecting straight lines possesses the focus Where is the focus ?
directrix property.
9
CHAPTER
VII
THE PARABOLA 1.
The parabola
referred to
the vertex as coordinate axes (# 2
The equation directrix
*/
=
2
axis of symmetry
its
=
and
ttie
tangent at
4 ax).
iax (Chap. VI,
8)
can be put into the focus-
form y
in one
way only
:
viz.
Hence the Focus S Directrix
The
XM
is
of the parabola 0. the line x + a
following definitions are
=
common
is
the point
(a, 0),
to all the conies
and the
:
from any point (P) on the curve to the (i) The perpendicular (PN) axis is called the ordinate of the point P with respect to the axis.
The double ordinate through the focus (LSL') (ii) latus rectum.
is
called the
The length of the normal at a point (P) means the distance measured (PG) along the normal to the axis, unless otherwise stated. (iv) The length NT is called the sub-tangent, and the length NGthe sub-normal. (iii)
we?
K
THE PARABOLA
258 (a)
Since the focus
equation y*
When
=
is
the point
P is
LSL'
thus the latus rectum
The equation
L
end
at the
= 44S
2
=
a,
and the
is
of the latus rectum
SL
or
,
AS =
we have
(a, 0),
4 ax expressed geometrically
L2
(b)
the point
4 a.
form
in the focus-directrix
expressed geometrically gives
SP*
= PM
2
SP
or
,
Thus
the focal distance of any point on a parabola distance from the directrix.
The
focal distance of the point
(#,
?/)
is
x+
is
equal to
a.
2. Tangent, Normal, Diameter, Polar. The student should refer back to Chapter VI,
4,
and work out
the equations given below for the case of the parabola y* (1)
The equation of
its
the chord icliose mid-point
is
(x',
=
4 ax.
y')\
y'y-y
or (2)
Tlic
equation of the tangent at the point
(#',
y)
:
-2ax + yy'-2ax' = 0, yy' = 2a(x + x'),
or
Example
i.
If the tangent
at
P
meets the axis at 1\ then
SP = ST. = 2a(x + x)
Let P be the point (x' y'}\ then the tangent at P is yy' hence, putting y = 0, the point !Tis ( #', 0), and since S is (a, 0) we have ST == a -fa:', which is, 1, the focal distance 6P. t
;
Hence 6T = 6T.
Example
ii.
Let
P be
the point
(a?',
is
Y
y')
;
then the tangent at
is
(0,
=
7-
(3)
therefore the line 2ay
-f
J.
But y /2
y' (j?-a)
The equation of the normal
P is
2a(x + x'),
(0, \y').
SY is
the tangent at the vertex at Y,
the tangent.
yy'
therefore the point
P meets
tangent at
If the
SY is perpendicular to
at
The normal being perpendicular
=
(i)
= 4a#',
therefore the point
which
perpendicular to
is
(#', y').
to the tangent, its equation is
Y (i).
THE PARABOLA If the normal
Example. Let
P
be the point
hence, putting y = Thus SG *= a + #' (4)
0,
G
=*
SP.
y'),
is
P meets
the axis at G,
then the normal at
P
SP =
SG.
is
the point {(2a-f ^'), 0}.
20 y?n^
i.e. tho
Let
(a?',
at
259
M
be the mid-point of any chord parallel to the fixed line
(#', T/')
y = WM-.
Since the equation of the chord
/(7/-7/0 the condition that
it
is
=
2a
should be parallel to the given line
hence the locus of the mid-points of
my = 2 a,
(.*-*'),
all
is
y
chords parallel to y
=
2a 5
= wx
is
a line parallel to the axis. Hence all diameters of a parabola are parallel to the axis. Note. The diameter bisecting all chords parallel to the tangent at c/ v * z yy' ( 2a(x + x') is the line y = ;/, which passes through the !/')> i.e.
>
point of contact of the tangent. Thus the chords bisected by any diameter are parallel to the tangent at its
extremity.
Definition.
drawn at F,
If the chord
through a point P on the parabola, meets the diameter through
parallel to the tangent at Q,
PV is
called the ordinate of
P
',)
with respect to this diameter.
(5) The equation of the polar of the point (/, /). This takes the same form as that of the tangent
viz.
at a point
= 2 a (# +
(#',
y'\
4, IV.) ')' (Chap. VI, yij' If a the Note. parabola, the polar of any point Q(x', y') is point on 2 a (x + W, which is parallel to T(h, y') on the diameter through Q is \j\j the tangent at Q and to the chords bisected by the diameter conversely, the pole of any chord lies on the diameter bisecting the chord. R 2 :
THE PARABOLA
260 (6)
The equation of a pair of tangents from a point
(x',
y')
to the
parabola.
This equation (Chap. VI,
4) is
we
Retaining terms of the second degree only,
find that these
tangents are parallel to the straight lines i.
=
2 0. y'xy -f ax between the tangents is therefore given
a
e.
0'y
The angle
a tan 6 .
Thus the
The
Jy'*-4axf ---, x'+a
=
locus of a point, the tangents from which include a
constant angle
i.e.
by
0, is
locus of the intersection of orthogonal tangents
is
x+a =
0,
the directrix. 3. (a)
The equation of
the parabola referred to the axis
and
latus
rectum as coordinate axes.
When
=
4 or, the focus is (a, 0). the equation of the parabola is y 2 the axes axes to through this point, we obtain parallel Changing the required equation as 1
y (b)
The equation of
the
2
=
4a(x+a).
parabola referred
to
any diameter and
the
tangent at its extremity as coordinate axes.
Let the origin be
0,
and
let the
axes of coordinates be the diameter
Ox and the tangent
Oy. Since all chords of the parabola parallel to Oy are bisected by the diameter Ox, the ordinates of
points on the parabola, which have the abscissa, are equal and of opposite
same sign.
Hence the equation of the parabola must be of the 'form y 2 = 4A#; we have then to find the value of A. Let the angle between the coordinate axes be
PN, OE perpendicular
:
draw the ordinate
PF and
to the axis of the parabola. If the coordinates of 0, referred to the principal axes of the
parabola, are
(hj
Jt),
then tan w
=
~ K
THE PARABOLA Since
2 Jc
= 4ah,
P be
Let
we have h
the point
a cot 2
= PJf + 012 = # sin + = # sin AN AR+ OF-f FM = But PN* = a.AN] hence PjV
a)
ft
2
or
and k
o>
2 a cot w.
==
then
y),
(a?,
=
261
t/
a>
-f
= 4a#cosec 2
2a cot a>,
and consequently X = a cosec 2 a>. This is then the equation of the parabola referred to a diameter and the tangent at its extremity it is of the same form as # 2 = 4a#, consequently the equations of the tangent, polar, and diameter will be of the same form as those already found, but the normal which ;
involves the condition for perpendicularity will not be the same, since the axes of coordinates are now oblique.
The equation
PF 2 = 44Scosec 2 w.OF. SO = a + h = a + a cot 2 = a cosec 2 PV 2 = 4OS.OV.
Now
hence
Example If
of
S
translated into geometrical notation gives
To find
i.
the tangent at is
equal to
the coordinates
(see Fig., p. 260)
But
.ST.
ST = SO =
=a
-f a,
o>
the focus
of
meets the axis at a 4- h
cot 2 o>
;
of
the
parabola
T, the ^-coordinate
;
;
= a cosec 9 w.
The y-coordinate
is
equal to OT, hence
y sin
The
focus
=
OR
k=
therefore the point (a cosec
is
Example
6)
On
ii.
P'
2 a cot
o>.
8 a>,
the diameter through
-2acosec'a>cosa>}.
a point
OP. OP'
of a parabola are constant
prove that the four points of intersection of the tangents dratvnfrom P, P' lie on two taken points P,
so that the rectangle
is
and
fixed straight lines parallel to the tangent at
:
equidistant from
for axes of coordinates and Take the diameter and tangent at be (, 0) (', 0), where &' c\ points P, The equation of the pairs of tangents from P to the parabola is
it.
loo che
F
which reduces
a
to
9
8
4-
(y
- 2 ax)
aac
The equation of the pair of tangents from for by writing '
.
= 0. P to the 1
-f
(i)
7
parabola
is
obtained
THE PARABOLA
262
Thus, if (r, ?/) is one point of intersection of the tangents from Panel P', and % satisfy equation (i). The product of the roots of this equation is x 1 thus x 1 = ' = c2 Hence the points of intersection of tangents from P, P' to the parabola lie on one or other of the lines x + c c = these lines are parallel 0, x to and equidistant from the tangent at 0, viz, x = 0. .
;
;
Example
The
iii.
which
triangle,
is
lines joining
wid~points of the sides of a
the
self-conjugate tvith
to
resect
a parabola, touch the
parabola.
Let the vertices A, (**
J5,
C
of the triangle be the points (x^
?/,),
(.r 2
,
y2 \
2/s)-
The conditions that the polar the other two are
of each of these points should pass through
w/3
=
2 a (x 2
+ a?,),
Now, the coordinates of the mid-points of
-<4J5
and
-<4C are
and
.e.
The equation
of the straight line joining these points
x
y
x
y
is
I
4a
i.e.
=
0,
i.e.
This
I^ ^ l ) 1
(
is
the equation of the tangent
w ^" c ^
P roves
^ie
proposition.
to
the parabola at
the point
THE PARABOLA Examples VII The tangent and diameter
1.
is
point
a.
at a point are perpendicular only
when the
the vertex.
The point (,
2.
263
lies
1 or inside the parabola y
=
lax, according as >/ -4a is positive, zero, or negative. 3. Obtain the coordinates of a point the tangent at which makes an with the diameter through the point. angle 4. If the ordinate of a point on the parabola is 2a, find the abscissa. 77)
on,
outside,
2
Find the condition that the straight line /.r-f my = 1 should touch the 1 parabola y = \ax. 6. Find the equation of the pair of straight lines joining the intersec5.
=
tions of the straight line Ix+my -- 1 and the parabola f/ 8 4ax to the vertex. What are the conditions that they should be (a) perpendicular,
coincident?
(b)
Show that
7.
=
the tangents at (x\
y'},
(x* y") inteisect ,
on the diameter
+ y. Find also the r-ooordinate of their point of intersection in terms of \j and y". 8. The tangent at P meets the axis at T and the tangent at the vertex at Y. Prove that PF=- IT. 9. The subnormal of any point on a parabola is half the latus rectum. 3 2am 10. At what angle does the straight line nix+ //-am cut the
2#
2/'
parabola ? 11. The tangent at P and the ordinate of P meet the show that ^4A = AT.
T
axis at
and N:
r
12. Find the equation of a parabola, with latus rectum 4&, which touches the axis of x at the origin and has the axis of y for axis. 13. PG is the normal at /'; prove that the projection of PG on the focal radius of /' is half the latus rectum. 14.
P
Show
that the length of the focal chord bisected by the diameter at
is 4tfP.
15. The perpendicular from the Y: show that SY* = SA.SP.
16.
Find the vertex, axis, (x + 2)* -40 + 6;
focus,
focus
X
to the
tangent at
P
meets
at
it
and directrix of the following parabolas
:
(i)
2
(ii)
(t/-l) 2
(iii)
= 2*-7;
# -8o;-f2y= 2
0; 2
2
(a;~l) -h(i/-2) =:iGr4t/~lj and write down their equations when referred to their principal axes. show that 17. The tangents at P and Q to a parabola are at right angles (iv)
;
:
PQ
passes through the focus.
18. The tangent at P meets the directrix at /?: show that the angle RSP a right angle. 19. Express the coordinates of a point on a parabola in terms of the angle which the normal at the point makes with the axis.
is
20.
Find the equation of the directrix of the parabola
if
=
4rt# cosec
a o>,
THE PARABOLA
264
Find the coordinates of the points of intersection of the parabola with the straight line 3y-4#-f 4 *t 0. Find the equations of the y* tangents at these points, and show that they are perpendicular and meet on 21. >=
4#
the directrix. 22. Tangents are drawn to the circle x* + y* axis of x equidistant from the point (c, 0). 2 a 2 (0 x)'. intersections is the parabola cy
=
a2 from two points on the Show that the locus of their
=
=
that the normals to the parabola y 2 4ax at its points of intersection with the line 2ar 3*/-f 4a intersect on the parabola. 24. Prove that the distance between a tangent to a parabola and the
Show
23.
=
parallel
normal
with the
is
acosec $ sec 2 , where
$
Find the focus and directrix of the
25.
is
the angle which either makes
axis.
pai-abola
(t/
2#)
2
=
5;r-f 1,
and
draw the curve.
=
Two
2 4a#, on the same side points are taken on the parabola y of the axis, such that the product of their distances from the axis is 4 a 2
26.
.
Show (ii)
that the tangents at these points (i) intersect on the latus rectum intercept on the directrix a segment whose length is the difference of ;
their distances from the axis.
Obtain the conditions that the straight line lx + my + n = should be = 4 ax. Find the locus of (i) a tangent (ii) a normal to the parabola \f the middle points of the portions of (i) a tangent (ii) a normal intercepted 27.
;
;
between the point of contact and the axis. 28. Find the equation to the parabola whose vertex directrix the straight line
3#-4y-f
10
=
is
the point (1,2) and
0.
Determine its focus and latus rectum. 2 29. For the parabola y = 4ax show that the middle points of all chords lie on the straight line 3#-f8a = 0; and that parallel to 3#+4y 2 = tangents at the extremities of any one of these chords intersect each other
on that diameter.
A point P is such that the line drawn through it perpendicular to its 2 2 with respect to the parabola y = 4 ax touches the parabola a? = 4%. polar Show that P lies on the line 2ax + by + 4 a 2 = 0. 2 31. If the normal at a point P, on the parabola i/ = Sx whose abscissa 80.
t
9P =
cuts the parabola again at Q, show that A tangent is drawn to a parabola of latus
8Q/1Q. rectum 4 a and makes an angle $ with the axis prove that the sum of the radii of th6 two circles which pass through the focus and touch the tangent and the corresponding normal is 2 a cosec (1 cot c/>). 33. The tangent to a parabola at P meets the tangent at the vertex in T, the normal at P meets the axis in G, and the diameter through T meets
is 18,
32.
;
<
-f-
Prove that TG* = 4SP. SQ. Through the vertex A of the parabola y* = 4ax two chords AP, AQ are drawn and the circles on AP AQ as diameters intersect in R. Prove the curve in Q. 34.
t
that if
P
and
lf
Q
be the angles made with the axis by the tangents at and by AB, then cot 6l -f cot 8 -f 2 tan = 0. 2
an<*
(/>
THE PARABOLA Parametric coordinates.
4.
When
(i)
y form if t
the equation of the parabola is in its simplest form, the coordinates of any point on it can be expressed in the 2 at): for since t can have any value positive or negative,
= 4 ax,
2
265
2
(at
,
the ordinate of any point on the parabola is#',
= y'/2a
and since
;
y'
=
2
4a#',
we have by
we can
find
t
so that
substitution x'
=
at 2 .
any point on the parabola y = 4 ax can 2 2 at), and conversely it is manifest by (at substitution that any point whose coordinates are of this form lies on the parabola. The quantity t is called the parameter of the
Hence the coordinates
2
of
be expressed in the form
,
point.
When
(ii)
the curve 2
coordinate axes,
represented by (iii)
its
When
?/
2
is
referred to the axis and latus rectum as
= 4a (# +
(at
the curve
is
any point on the curve can be some value of /.
),
a, 2at) for
referred to a diameter
extremity as coordinate axes, we have y a = a cosec 2 a>.
We can then
=
2
and the tangent where
at
0(x
use (at 2 2at) to denote any point on the curve. ,
results apply equally to this case except when we the axes to be rectangular, e. g. when we are using the
The following assume
condition that two lines should be perpendicular or the condition that an equation should represent a circle. 5.
I.
To find the parameters of the points and the parabola y 2 = 4 ax.
of. intersection
of any
straight line
Let the straight line be If
2
any point
(at
,
lx
+ my+ 1 =
0.
2 at) of the parabola lies on this 2
lat -h
2 mat + 1
we have
= 0.
(i)
quadratic in t and gives the two values of the of the parameters points where the straight line cuts the parabola.
This equation
is
Cor. i. The points of intersection are real and distinct, coincident, or imaginary according as the roots of the equation (i) are real, coincident, or 2 imaginary, i.e. as aw is >, =, or < /. In particular the line touches the parabola when am* = /.
Cor.
ii.
Let
t
l
,
/
2
be the roots of the quadratic /
=- 2m and
thus If
+t
we
substitute these values of
I
t
t
(i),
then
= -
m=
-
and
m in
-f*
the equation of the straight
THE PARABOLA
266
we
find that the equation of a chord of the parabola joining the whose points parameters are ^ t2 is line,
,
y(i + 0-2a?2a/ 1 fa Cor.
direction of the chord joining the two points whose paradepends only on the value of ft-H 2 )' the axes are rectangular the angle ^ which this chord makes witL
The
iii.
meters are
When
t
l9 t 2
\ ft -f f ). given by cot ty if the direction of a chord Consequently,
the #-axis
is
,
sum
the
is
constant, the
sum
of the
constant, or since the ordinates are 2at 1 and of the ordinates of its extremities is constant. Hence the
parameters of its extremities 2a^ 2
(ii)
is
this gives us another proof of the its mid-point is constant property that the locus of the mid-points of parallel chords is a line parallel to the axis.
ordinate of
:
Cor. iv. The focus 2
points (a^ , 2a^), (a? 2 focus if 2a 2a^^ 2
=
Hence,
if
t
l
is
the point
is 2 ,
,
2r
2 ),
or
t
l
t
(a, 0)
2
=
hence the chord joining the two = 2at l t 2 passes through the
;
+ t 2 )-2x
viz. y(t l 1.
the parameter of one end of a focal chord,
-
-
is
the
M
parameter of the other end.
II.
The length of a chord in terms of the parameters of its
Length
= V ((at
2 l
extremities.
-aL*
Thus the lengths
of parallel chords ft + 1 2 =-. constant) are proportional to the difference of the parameters (or ordinates) of their Also for a focal chord, since extremities. 1, we have
^=
length
Example.
To find
=
the locus
a
of
2 (t l
the
2)
.
midpoints of chords of constant
length.
Let the constant length be c and let f u t 2 be the parameters of the ends of any one of the chords. The coordinates of its mid-point are given by
= a(f + *
2or
and by hypothesis
Now
(/!-
Hence
c
1
=
l
c
2 ft* +
2
*
9
=
a
8
f *), y -aft-M,), 8 a -f ^ 8 ) {(t + 7 2 ) 2 + 4}. l
*)-(f l +
g
(lax-if) ]^ 2 -f 4[ and the required locus ,
is
THE PARABOLA To find
III.
a point
the equation
ivhose parameter is
The equation t l9
is
t.
2
with
t
(i.e.
t
l
2
=
=
/)
whose equation
The tangent /
#
-
-
)f/-f
A/
\
t
= at = + x 0. x
=
-
A
4a#
at
A
/
.
is
or-
- B\
A/ly
(
\
2 ,
can have any value,
of the form
is
12.) (Chap. II, to the parabola which
C
- B\
(
t
Since the parameter
the parabola.
=
get for the equation of the tangent
(tfi
straight line
2
.
we
or i.
parabola y
t.
ty
Cor.
to the
of a chord joining the points whose parameters are 2x If the points J u t 2 are coincident 2at l t 2
#('1 + ^) t
of the tangent
267
is
aB* -f
A
-TJ-
x
ty
+
at*
parallel to
=
A 0,
follows that
it
=
is
any
a tangent to
Ax + By+C =
,.
Q or
f
,
the parameter of whose
point of contact is (-B/A). This result is often useful when the middle point of a chord is required for the diameter through the point of contact of the parallel tangent ;
bisects the chord:
thus the chord whose equation
=
bisected by the diameter y
is
Ax By -f C = -f-
is
2aB/A. The geometrical meaning of the parameter
t follows from the form of the equation of a tangent; the tangent at the point (a* 2 2at) makes with the axis of y such that t = tan an angle 2 is also the angle which the normal at (a* 2at) makes with Evidently
Cor.
ii.
,
.
,
(f)
the axis of x. as the parameter of a point It is often convenient to take tan coordinates of this point are then (atan 2 , 2atan). The equation of the tangent at this point can be put in the form x atan 2 i/ 2atan
IV. Since the equation of the tangent at the point at
2
ty
+x
=
1/j)
,
2 at)
the
is
0,
the condition that this tangent should pass through
point (x l9
2
(a<
;
some
specified
is
o
2
-y
1
+^ = 1
0.
(i)
Conversely, then, the parameters of the points of contact of tangents from the point (
xu
#1) to
the parabola are given by this If t ly / 2 are the roots of equa-
equation. tion (i), then the tangents at the points whose parameters are t l9 t 2 pass through the point (x, y^. Cor.
i. Since the equation (i) is quadratic, furnishes a proof that two tangents can be drawn to a parabola from any point these tangents are real, coincident,
it
:
THE PARABOLA
268
or imaginary according as t/1 2 ~4aa?1 is positive, zero, or negative, inside the curve. (a?|, yt) lies outside, on, or
Cor.
If
ii.
t
lt *a
are the roots of equation
(i)
as
i.e.
we have
the point (xl9 y^ is (a^fg, aft-M^)}. This gives the coordinates of the point of intersection of tangents at the points t l9 fa ; they can also clearly be found by solving the equations of i.e.
the tangents at these points. It follows also that (#lf y^)
lies
on the diameter bisecting the chord
joining ft,*,). Incidentally, if the points
shown that
t
/ l a
=
are at right angles,
V. To find
are the extremities of a focal chord
we have
=
and
the lengths
y* = 4 ax.
parabola
tlt *a
Hence, tangents at the extremities of a focal chord 0. intersect on the directrix a? + a
1.
of the tangents from any point 0(xl9 y^
A
straight line through the point with the axis of x, is
0(x l9 y^, making an angle
^1 = ^1^. cos
to the
(i)
sin
Let this straight line meet the parabola in the points P, P' then, r has the value OP (or OP'), the point P (or P') has coordinates ;
if
{rcosfl-j-tfj, rsinfl
the condition that this should
which If
is
we
+ yj;
on the parabola gives
a quadratic equation whose roots are OP, write w t E: y\ r* sin
Now
lie
2 --
2
^ax ly
+ 2r(y
l
this equation
sin
OP
the line
(i)
.
becomes
0- 2a cos 6) + u =
a tangent to the parabola, of and the this roots cide, equation are equal. if
7
is
(ii)
P
and P'
coin-
In this case
- 2 a cos 0) 2 =
(y l sin
which reduces
sin 2
(y^
- 4 axj,
to
a cot2
-y
l
cot
+ xl =
0.
(iii)
This equation is quadratic and gives two values of 0, viz. those corresponding to the directions of the two tangents OP, OQ, which can be drawn from to the parabola. If
has either of these two values, then equation (ii) will have to the length of the corresponding tangent
two roots each equal from 0.
THE PAEABOLA if
Thus,
drawn from
the length of the tangent,
is
I
parabola, whose
269
direction is
-..
), "
l ^
Bin 2 6
But the values
0,
are given
of cot
a cot 2 Hence, eliminating cot
a*Z*-u 1
0,
#x
we
i.
=
e . C ot 2
by
cot 6 -f x l
=
0.
get
-2as + 2a*)P + V{(**i-a) 2 + 2/i 2 } =
a
(y 1
to the
we have
l
0,
P giving the squares of the lengths the tangents which can be drawn from the point fa, y^ to the
which of
a quadratic equation in
is
parabola.
Since the roots of this equation are
a2
and
OP 2 OQ 2 we ,
,
have
OP 2 OQ2 = u^ .
.
the latter result can also be written
or
a.OP.OQ
Ut.OS.
Two tangents TP, TP' to a parabola meet the tangent in Prove that the radius of the circle TQQ' is at the vertex Q, Q'.
Example
i.
f2
1/iVa 4 where &, ,
are the focal chords parallel to TP, TP'.
Let P be the point whose parameter is t lt P* the point t 2 Since the tangent at P is f y a; a^ 2 = 0, the parallel focal chord .
,
=
0.
Therefore
if
tty
x+a
substitution
X
2
2aX)
is
~2# 1 X -1
=
(aX 2
,
is
either extremity of the focal chord we have by which gives the values Xj, X a of the para-
0,
meters of the ends of the chord.
Hence
\l
+ Xa
= 2< ,X l
1
Xa
Length of
Thus
/ = 4a(l4-f
Now
the intersection
I
and
focal chord
2 1
= ~1
and
);
Q
=
a(X 1 ~X a )
/ = 4a(l + f a
2
4a(l+tf).
2 9 ).
P
of the tangent at
with the tangent at the
is (0, a^), and Q' is (0, a* 2 ). vertex x = If 6 is the angle between the tangents at the points P,
sin 6
-
-
-
k-A ----------
+ <j j ) (1 + tt>)
But the radius of the
circle
^(1 +
TQQ' "
QQ' 2 Bind
=
T ,')
(1+
P
/
O
,
THE PARABOLA
270
Example
ii.
Show
inscribed in the conic
scribed to the parabola
Also
that, if P, Q,
the centroid
ofPQR
It is evident that
with two of
its
#
an infinite number of triangles can be Sax 12 + 9/3 2 --12aa = 0, and circum-
that
8y
2
B
%
j
= 4 ax.
2
any such
arc the points of contact of
triangle^
is (a, /3).
= lax
a triangle can be circumscribed to the parabola y a
vertices on the conic 2 2 8y -8aa-12/ty + 9/3 -12aa
=
0.
Suppose that the vertices of such a triangle are
Let X-f/i-fv=6 and XM^ conic
when
t is
=
equal to X or
p
;
then the point
| \
,
*
a(s
)
Hence
p.
-
0,
8a 2 ,43 + 4a(3^-4as)^ 2 -i-(8a 2 52 49/3 2 ~12aO(-12a3s)f-8a 2^
or
Two
of the roots of this cubic are therefore X and
three roots
is
p
(i.e. X/ui>),
on the
t)\ lies
=
0.
(i)
the product of the p. hence the third root of the cubic is v. This shows ;
a triangle circumscribes y2 = tax, and has two vertices on the given conic, then the third vertex also lies on the conic. Any number of such triangles can be drawn for if any tangent to the
that, if
;
parabola cuts the conic at A and J5, and tangents from A and B to the parabola intersect in C, then we have shown that C also lies on the conic. 2 points of contact, P, Q, R, of the sides are (aX 2aX). The centroid of the triangle PQR is 2'iv).
The 3
(a*
,
2
(a/i
,
Since
X,
/A,
v
are the roots of the cubic
o
x , t>A 4- X/i
Hence
X3
a
-f^t -f ^
2
,
(i),
33
3)3
=
-
(X
+ /i-f ^) 2 ~2f/x/-f vX-f X/t)
^3a a
The coordinates of the centroid are therefore VI.
y
When
the focus
is
(ft,
3).
the origin, the equation of the parabola
2
is
4a(x+a). shown, by a method similar to that used in I and II, 2 that the equation of the chord joining the points {a^ a, 2a^}, 2 = that the and is 2a equation a, #(i+ 2 ) 2(#-ha)-h2a 1 2 {a 2 a} It can be
>
=
2 a. 2a) is ?/ of the tangent at the point (a tf-ha + a modification is illustrated in the following example.
A
2 .
THE PARABOLA
271
Example. Two parabolas of the system sy*a(jc where
lt
and
l
z are variable, touch
Z^,
2 ,x'
=4a(?/
Z
2 ),
one another; find the locus of their point
of contact. Let the parameters of the point of contact
be
and
t
(x, y] for
the two parabolas
s.
Then the coordinates
of this point are + ot7 2 at} or {2 as.
{fi
,
J
2
4 a* 2 ).
Hence x = ^ + at'* = 2as y = 2a = /2 -f as2 The equation of the tangents at the point to the two parabolas are .
;
l^-at* 2 a -as
and, since these are identical,
Hence
= 23 .2 = 4a 4
ocy
(x, if) is jry
=
=
0,
=
0;
te =- 1.
4a 2 sf
=
4 a2
,
i.e.
the locus of the point of contact
2
.
Examples VII b. In the parabola */ 2 = 6#, chordb are diawn through the fixed point 5). Show that the locus of the mid-points of these chords is the parabola
1.
'9,
y -5y-3.r-f 27 = 0. 2. Show that the locus of the middle point of a chord of a parabola which subtends a right angle at the vertex is another parabola of half the latus rectum. 2
3. Show that the angle between any two tangents to a parabola is cos- ^/;*,) where r 1? r2 are the respective distances gf their point of intersection from the focus and directrix. 1
From the point (or, /3) two tangents are drawn to the parabola 4a#: show that the square of the area of the triangle formed by these 3 2 2 4aO*) /4a tangents and their chord of contact is ( 4.
2
t/
=
.
taken for origin, show that the equation of a tangent to the parabola can be thrown into the form x cos a -f y sin a -f a sec 0* = 0. 5.
If the focus
is
Two tangents, t l and f 2 are drawn to a parabola; h is the internal Show bisector of the angle between them and t the tangent parallel to h. that the product of the perpendiculars from the focus to t l and / 2 is the ,
same as that of those drawn
to
t
and
h.
that the envelope of the chords of the paiabola y 1 subtend an angle of 45 at the vertex is 6.
Show
#a 7.
The tangents
at P,
-f-
Q E }
2 8y-~24oo: + 16a
==
4#
0.
of a parabola intersect at the points
find the ratio of the areas of the triangles
PQR
which
and
P Q 7
,
',
K
:
P Q K. f
f
locus of the middle points of all chords of a parabola which pass a through fixed point is another parabola. 9. Tangents to a parabola cut off a length on a fixed tangent which subtends a right angle at the vertex show that their inteisections lie on 8.
The
:
a fixed straight
line.
THE PARABOLA
272
10. Two tangents to a parabola make angles fa fa with the axis prove that their lengths measured from the points of contact to their point of 3 a intersection are a sin (fa *> a )/sin fa sin fa ; a sin ((f) l <*> 2 )/sin fa sin fa :
,
<
.
=
2 4 ax make angles 0, <j) respectively tangents to the parabola y with the axis of y. Prove that the equation of their chord of contact is
Two
11.
=
ysin(0 + $)
<)
a)cos(d
(a?-f
+ (a? - a) cos (0 -f
The envelope of chords of a parabola, the tangents
(f>).
which
at the ends of
include a constant angle, is in general an ellipse. What are the exceptions ?
The envelope of the chords of the parabola whose mid-points
12.
x
ss
nty + c
2
is (y-f
lie
on
=
8a(a; c). 2a?w) 13. The locus of a point, the tangents from which to the parabola 1 y =3 4, ax make equal angles with y = x cot 6 -f c, is y (a -a?) tan 20. 14. Show that the ratio X of the lengths of the tangents drawn from any
=
point on the latus rectum produced to a parabola where y is the ordinate of the point.
Tangents are drawn abscissae are in the ratio M 15.
:
the parabola y*
is
8 given by aX
y\
-f
a
= 0,
the parabola y a = 4a# at points whose Show that the locus of their intersection is
to 1.
= (pi + n't)* ax.
A triangle circumscribes y = 4a# and two = 4 a (a?-f I) find the locus of the other vertex. y* 1
16.
of
its
vertices lie
on
}
is the orthocentre of the are tangents to a parabola and OT Prove is that bisected directrix the of the parabola. by triangle TPQ. 18. An equilateral triangle circumscribes a parabola show that the join
TQ
17. TP,'
:
of the focus to each vertex passes through the point of contact of the
opposite side. 6.
The equation
The equation
of the normal.
of the tangent at the point
ty-x-at Since the normal point
2
(at
,
2at)j its
is
,
= 0.
2 at)
is
is
at
3
t
-f
=O x)-y = 0. at 3
2at
y-f tx
or
2
(at
the perpendicular to this line through the
equation
i.e.
2
(2a
Similarly, the equation to the
normal
(i)
(ii)
at (a tan 2 fa 2 a tan
)
can be
written
x
atan 2
cos(f>
Note
i.
__ ""
y
2atan
The condition that the normal
through a particular point (xl y^ ,
...
is (see
at the point
equation
af + ttfa-xj-y^
(ii)
(t)
should pass
above)
0,
and conversely this equation gives the parameters of the feet of the normals which pass through (xl9 t/J. Now, since this equation is a cubic in t, it
THE PARABOLA follows that from these
may Note ii,
be
normals can be drawn to a parabola
three
any point
all real
273 ;
or one real and two imaginary.
The condition that
three
normals should be concurrent.
normals at three points whose parameters are t l9 ,, f s are concurrent at the point (a?t> t^), then t l9 fa t 3 all satisfy the equation If the
,
a*8 i.e.
fj,
2
,
t9
* t(2a-xl )-y l =
We have then
Hence
0,
are the roots of this equation.
^ + *a + tt
=
=
0,
the necessary and sufficient condition that the 2 t$ should be concurrent: the remaining two conditions give the values of the coordinates of their point of intersection in terms of these parameters provided that the above condition is satisfied. / L -f / a
-f^ 8
normals at the points
Now
is
t l9
i
if
ti
+ tt +
ts
2a/ +2a? a + 2af 3
also
1
hence
if
= =
0, 0,
the normals at three points on a parabola are concurrent the
of the ordinatea of these points
is
zero,
sum
and conversely.
Note
To find the condition that the normals at two points should interili. on the parabola. Suppose the normals at two points whose parameters are t ly a intersect at a point P(xl1 1^) on the curve whose parameter is X. Put xl = aX a y, = 2aX the parameters of the feet of the three normals
sect
;
,
meeting at
P are
given by a*
or
3
+ *(2a-X 2 )-2aX 3 2 * + *(2-X )-2X
= =
0,
0.
Evidently one of these three normals is the normal at the point P itself the equation may be written (*-X)'(* 8 -f A-f 2) = 0, so that t l9 / 2 are given by t* + 1\ + 2 = 0, and the required condition is t l t z = 2; thus the product of the ordinates of the two points is 8a 2 ^ind, further, the chord joining :
them
passes through the point ( 2 a, 0). Further, the values of f n *a are given by
= i(-X v X^r 8), /
^
and consequently the two normals which can be drawn from a point 8 (aX 2aX) on the parabola other than the normal at this point are real, a coincident, or imaginary according as X is >, =, or < 8, i.e. according as ,
the absciasa of the point
Note iv. To find the a parabola are coincident.
is
>,
it
.
locus of
If the point is (xl9 yj, the
from
=
or
<
8 a.
a point, two of
the
normals from which
to
parameters of the feet of the normals drawn
to the curve are given by
THE PARABOLA
274
Now equal.
if
two of the normals are coincident, two roots of t then we have lt t l9 1 2
Let the roots be
this equation are
;
or
'.= -2^,' the roots of the equation are
i.e.
Hence
-3/,
2
=
a
t
lt t lt
~ x\
or
2^.
xi~
=
3/2
and Eliminating
i.e.
t
l9
we obtain
the equation of the locus required 27 at/ 2
=
is
4 (a? -2 a) 3
.
This equation represents the locus of the intersections of consecutive normals and the curve is called the Evolute of the parabola. Now at a point (if any) where the evolute meets the parabola the two normals which can be drawn through it other than the normal at the point itself must be coincident if this point of intersection is the point :
2
(aX 2aX) of the parabola we have shown (Note of the feet of the other two normals are given by ,
*
2
These are coincident when X the parabola and
its
+ fX + 2 2
=
evolute are
=
iii)
that the parameters
0.
8: hence the points of intersection of
(8, 4>v/2a) and
(8a,
~4 v/2):
the
s c
student can verify this by substitution in the equation of the evolute. evolute meets the axis at (2 a, 0); its graph is shown in the figure;
The
PR, PR,
PG
are the normals meeting at P.
THE PARABOLA We
Cor.
have seen that the normals meeting at (# n
=
at^ + t(2a- T )-i/ l t
Now
275
l
?/,)
are given by
0.
Hall and Knight's Higher Alyebm, 579) there is always one real root of this equation the other two are real, coincident, or imaginary (vide
;
according as 27a// t
2
>, =, or
is
<
4(^~2rt)
3 ,
according as the point (a? 1? yj lies inside, on, or outside the evolute thus the evolute divides the plane of the parabola into two parts such that from any point (e.g. Q) in the one three real normals can be drawn to the
i.e.
:
parabola, and from any point in the other only one real normal can be
drawn.
=
A
chord of the parabola y* 4 ax passes through the Example i. point (A a, 0): prove that the normals at its extremities intersect on the A 2 a (x A a 2 a). curve y*
=
f
t ,
Let the parameters of the extremities of any chord through (Xa, t the equation of the chord is 2
0)
be
:
2Xa + 2?
hence or
l
f
a
^/ 2
Now by
a*
3
if
the normals at
+ *(2a-;r)-t/
Thus
f1
+ f,-f< 3
=
=
0,
t
ly
t
2
>
= =
meet
tz
V
a
+
Vs + '3^=
ti
+
t2
1
^^ = X-f^ Example
3
ii.
the point (X,
at (x, y), these parameters are given
J and
3
2
and
a
^
X.
0.
Hence, by substituting
Eliminating
0,
5 ,
?jf 2
'iVs^'J-
=
=-Xf
a
w^
^>
3
have
.
we get the equation
of the locus required, viz.
Normals are dratvn
to the
,
Y) triangle formed by :
shotv that the equation
parabola y
of
1
ax
=
the nine-point circle
/rowi
of the
their feet is
Let the feet of the normals be A, B, C and the parameters of these the nine-point circle passes through the mid-points of 3 points be f lf f 2 ,
:
the sides of the triangle ABC. Since the normals at A, B, C meet at (X, Y)
thus
/
l
+ *8 +
f
3
=
The mid-point
0; /^ 2
of J?C
+ Ms-*^3'i= is
2
{Ja(? 2
X
2a
+ f3
7 2 ),
a(
J
y
t
l9 t t
,
f
3
are given by
THE PARABOLA
276
Now
=-
tt -f tt
f,
and
,
t^ + tt t3 + ts t =
.-.
,,,,,-.
...
But
t**
t t t 9
t
-t*,
?t*.
+t
,
hence i.e.
BC
the mid-point of
is
and the coordinates of the mid-points of CA, metrically by writing f a *3 for ^ respectively.
AB
can be found sym-
,
Now
let
the nine-point circle be
Since each of the mid-points
{-XT-2a-|at which reduces to
2
2
}
Since the coefficient of
*
lies
on
this,
3
is zero,
the
sum
but the sum of the three roots root must be zero.
is
zero
t
1
,
t
2,
;
tl9
which, having roots
satisfy the equation
,
=
0,
-1-4
{3a-X-g} *-8/=
0,
!
t
3
of the four roots of this equation t is zero, therefore the fourth 3 2
2 Hence (X-2a) + 20(X-2a) + c and the equation reduces to
at8
and
+
l9 t 29 ^ s
,
is
(i)
dentical with
12-4X-4^ = 2a-X,
Hence and
8/=
But) substituting for g in
(i),
F.
we have
Hence, substituting in the equation of the circle for the nine-point circle
and
c,
we
get for
7. Relations between the coordinates of the points of intersection of the tangents and normals at any two points on a
parabola.
Let
(xy y)
and (,
77)
be the point of intersection of the tangents at two points
that of the normals.
The parameters meet at
(#,
of the points' of contact of the tangents which
y) are given
by
=Q,
(i)
THE PARABOLA and the parameters of the from (, rj) are given by
Hence two equation
277
normals which can be drawn
feet of the
of the roots of equation (ii) are the same as those of These two roots therefore also satisfy
(i).
*-yt + x} this equation is therefore identical with
a
= 0,
and we have
(i),
x
y
now
that the points of intersection of two tangents under certain conditions lie on some locus /(#, y) 0, the corresponding
Suppose
=
locus of the intersections of the normals at their points of contact will be
(,
77)
= 0,
and the equations
The converse have from
obtained by eliminating x and y from f(x, y)
(iii).
if
more
proposition can be stated
directly
;
thus
we
(iii)
,~a.
*--.+.., and
=
the locus of the intersections of the normals at the ends of
=
a chord moving under given conditions is /(, t/) 0, the equation of the locus of the intersections of tangents at the ends of the chord is
2}
_*
Example. I'D find the locus of the intersection of the normals at the ends of a focal chord.
We know
that the locus of the intersection of the tangents at the ends of is the directrix x -f a = 0.
a focal chord
The Xj
locus of the intersections of the normals
y from
this equation
2a_ ~~
y__#-f ~" a y a
TT
jience *
y =
n,
is
obtained by eliminating
and
_
~~3a
y
/;
x
y __ ~"~
? ~*
a
y
and
^-atf-Sa); or the required locus
is
y*
= a (x - 3 a).
Examples VII
c.
Find the equation of the normal at the point (-a-f a* 2 2at) of the 2 parabola y = 4a(a?-f a). 2 2 a*) meets 2. Find the parameter of the point where the normal at (a< 1.
,
,
the parabola
2 r/
=
4o#
again.
THE PAEABOLA
278
Find the equation of the normal at the point (eft 2 , 2 dt) on the parabola 2 4df#, where dsin o> = a (oblique axes). Show that the normals to a parabola at the points (a^ 2 2 a^), (af 22 2at z )
3.
y
=
2
4.
,
,
intersect at the point {2 a -f a (t? -f ^ f 2 -f / 2 2 ), - at l t 2 (t l 2 5. Normals at pairs of points on the parabola y
-M 2 ) } = 4ax meet on the
=
x
.
line
Find the locus of the intersections of tangents at these pairs of Also when the normals meet on y = k. points. 6. Tangents are drawn to a parabola from points on a line parallel to the axis prove that the normals at their points of contact intersect on a fixed h.
;
straight line.
Show
1 that, if a variable chord of the parabola y at the its extremities meet x, parabola y tangents
7.
2
2 f/
=
16,r,
=
and
find the locus of the
= 4#
touches the
on the parabola
meets of the normals at
its
extiemities.
P of
a parabola cuts the axis in G and meets any point If the normal makes an angle Q with the axis, prove the curve again in Q. that G#sm 2 2^Gcos<9.
The normal
8.
at
=
that the equation of any normal to the parabola ?/ 2 = 4& (x + c) s may be written in the form y -f nix (2b c) m 4- 6m 2 2 am) to the parabola y = ax. 10. Normals are drawn from the point (am 2 Show that the feet (^ 2a^), (at*?, 2at 2 ) of these normals are given by
Show
9.
=
.
1
,
,
+ mt + 2 = 0. Show also that the product of their lengths is 4r< 2 (l -f m')\ 11. Show that the middle points of the sides of a triangle formed by ^ax lie on the parabola 2# 2 -f ax = tangents at P, Q, R to the parabola if 2
/
the normals at P, Q, R are concurrent. 2 12. If chords of the parabola / = 4ar pass through the foot of the at their show that normals extremities meet on if1 = a(x a). directrix, if
the chord PQ of y2 == kax passes through -2a, 0), the normals at on the curve and contain an angle equal to PAQ. meet Q
13. If
P,
14. If
(
the normals at
P P P {
2
,
8
,
are concurrent, the centroid of the triangle
PjP P3 lies on the axis. If Pj, P coincide at 2
2
15.
meet
If the
SN
at N, then
16. If
then x^ 17.
P P^ is xjx' y/y' = 2. (.*/, >/') the equation of normals corresponding to the tangents drawn from T(h,
A
2 :
ST*
}
=
the tangents at P,
SP + SQ and
Xi
2
4- A;
=
k)
2
2
Q meet ay^
-\
:
r/
.
at (x l a:
l
yl
,
yj and the normals
at (r 2
,
y 2 ),
.
is inscribed in a parabola and the normals at its vertices show that the perpendiculars to its sides at the points where
triangle
are concurrent
:
they meet the axis intersect on the tangent at the vertex. 18. TP, TQ are two tangents to a parabola; PJV, QN are the correis the mid-point of TN. sponding normals ;
M
TM
subtends a right angle at the focus. 19. If P, Q, R are the feet of the three normals from a goint on the line 2 x 2 a + c, the intersections of the tangents at P, Q, R lie on f/ = a(x + c). 2 = 1 20. If a tangent to y = 4a(#-f a) meets a normal to t/ 4b(x + b) at Prove that
right angles, the locus of their intersection is a parabola. 21. A 9 By C are the feet of the normals which meet at (OK, 0)
;
prove that
THE PAEABOLA
279
AB
the straight lines bisecting BC, CA, at right angles are normals to the parabola if = 8a(4a-fa-#) at the points whose ordinates are equal to those of A, B, C. 22. The normals at P, Q, the ends of a focal chord, meet the curve again
Show that P'Q' is parallel also that the envelope of P'Q' eight times that of the given parabola. at P', Q'.
Show
and equal is
to
3P0.
a parabola whose latus rectum
is
2S. Find the orthocentre of the triangle formed by the feet of the normals fiom (X, Y) to the parabola. 24. The three normals from a point P to the parabola y' = 4jr and the line through P parallel to the axis form an harmonic pencil: show that P lies
on 21 a^
=
2
(a?
-2 a) 3
.
The parabola and the
8.
circle.
We a
propose firstly to discuss the intersection of a circle and in the next parabola by means of parametric coordinates ;
we
section
discover the forms of the equations of circles and other curves which are variously related to the parabola. Some of the work overlaps the student will learn by experience which shall
;
method is the more appropriate The general equation of a circle a?
for a given problem. is
+ y* + 2gx + 2fy + e=
0.
(i)
2
any point (af 2 at) on the parabola lies also on this circle, the parameter t of the point must satisfy the equation obtained by 2 at in (i), viz. a 2 y substituting x If
,
=
,
= 0.
a2 /4
Since the
+ (4a 2 +20a)J 2 + 4/* + c parameter of any point common
parabola satisfies this equation, the values of the parameters of circle
it is
all
(ii)
and the
to the circle
evident that this equation gives the points of intersection of the
and the parabola.
Cor.
i.
The equation
parabola in four points or all imaginary.
Cor.
ii.
If the
coefficient of
t*
is
;
is
a quartic inf;
these,
may
be
hence every circle meets the two real and two imaginary,
all real,
four roots of equation
(ii)
are
we have = 0. fi + fa + f 8 + f 4 sum of the parameters (f lf
t
ly
t
2
,t s
,t^
since the
zero,
Conversely, if the the parabola is zero, these four points c so that
20 4- 4 a
=
lie
on a
?
2
3
,
circle
;
,
t4 )
of four points on
for if
we
find
<;,
/,
and
THE PARABOLA
280
then ID f*, * 8 * 4 are the roots of the equation (ii); but the condition that 2 t l should satisfy this equation is also the condition that the point (a^ 2at ) should lie on the circle #2 4- /2 + 2gx -f 2/# 4 c = 0. ,
,
}
Thus the necessary and sufficient condition that four points on the parabola should be concyclic is that the sum of the parameters, and therefore the sum of the ordinates, of the points should be zero. Cor.
If the four points
iii.
we have ^ +
ABCD
a + *s + **
whose parameters are
But
=
t
l
,
t
29 tst
4
are
0*
concyclic, These four points can be joined in pairs in three ways, AC, BD. The equations of the first pair are tf
AB, DC-, AD,
BC
;
+ ^4), hence these chords are equally inclined to the axis The same is true for the other pairs. Hence, The common chords of a circle and a parabola are in pairs 7j-f
t,
2
('3
of the parabola.
equally inclined to the axis. Cor. iv. When two of the points of intersection coincide (e.g. C and D] and the common tangent the circle touches the parabola at C. Also at C, being a pair of common chords, are equally inclined to the axis.
AB
Cor. v. Three of the points (e.g. B, C, D) may coincide; in this case the circle both cuts and touches the parabola at B. The circle is then said to osculate the parabola, and is called the osculating circle or the circle
of curvature
at the point B.
AB and
the tangent at B are a pair of common chords, they are equally inclined to the axis of the parabola. The properties of the circle of curvature can be at once deduced from the
Since
2 2 equation (ii) a*t'+ (4a + 2
.
Thus
S^ + fg
(a)
or
*
2
O,
=
-3*!,
2 BO that the circle of curvature at the point (fl^ , 2a^) meets the parabola 2 again at the point (9a^ ~6a,) the equation of the common chord of the ,
;
parabola and the circle of curvature at the point
(b)
The sum of the products .-.
The sum
+4
t
l
is
of the roots two at a time
=
20 4 4-
;
S^ + Sfjf, = 8^-9^ = -6^;
of the products of the roots three at a time
= ---
;
THE PARABOLA The product of the
roots
281
C
=
2
The centre of the circle, i.e. the centre of curvature, ~2af 1 3 ). The radius of curvature (p) is given by
is
therefore
;
(c) If the length of the radius of curvature is given, the parameter of the point of contact of the corresponding circle of curvature is given by
There
is
only one real cube root of ~-^
hence there
y
only one real value
is
This value is positive provided that p > 2 a, in which case equal and opposite values. Thus the minimum length of the radius of curvature is 2a. of t*.
tl
has two
Two circles of curvature, symmetrically placed with respect to the axis of the parabola, correspond to any value of the radius of curvature greater than 2 a. (d) It follows
from the
results
found in
of curvature at the point (a* 2 2a/) ,
?
a?
(b) that the
equation of the circle
is
+ t/2 -2(3a*2 + 2a)# -f 4f y-Sa2 * 4
=
0.
R
are three of the points of intersection of a circle and a parabola, then, when and R coincide with Q, each of the chords PQ, to both the circle and the parabola. a becomes QR tangent (e) If P, Q,
P
These two tangents are coincident, hence the corresponding normals are Since these coincident normals are normals to the circle, they
coincident.
intersect at the centre of curvature.
Since they are consecutive normals of the parabola, they intersect on the evolute of the parabola. Hence the evolute is the locus of the centre of curvature.
We
have shown that the coordinates of the centre of curvature corre-
sponding to the point
(at*,
2 at) are given by
y Eliminating
f,
of the evolute,
is
= -/
the equation of the locus of the centre of curvature, 27 ay* = 4(.r-2a) 8 .
i.e.
Cor. vi. If equation (ii) has two pairs of equal roots, the circle touches the parabola in two points and is said to have double contact. Let the two points be flt Ja ; then the roots of the equation (ii) are *i
>
*i
The
*a
*a
i
but 2
*
=
;
hence
*a
-^
.
points of contact are therefore symmetrically placed with respect to
THE PARABOLA
282
the axis, and the common chord of the circle and the parabola dicular to the axis.
The centre is evidently on the axis the abscissa of the centre is given by
-a =
and
minimum
its
value
perpen-
-2^
1^
2 ,
at^ + 2 a,
2.
therefore
is
and, since in this case 2
;
is
i. A circle cuts a parabola in four points: if the normals of these points are concurrent, prove that the circle passes through vertex, and find its equation if the normals meet at (f, ?/).
Example at three the
Let the parameters of the four points of intersection be
t
l}
t
2
,
f
3
,
4
.
Since the points are concyclic 'i
and
if
the normals at
t
lt f a
,
+ '2 + 's + *4
=
are concurrent
t3
= 0.
+ * a + 's
*i
Hence ti = 0, i.e. the circle passes through the vertex. Since the normals meet at (, ;) the parameters of their leet are given by at*
Let the circle be
x~
+ (2a-)t-r)
=
2
+ y + % gx + 2/y
=
0,
(viz.
t
lt
f
a
0.
/
,
3)
(i)
then the parameters of the
points of intersection of this circle and the parabola are given by at3
+ (4 + 20) t + 4/=
Since these are by hypothesis ^,
Thus
20
t
2
f
,
3
,
0.
equations
(ii) (i)
and
(ii)
are identical.
= -(f-r2a),
2/=-H Hence the'equation
Example real points,
A
ii.
from
If
from
L L
is
is
straight lino cuts the evolutc
of a parabola in
drawn.
is
normals are
any point on the
to the parabola, other tJian the SJiow that the centres of curvatures at
collinear.
e volute,
two of the normals which can be drawn
to the parabola coincide with each other. If P(at l z t 2a^) is the foot of these coincident normals, then
centre of curvature of the parabola at P. If the third normal from L since the normals at }
2f 1 + i.e.
the parameter of
We
three
each of which the normal
radius of curvature, the feet of these
of the circle
Q
is
f
a
Q t
(at^, 2
lt t l
,
*
2
a)
is
L
is
the
the foot of
are concurrent,
= 0,
-2^.
have therefore to show that
if
the centres ot curvature at three
points whose parameters are * lf / a >*s are collinear, then the centres of curvature at the three points whose parameters are -2^, -2f a -2* 3 are ,
also collinear.
THE PARABOLA
283
The centre of curvature at the point t is [2a + 3 2 -2aJ 3 } the centres of curvature at the point ^, * 2 ? 3 are collinear if f lf ? 2 , *3 satisfy an 2 3 0. equation of the form p(2a + 3a/ ) 2a#/ -f r = 0, or 2 (!/<)= 0, and the The necessary condition for this is'that 2 ;
,
',
=
2^
condition
evidently sufficient. the condition that the centre of curvature at the points 0. But if 2 (1/0 0, -2^-2/2, -2f s should be collinear is 2 (-1/2 is
Similarly,
=
=
then obviously 2 (-1/2/)
=
0.
Examples VII d. Find the radius, centre, and circle of curvature at the extremity of the
1.
latus rectum.
A
2.
circle
A
3.
touches the parabola at the two points of intersection of the
x
curve and
3a
circle
find its equation. described on the chord of a parabola whose equation is as diameter; find the equation of the other common chord
is
Ax + By + a
;
of the circle and the parabola. 4. The circle of curvature at the vertex meets the curve in four coincident points.
The extremities of any two chords of a parabola which
5.
dicular to the axis are concyclic. 6. Find the points at which the radius of curvature
are perpen-
16.
is
The common chords of the circles of curvature at (x */,), (#2 prove that tively and the parabola intersect at the point (, ?j) 7.
v ,
,
respec-
// 2 )
;
P
The
8.
on a parabola circle of curvature at a point If p^ p.2 are the radii of curvature at
P
again in Q. $ P\
Pz i 8 constant. If IM,, w 3 are the roots of the equation 2
9.
?
,
show that the points 2 circle .r 4 /r + (
2
(ani^, 2^Wj),
(rtH 2
,
2rr>;; 2 ),
w
meets the parabola and Q, prove that
3
2
(o>' 3
,
4
pm*
2ani 3 ]
-)
qm lie
-\-r
0,
on the
ax + I (r pq) ro/ tfpr 0, and deduce the length 2 of the radius of curvature at any point of// = ax. 10. A circle passes through the vertex and three other points P, $, It of a parabola. The lines joining P, Q, R to the focus meet the curve again at 2
>r
P',
<J)',
9.
7i
Prove that the centres of curvature at
.
Forms
t
P',
R' are collinear.
of Equations.
In this section
u
4)
we
shall use the following abridged notation.
P = 0, the equation of any parabola. = 0, v = 0, the equations of two chords of the parabola. = 0, = 0, the equations of any two tangents to the parabola. C = 0, the equation of a circle. t'
Throughout k
is
used for an undetermined constant.
THE PARABOLA
284 (i)
P = kuv.
Now
= 0, v =
P=
in the points -4., B, (7, D. cut the parabola the coordinates of any one of these points satisfy the equation
Let u
P-Jcuv For the point A
P
substituted in
sequently,
Now P uv
when
0.
its coordinates, therefore, on P = and u = and conand u, make these expressions zero
lies
;
:
substituted in
P
Jcuv,
Hence
make
it zero.
and since u and v are
of the second degree,
is
of the second degree.
they
linear,
Pkuv is of
the second degree the equation P Jcuv consequently represents a conic passing through the four points A, B, C, D. The constant Jc is still at our disposal, so that the conic may be made to satisfy one other condition is
by giving
a suitable
Jc
For example,
value.
parabola, or it may pass through In two cases the equation
some given
P=
lines, viz.
AC,
BD
represent a circle, that u
= 0,
v
Example. the points
and
=
The equation
it
may
be another
fifth point.
represents a pair of straight
AD, BC. The equation cannot in general since two conditions are necessary we have seen ;
:
must be equally
To find
inclined to the axis of
the equation of the
of intersection of y
x+2y =
Jcuv
;
2
= 4# and
P=
0.
parabola which passes through the straight lines Sx + iy 6
=
3.
of the parabola must be of the form 4 y -4a? + fc(3a? + 4y-5)(3? + 2y--3)
The condition that the curve which be a parabola is 3fc(l -f 8AO-25&1 = 0, Hence the required equation is
=0.
the locus of this equation should i.e. k = or 3.
is
0.
THE PARABOLA
P-ku = 2
(ii)
0.
SD become
= 0,
u
If the straight lines
AC,
285
=
v
coincide, the pairs of points
coincident: the conic
the parabola at the points
A
Pfat?
=
therefore touches
and B.
AC
B
Thus
Pku = 2
the general equation of a conic having double contact with the parabola, u being the chord of contact. is
=
The equation can only represent a pendicular to the axis
circle
for one value of k
:
straight lines, viz. the tangents
A
and
when u
it
=
is
per-
represents a pair of
B to the parabola.
A
conic has double contact urith a parabola, one of the Example. Show points of contact being the vertex, and passes through its focus. that the locus
of its centre 2
a parabola.
is
= 4a#.
Since the chord of contact passes through Let the parabola be y the origin, its equation is of the form lx + my = 0. The equation of the conic is then
k (y 1 - 4 ax) 4since it passes through the focus
4a2 fc
.-.
The equation of the conic J
4Pa?
or Its
centre
is
2
is 2
2
(y
=
(Ix
0)
(a,
a2 /2,
+ my)*
=
0,
;
i.e.
k
= \l\
then ~4aa?)
+ 4(te + my) 2 =
+ 8Jwa?y + (4w + P)y2 -4aP#= 2
0, 0.
given by 4J2 # + 4fwy-2aJ
2
=
0,
or 2lx +
4lmx + (4ma + P)y
and
" and the required locus
I
2a is
_ = m
y
^
a
__
=
2my-al =
0,
0;
Ix -f my
a(2#-a), which
is
a parabola.
THE PARABOLA
286 (iii)
P-kut =
When
0.
the line v
the points
(7,
D
=
is
coincide
=
a tangent to the parabola (viz. t 0), and the conic touches the parabola
at C.
Thus
PJcut=0
represents a conic passing through the intersections of P and u and 0,
=
P=
touching of
t
It
u=
=
at the point of contact
= 0. can only represent a
=
0,
the axis of
circle
when
are equally inclined to
P=
0.
For one value of k
it will represent the pair of straight lines CA, CB. Note that, if the line u is not
=
given, the general equation of a conic at the touching the parabola
P=0
point of contact of
t
=
is
=, 0.
We
have
now
three undetermined constants, and the conic can
therefore satisfy three other conditions.
Find
Example. y* =
the
4#-f 4 at the point
The tangent
equation of the circle touching the parabola (8,
6) which passes also through the focus.
at the point (8, 6)
is
10
The equation
Since
it
of the circle
is
0.
of the form
passes through the focus
-4fc
=
=
(i.e.
the origin)
10 or
fc
= -f
Hence the equation becomes 2
5(#
-4#-4) + 2(#-3y + 10)(te + wf/ + 1) =
The conditions that
this should represent a circle are
and .-.
The equation
of the circle
is
5-6m = 2J, m = 3J; = i,m = |. ;
therefore
a
10(y ~4a?~4)-f(a?-3i/4-1 i.e.
#a + 1/2 -26# +18y
=
0.
0.
P
(iv)
= 0,
kut
THE PARABOLA when u = 0, t = intersect
If three of the points, for
example A,
C, I),
287
P=
on
coincide, the conic
=
touches and cuts the parabola at A. The conic PJcut are then said to have three-point contact parabola P = *
tact of the second order
To find
Example. the
of curvature
circle
(aX
2
'.
at
The
tangent at the point X
and the '
or
*
con-
ACG.
the equation
the
2a\) of the parabola #
,
0.
of
point
= 4a#.
2
is
and any chord through the point of contact
is
H
where
= x - a\ *n
is
2
=
+ w(t/-2aX)
0,
a constant to be determined.
The equation of
the circle of curvature
The conditions that
X
1 -f
-m =
2
(l4-X )(y which reduces to
circle is
k\m
;
v
The equation of the
therefore of the form
this should be a circle are
k
and
is
-1
,
then
2
0,
=
0.
P-kt 2 = 0. When the four
(v)
u=
0,
v =
Pkt* = is
points A, B, C, 7> coincide, both the chords 0. The conic coincide with the tangent at A, viz. t
=
then meets the parabola in four coincident points and
said to have
The
*
four-point contact
conic can only be a circle
when
f
or t
=
'
contact of the third order is
'.
the tangent at the vertex.
THE PARABOLA
288
=
P-ktt'
(vi)
When
0.
the chords
u
S
= 0, v = D coincide,
both become tangents, the pairs and we get another form of the
and (7, of points A, equation of a conic having double contact with the parabola points of contact are those of the tangents.
:
the
2
Thus
#
and
#
2
both represent a conic having double contact with the parabola at the points (a\ 2, 2aX),
(vii)
P-kC = 0.
By similar reasoning this represents a conic passing through the and the circle four points of intersection of the parabola P C 0. For certain values of Jc it will represent the common
=
=
chords of the circle and parabola. Note. It is evident that the seven forms here discussed would give similar results if we used S = 0, the equation of any conic, parabola, ellipse, or hyperbola, instead of P = 0. It is therefore important to understand these forms we have used the parabola because the student is now familiar with :
the form of its equation.
Example. Prove that the equation of the parabola which passes 2 4o# at through the origin and has contact of the second order with y
=
the point (a/x
2 ,
2a/i) is
2
2
(4^-3fiy) +4afi (8^~2 /iy)
=
0.
is the The required parabola is of the form P~kut*=Q, where t = 8 = this to of is the w the 2 and at point, i.e. origin join a/*) (a^ tangent ,
0.
THE PARABOLA Since the curve
289
to be a parabola,
is
.e.
.*.
the required equation
is
p?(y*-4cax)-8(ny-x-an ) (2#-/zt/) = 0, 16o?2 -24^^ + 9^y 2 + 12a/x2 a;-8a/i3 y = 0, 2 2 (4#~3/x?/) + 4a M (3.r~2/zy) = 0. 2
i.e.
or
Examples VII
e.
Find the equation of the other parabola which passes through the four common to y 2 = 4a# and #2 -f t/2 -f 2gx + 2fy + c = 0. 2. Find the equation of a parabola which has contact of the second order with y 2 = 4 ax at the point (at 2 2 at) and passes through an end of the latus rectum. 3. Find the equation of the rectangular hyperbola which has contact of the third order with the parabola at the point (at 2 2 at). 2 4. For what values of \ does the equation 4ax + A (x 2 a) (# 3 a) = y 1.
points
,
,
Illustrate these lines in a diagram.
represent straight lines. 5.
any point T on the line x = 2 a the origin. passes through circle has double contact with the parabola t/ 2 =. 4 a (x -fa), and the
TQ
TP,
are tangents to a parabola from
show that the 6.
A
point whose abscissa its centre. 7.
y
2
:
TPQ
circle
is
8a
is
one point of contact.
Find
its
equation and
passing through the vertex of the parabola and cutting the parabola orthogonally at the other point of inter-
Circles are described
= 4a#
section.
Show
that their centres 2
2
lie
on the curve
=
-
2
ax (3#-4a) 12 ax) (2y + x 1 8. Find the equation of the circle which touches the parabola y 2 at the point (am 2am) and passes through the focus. 2f/
1
.
= 4 ax
,
Prove that three such circles can be drawn to touch a given line at the focus and that the tangents at their points of contact form an equilateral triangle. 9. From a point T(x', y') tangents TP, TQ are drawn to a parabola: show that the other common chord of the parabola and the circle TPQ is
the polar of (2a-#',
7
-y )-
=
h tangents are drawn to the parabola points on the line x round the triangles formed by each pair are described circles y and their chord of contact. Find the locus of their centres.
From
10.
2
=
4a#, and
11.
y* ==
A
circle of variable radius
4ax at
P
and
Q.
Show
whose centre
is
(0, b)
meets the parabola
that the locus of the intersection of the
2 tangents to the parabola at Pand Q is y*-2axy + 4a*y-4a & = 0. 12. PQ is a focal chord of a parabola. Two circles are drawn through the focus to touch the parabola at P and Q respectively. Show that they cut one another orthogonally, and find the locus of their second point of
intersection.
THE PARABOLA
290 13. Circles are
drawn through the focus
to touch y*
=
4a#.
Find the en-
velope of those common chords of the circles and the parabola which do not pass through the points of contact. 14.
The tangents
of the circle
= 4ax
1
to y
P
at
on the parabola. TPQ Show that the locus of T is y (a-#)2
Method
10.
and Q meet
in
T,
and the centre
lies
=
16a* +
8aV-
of reducing the equation of a given parabola
to its simplest form.
When
the general equation of the second degree
takes the form
it
(lx
=
=
+ my) 2 + 2gx + 2fi/ + c
0.
X= Y=m
Then
I
and the equation of its axis (aX + hY = I'2 and h = lm
since a
=
Chap. VI,
0,
7)
becomes,
y
_
+ wy +
te
,
or (x-f wi^-hw
=
.
._
0,
if
..
we
write
+ __ =0 to
fn/
A ,
+ mf M= Iq ^ ^W-^ ~T"
The equation which reduces
of the parabola can be written
at once to
+
(te+mjf + n) If
we
take Ix -f
my -h w
2(J^,!)(M:-Zy)
=
-
z
- r
?
new
axes of ^ and
Ja<^ r^c/wm and
JF'iwd
Example.
=
A
the equation
X=
also
a25,
^
=
60;
of the axis and
of the parabola
-25= In this case
0.
ffm-fl the equation becomes
^/,
the tangent at the vertex
=
and 2
as
+ c-
25# -f 60y - 78
.'.
\
=
T^
5 (5*
4-
12y) -73,
0.
THE PAEABOLA The equation
of the axis (5 x
i.e.
is
291
therefore
+ 12y) (25 + 144) - 365 + 534 = 5#-f 12y + l = 0.
,
The equation of the parabola can be written
we take 5#+12y +
If
1
1 equation becomes y = x, Rence the latus rectum
at the vertex are
=
0,
is 1,
bx+ 12y + 1
12x~5y + 2 =
as axes of
X
f.
Find the latus rectum of 9a; -f-16y -f 24a#-4y-a? + 7 = Reduce to their simplest form and draw the graphs of 2
2
1.
T, the
and the equations of the axis and the tangent 0, and 12oj~5y-f 2 = 0.
Examples VII 2.
and
0.
:
(i) (ii) (iii)
9#2 + 16y2 + 24*y-34#-f-384r-f
1
=
0;
(iv) 3.
Show
that the parabola
has the same axis and focus for all values of X. 4. Prove that the two parabolas which can be drawn through the four common points of ax* -f by* = 1 and #2 -f-ya + 2## + 2/y + c = have their axes perpendicular, and that their latera recta are equal if bf = ag, 5. Find the equation of the parabola which cuts the axes at the points (a, 0) and (0, b) and has its tangents at these points parallel to the axes oft/
and x respectively. 6.
A
parabola has for focus the point (,
#+%+c
=
Show
q)
and
for directrix the line
Ax + By+C=Q
is a tangent to the + + B^ b)(A C) = 0. parabola 7. Show that the locus of the intersection of normals to a parabola which are at right angles to each other is a parabola. Find its focus and vertex. 2 8. Find the focus of the parabola (oo;-f fo/) = 2y 9 and show that the a 2 a 8 = is 2 a (a + & 0. of its latus rectum ay) + a equation ) (bx
if
0.
that the line
2 (A + B*)(a( + b n + c)-2(Aa +
Z>
9. If
nates of
11.
aar*-f its
2 hxy + by* + 2gx
=
represents a parabola, find the coordi-
vertex.
The equation of a parabola
referred to
any pair of tangents as
coordinate axes.
Let the parabola touch the coordinate The chord of contact is therefore
T 2
axes, at
A
(a, 0)
and
B (0,
b).
THE PARABOLA
292
We have seen in chord of contact
is
9 that the equation of a pair of tangents whose
u
*-=
is
form
of the
Pku* = 0.
In the present case therefore 2
or
In order that
should be a parabola, k must equal } ab
:
the equation of the parabola
is
therefore
ab
The
Parametric representation.
coordinates of any point on 2 2 this parabola can be expressed in the form 6 (A I) }, for [a A, these values of #, y satisfy the equation of the parabola and they ,
can have any positive value we please since A may have any value. This point will be referred to as the point A on the parabola. (i)
To find
the equation
of the chord joining
the points A,
Let the equation of the chord be
then, since the points
By
A,
/u,
are on
it,
cross multiplication
B
A hence
therefore the equation of the chord
-2)
-
is
(A+ M)
= 2AM -A-
//.
THE PARABOLA (ii)
To find
the equation
298
of the tangent at the point
A.
=A
This follows from that of the chord by putting p the tangent is
(iii)
To find
the locus
of the intersection of perpendicular tangents.
Suppose that the tangents equations are
(A
is
A and
at
1)
their point of intersection
^
A
=
/x
A
are perpendicular
(A
1)>
therefore given
by
The condition that the tangents should be perpendicular
fr-Dfra
1)
+ *J + Zr
2
x
it
+
"21 2
a 6
122
+
^r/>
(iv)
is
&
-^x
y
a
a
b
1
a6
cos w
(6
+ a cos to)
=
ab cos
w.
therefore the equation of the directrix.
To find
the equation
of the tangent
at the vertex.
If the tangent at A,
is parallel
to the directrix
we have A
1
Hence, substituting
_
_
= 0,
is
x(a + 1 cos w) -f y
This
M(^-iHMM-D coS(o = 2x
Hence
or the required locus
_
6-facosco
thus
;
for A in the equation of the ta ,
__
a + 6cosco
a2
'
is
;
their
THE PARABOLA
294 (v)
To find
the coordinates
of the focus.
The perpendiculars from the focus pn the coordinate axes the tangent at the vertex, since the axes are tangents. be the focus, the feet of these perpendiculars are
(+ Hence the equation
COS
7]
0),
(0,
0),
of the tangent
T]
ajt
lie
on
If (,
rj)
+ f COS O)). the vertex
is
t
-ff cos
17
Comparing
this with the equation already found in coso)
&
co
_
T]
-f
f cos co
(iv),
we
get
a&
__
+ acosco
whence immediately
(vi)
The
latus rectum is twice the perpendicular
from the focus
to
the directrix. .
4a 2
.
2 fe
sin 2
co
Its value is
(vii)
To find
?#-fmt/ +
the condition that
w
=
parabola.
Any
tangent to the parabola
L a
Comparing
is
of the form
(A-l)-f6 A =
A(A-1).
this with the given equation
A-l _ ""
A
6m
al
...
X=-^
a?;
,
_ ~" A(A-l) n
'
A-l -+=5.; om
equation of the parabola can be written
should touch the
THE PARABOLA
295
form
It is often referred to in the
Evidently for points in different positions on the curve, different The equation is true with signs for the radicals must be taken. for the the of curve between the points of contact positive signs part A and B. For the rest of the curve we must use
-
&-"
Ji-
The equation fx /y = ./-+./,
of the tangent at the point (/,
and
.
/
-,
on the parabola
/*~_i '~
have those signs which
+
V/ It should
1
W
fi+y *
aV */
yf)
10r>
can be put in the form
!
x.
where
N /!+V*-
satisfy
& VA^i
be noted, however, that
when
(#', y*)
does not
lie
on the
parabola, this equation is not that of the polar of (#', y*). 4 It is sometimes stated that (a cos4 0, b sin 0) can be used to this is only true, if 6 is real, for the denote a point on the curve portion between the points of contact, i. e. when x < a and y < b. The notation given above covers the whole curve and is to be :
preferred.
f
y-
Example
A
i.
tangent
to
tfie
/-- 4-
parabola
.
/
,-
=
1
meets the
axes of coordinates in P, Q, and perpendiculars are drawn from P, to the opposite ares : prove that the locus of tJieir points of intersection
-cos
+x y ?_
cos co ----------
co
j.
Let the tangent be -(X-l)-t/r points
Let
P(X,0),
P, QM
the point
(0,
cos
a
o
=
X
(X-l)
;
this
Q is
co.
meets the axes at the
be the perpendiculars on the axes of y and x\ then
a\coso>) and Af the point [&(l-A)coso>,
The equation of
PL
is
-^ + a\
-^- -
=
1
.
0].
2. is
THE PAKABOLA
296 The equation
of
QM
is
x
_JL_xv +'r/1
&(l-X)coso>
= .-.
X cos CD
=
r-f
j
_
i> A
fc(l-X) X) cos
(1
a)
;
their point of intersection lies on the straight line
x cos a) -f y
# + y cos o>
"
"T
a
Example
A
ii.
and on
tangents,
(i)
variable tangent to a given parabola meets two fixed
segment as diameter a circle is described : a conic touching the two fixed tangents in the
circles is
points where they are met
Take
COS
the intercepted
of the
the envelope
__ ~~~
i
b
the directrix of the given parabola.
~by
and
the fixed tangents for axes,
let
the parabola be
b
Any tangent -(X-l) points P(aX,
E (x,
If
y)
|x
ab
]
= X(X~ 1)
meets the fixed tangents at the
0), Q (0, b-b\). be any point on the circle described on
PQ
as diameter,
we
have, since &2 (1
- X) 2 -f 2ab\ (\- 1)
cos
a>,
=
0.
which reduces to ab coa
a>X
2
X [x
(b cos
w
a)
-I-
y
(b
ab
a cos
cos a>} a>) -f 2 2 cos -f 2 (a? -f
#y
y
Since X
w
an undetermined constant, the envelope of
is
bx cos a>)
fey
this circle
is
(Chap. VI, p. 253)
{x
(b cos
o>
a)
-f
y (5
a cos
a>) -f
-f
This
may
2
ab cos
4 a& cos
a>} a>
(#
2
+ y 2 -f 2#y cos
&>
for
cos
a>
&y)
0.
also be written
{x (a
-f
b cos
o>) -f
y
(Z> -f
a cos o>)
a& cos
2
a>}
=
4abxy sin
2
o,
which represents a conic (equation of second degree) touching the x = 0, y = 0, the chord of contact being
x(a + 6coso))-f y(6 + a cos a>) i.e.
=
the directrix (because the form
is
uv
=
a& cos a>
==
lines
0,
kw*).
Examples VII
g.
A
variable tangent to a parabola meets two fixed tangents at the points P, Q. Find the locus of the mid-point of PQ. 2. OA, OB are fixed tangents to a parabola and P any point on the curve. 1.
The harmonic conjugate of gent at
P in Q
:
OP with
find the locus of Q.
respect to
OA and OB meets
the tan-
THE PAKABOLA 3.
A variable
P and 4.
297
TA TB
tangent to a parabola meets two fixed tangents Find the locus of the centre of the circle TPQ.
Q.
Show
that the normal at X to the parabola (x/a
x (a\ + b\ - 1
cos
o>) 4-
y (a\ cos o>
-f
&X
H-
t
- 1) = 4xy/ab 2
y/b
2
+ (y-y') 2 + 2(ar--a;')(2/--t/')cosa>
where a + # =
is
,1)
=
5. Find the focus and directrix of the parabola (x/a + y/b -1)* by comparing the equation with
(#-a?')
at
=
(x cos
(4xy/ab)
a -f y cos - p) 2
,
a>.
N.B. Use the identity (# cos o>
2
-f
-f
y)
(y cos
a> 4a?)
2
- 2 (# cos a> -f
/)(/ cos
o) -f a?)
cos a
(#
A
+
o> 2
.t/
+ 2 #) sin 2
.
parabola touches OA, OB in -4 and B show that the portions of any chord, which has its middle point on AB, intercepted between OA, OB and the parabola are equal, 7. Parabolas are drawn which touch the axes Ox, Oy, inclined at an angle 6.
:
and whose directrices pass through a fixed point touch the line #/(& + & sec o) + y/(A?-ffc sec ) = 1.
o>,
(h, k)
:
show that they
all
AB, CD, two
8.
fixed segments of straight lines, are divided similarly at that and CD. prove PQ envelopes a parabola which touches Show that if a and b are variable and h/a + k/b = 1, the directrices
P and Q 9.
of */x]~a 10.
A
AB
:
+ \/y/b
=
1 pass through a fixed point. touches two given straight lines OA, parabola
and a variable tangent meets OA,
OB
at P, Q.
Show
OB at given points that the circle OPQ
passes through the focus. 11. If the chords of contact of parabolas touching two fixed lines are concurrent, their directrices are also concurrent. 12.
\/x
-f
13.
The
parallels
through the origin to the tangents from
(x', y')
to
\/y =
The
0. 4/c are the lines cxy + (x y) (xy' x'y) the of the to from equation tangents parabola (x', y')
+ (*y-xy') {(x'-x)fa-(y'-y)/b\ (x'-x}(y'-y)
A
=
is
0.
variable tangent to a parabola meets two fixed tangents, and is drawn touching the fixed tangents at these points prove that the envelope of its directrix is a third parabola touching lines drawn at right angles to the fixed tangents through their intersection, in 14.
another parabola
:
the points where they are met by the directrix of the given parabola.
Illustrative Examples. (i)
If
triangle
the tangents at
PTQ
lies
P and Q
meet at
T and
the orthocentre
on the parabola, show that either the orthocentre PQ is a normal to the parabola.
the vertex or the chord
Let
Pbe
of the
the point (aX 2 , 2aX), Q(ap*> 2afi), then Tis the point
is
at
THE PARABOLA
298
The equation of TP is \y-x-a\*
Q on
X(#-a/u )-f y-2a/u = 2
it is
Xo?
+y
^itt
Solving
(i)
TPQ,
triangle
(ii)
rce
have
x Since this point
lies
(0, 0),
=
3
+ 2aj*.
(i)
+ 2aX.
(ii)
is 2
a/nX
to obtain the coordinates of the orthocentre of the
-aX/i-2a, y = a on the parabola,
a 2 (X
Hence
and that of the perpendicular from
0,
= aX/A
Pon TQ
So the perpendicular from
and
0,
+y
=
i.e.
+ /i) 2
(X/x
=
+ 2) 2
(X
-f /i) (X/x 4- 2).
- 4 a2
(X/i -f 2).
which case the orthocentre
either X/i-f-2^0, in
is
the vertex
or 2
(X+ M ) (X,x-f2) a (X + X/x + 2)(/
i.e.
X 2 -fXfi
Hence
+2
= -4,
(i)
^--0 or
Therefore Q is the point where the normal at P cuts the parabola, or the point where the normal at Q cuts the parabola.
Prove that the tangents of the angles at which y 2 ax are given by parabola y
(ii)
the
= mx + n
P is cuts
=
tan 2 0(w +
2am + aw 3)-f 2tan 6 (a
and deduce conditions
Une be
that the
mn) + nt(nm a tangent,
(i)
(ii)
a)
=
0,
a normal
to the
parabola.
Suppose that the
line
the condition for this
y
= w#
4-
n cuts the perabola
at
2 (<7/
,
2aO
:
is
mat*-2at + n
The tangent Hence
at the point
Hs #y =
x -f
a*
=
0.
(i)
2 .
-1
tw A
tan ^
-
Hence Substituting this value in
(ii)
/m
=
t -f
1
,., x
m
(ji) v }
we have
(i)
m tan 0) (tan - m) n (tan Q - w) a tan*0 (am 3 + 2 am 4 n) -f 2 tan 0(a ~ww) 4 m (nw -a) = 0.
w?a (1
In
m
1-fmtan^ tan 6 - w
TT
Thus
-
^
=
/i
+ m tan
6)*
+ 2a
(1
-f
-f
we could equally well take the supplementary 1
*^ f
which gives the alternative
sign.
angle,
i.e.
THE PARABOLA (i)
mn =
If the line touches the curve, both values of 6
must be
zero,
hence
a.
If the line is a normal,
(ii)
one value of tan 6
.'.
299
one value of
is infinite
and
90.
is '.
=
0.
Miscellaneous Examples, VII. 1. Prove that the orthocentres of the Wangles formed by three tangents and by the corresponding three normals are equidistant from the axis of
the parabola. 2. Prove that
2
if a >85 a point can be found the two tangents from which to t/ 2 = 4arr are normals to x* = 46y. 3. Find the equation of the common tangent to the parabolas represented 2 by f/ == ax and a? = 4%. 4. A system of chords is drawn so that their projections on a line inclined at an angle a to the axis of a parabola are of constant c 2
length
that the locus of their middle points a
(y
- 4 ax)
(y cos
a
-f
is
:
prove
th( cuive
2 a sin a) 2
-f
a'c*
=
0.
Prove that the locus of the intersections of the tangents at the points 2 {aBinh ^^/!), 2asinh(a fl)}, where Ot is variable and ft constant, is 5.
a parabola having the same focus as y 2 = \ax. a 6. A is the vertex of i/ = 4f/*r P is any point on it, and the circle on AP as diameter meets the parabola again in Q and 7?. Show that the normals to the parabola at P, Q, R nnet at a point on the parabola 7.
The normal
at
P
parabola again in Q, R. meet at P.
R
8.
If
P
is
meets the axis at G\ the circle APG cuts the Show that the normals to the parabola at Q and
such that when
parabola in Q, a parabola.
R
PQR
the rectangle
is
drawn in a
PQ PR .
is
fixed direction to
meet the
constant, the locus of
P
is
P
9. Two parabolas touch at and intersect at Q, R. Prove that PQ, PR are harmonically conjugate to the diameters of the two curves at P. 10. Prove that, if the normal at meeta the curve again in Q, and if the circle on PQ as diameter ciita tne curve in 7?, the locus of the middle point
P
a a 4 is the curve f/ (i/ ~4oa:) + 64a = 0. The normal at P to a parabola, whose vertex is -4, meets the curve again in Q: show that the locus of the centre of the circle circumscribed to
of
QR
11.
APQ 12.
is
a parabola.
Through any point not on the axis of a parabola the two straight drawn which are conjugate with regard to the parabola and
lines are
perpendicular to one another. equidistant from the focus.
Prove that they meet
tlte
axis in
two points
THE PARABOLA
300 13. If its
drawn to pass tP rou gh the vertex of a parabola and to have on a fixed diameter of ie parabola, show that the orthocentre of
a circle
centre
is
t'.
the triangle formed by the tanf? ei*ts to the parabola at the other three xed. points where the circle cuts it is f
=
8 ^ and x* = ay, and prove that they cut each other at rig 1 * angles at the origin and at an angle whose tangent is -6 at their other P oint of intersection. 15. Find the locus of a point su'k that the angles between the connectors of the vertex of the parabola y2== * ax witn the points of contact of the 14.
Show
fabolas diagram the pa
in a
3
!/
tangents from the point may havr a given pair of bisectors. * 2 touch at two points, 16. Prove that the curves y* = ^ x + y -3#-fl = and find the equations of their c o mmon tangents. Show also that each of these curves touches in the sanV two points any curve whose equation 2 2 2 = pr all values ofX. is # + t/ -3# +
l-fX(y ~;c) Find the equations of the circles which touch the directrix of the 2 a parabola t/ = 4#, and pass thro &h the points of intersection with the 17.
parabola of the straight line y
=*-!
2 Normals are drawn to th< parabola y = lax to touch the circle t ~ R t the radius of the circle the values of r c y + y* Find for differ* (x
18.
t
locus of their points of contact. on a parabola meet at a point 19. If the normals at three poirts P, Q, whose abscissa is x, prove that t*16 centroid of the triangle PQE is on the
R
axis at
drawn 21.
a distance from the vertex equal to %(x-2a). ;
Show
20.
that the locus of the ntersections of equal chords of a parabola
in fixed directions
Tangents
of the circle
is
a straight
line.
TP TQ are drawn to the parabola y2 = lax; t
find the equation
TPQ.
whose diameter is a chord of the ^ n a ^ point on the axis of #, and show xe( lax passing throuf parabola y nt the that for one position of the po envelope reduces to a circle and 22.
Find the envelope of the 2
a straight
circle
==
line.
Three normals of which ttf lengths are w t w a ;? 3 and two tangents of which the lengths are * lf t^ are drawn from the same point to a parabola with parameter 4 a. Show that i\ n i n * = ^iV 24. Find the equation of that rectangular pair of conjugate lines with 2 whose intersection is the point (h, k). regard to the parabola y = 4ax C is a point on the latus rectu111 anc^ P a point not on the latus rectum such that PC is equally inclined to the rectangular conjugate lines which intersect at P. Prove that the l/ cus of P is a circle. 23.
,
,
to the parabola i/ 2 = lax from a point Tangents OP, OQ are draw? -*<*s show that the envelope of the circle lying on the straight line x = OPQ is the curve if(a + x) = a? (! + ) (5-a:). on tl* = *ax, and radius 2SP cuts the 26. A circle centre P, a
25.
P
diameter through 2
poinl in Q, Q' : sfr>
# -f4a#-f 8a 27.
Chords of
2 t/
2
=
w
that the loci of
and 3y
a
-4fl.r
Q and
+ 8a
lax pass ttf ou g n a fixed P oin t
2
= 0.
(
0)
Q' arc
:
show that the
THE PARABOLA locus of the orthocentre of the triangle formed by the tangents at its extremities is
2x* + 4ax-20(x-py 28. If p l9 p a are the
radii of curvature at
a parabola from a point
=
301 any such chord and
0.
the feet of the normals to
P
on the curve, show p5 pa pp where p and p P and at the vertex. 29. Tangents TP, TQ are drawn to a parabola such that PQ is the normal at P. Show that the area of the triangle TPQ is 4aa sec 3 0eosec 3 0, where B is the acute angle which the normal makes with the axis. 30. Prove that a circle whose diameter is a chord of a parabola such that the distance between the diameters through its extremities is double the latus rectum will touch the parabola.
are the radii of curvature at
31.
The
circle of curvature at (a/ 2
2at) cuts the parabola again at the 1 3 4 angle tan" {8* /(3* 6*2-1)}. 32. If the normal at P makes an acute angle ty with the axis, and the normals at Q, E each make acute angles i^-^ with the axis, the three normals form a triangle of area 2 (tan 8 cot^). ,
\//>
Through any point on a given line through the focus three normals are drawn to the parabola show that the sum of the angles they make with 33.
:
any fixed direction is constant. 2 34. If the normals to t/ = 4 ax at the points
(x, y), (x\ #'), (#", y") form an 2 2 2 2 2 2 6 0. equilateral triangle, prove that (3y -4a ) (3?/' -4a ) (3y" -4 )-f 64a 35. Four points on a parabola are coney clic and the orthocentre of the
=
show that triangle formed by three of the points is joined to the fourth the mid-point of the joining line is the same whichever three points are chosen. Also the line joining this mid-point to the centre of the circle :
i*
bisected
by the
axis,
and the length
of its projection
on the axis
is
the
latus rectum. 36. jVj, NI,
#3 are the lengths of the three normals drawn from a given and n l9 2 3 are the lengths intercepted between the
point to a parabola,
curve and the axis
N^ AV'h n a 37.
Show
4-
,
prove that, with the usual convention as to signs, r n v + 2 ( NJn 1 + 2V2 /n 2 4- 3 /n 3 ) 4-3=0. JV2 A 3 /n a n s + z NJn z :
^
N
that the area of the triangle formed by #cosCX-f ysin OLp = at its extremities to y2 4ax is 4a'^ (a tan 3 a -f jpsecft)^.
=
and the tangents 38.
Show that
39.
A
the length of the normal (other than the radius of curvature p) drawn from the centre of curvature to the parabola is of length
chord
tangents at
with the 40.
PQ
a{3-(,,/2a)*} {4 ( P /2a)* -3}*. makes acute angles a and show that it makes an acute angle
of a parabola
P and Q
:
with the
axis.
Show
that the locus of the poles of the axis of a parabola with respect
to its circles of curvature
is
>-ax + a*)*.
THE PARABOLA
802
is the centre of curvaThe normal at P meets the axis in and show that no other normal intersects PG at a distance from G on the side opposite to which is >^ GO.
41.
,
ture at Pi
42. In a parabola the normal at P meets the curve again at Q. If R H a point midway between the centres of curvature at P and Q prove that R'a distance from the tangent at the vertex is least when P's distance from y
a
it is
\/2.
Show
43.
ratio X
:
1,
that chords of y* = 4a#, which are divided by (x\ y) in the have for their equation
44. P6r, the
Q
to
so that
normal
at
P to
a parabola, cuts the axis in G y and is produced that the other normals passing through Q
GQ = \PG: show
intersect at right angles. 45. Through a fixed point
on the polar of ('^a, 4 a), with respect to a parabola y3 = 4rt#, a chord of the parabola is drawn. Prove that its length will be either a maximum or a minimum when it is inclined at an angle \ir to the axis of #.
a
Is it
maximum
or
minimum ?
rectum of the parabola which touches the four common tangents of two circles whose radii are a, 6, and the distance between whose centres is c, is 2(a2 -fc2 )/c. Find the locus of 47. The normal at P is produced outwards to K. 46. Prove that the latus
K
when
(i)
48.
lines 49.
PK
=
PG
(ii)
;
when
PK =
the radius of curvature at P.
Find ths equation of the parabola which touches the four straight
x/ay/b = 1, x/a' y/b = 1. A chord of a parabola is drawn
as diameter a circle
parallel to a fixed direction
and on
it
Prove that the polar of the vertex with circle fixed parabola. this another to envelopes respect 50. Find the locus of the foot of the perpendiculars from (h k) to is
described.
t
tangents to a parabola, and show that it lies inside an infinite strip perpendicular to the axis of width equal to the focal distance of (h, k). 2 51. Show that five common normals can be drawn to y =4aa? and 2 a?
4 fry, and that
if
they are inclined at angles
lt
a
,
#8
4
,
of either parabola then
tan (0,
+
2
+
3
-f04
-f
B)
=-
tan^
.
tantf,
,
tan08 tan .
4
.
,
6
to the axis
CHAPTER
VIII
CENTKAL CONICS THE ELLIPSE AND THE HYPERBOLA 1. We have seen ia Chapter VI that the equations of the ellipse and hyperbola, when referred to their axes of symmetry, take the forms x */a 2 + y V6 2 = 1 and x 1 /a?y*/b'z = 1. The ellipse makes intercepts 2 a, 26 on the coordinate axes, and these are the lengths
of the axes of the conic.
It is conventional to take the
along the axis of x, so that
we have a >
The hyperbola does not meet the are
intersections
b\/
(0,
It
1).
the
imaginary '
the length of the other or either greater or less than a. Equilateral or Rectangular.
The
central
=
2
axis of y in real points whose coordinates
;
points
common
however,
is,
major axis
6.
to find
6 referred
conjugate' axis; evidently 6 When a = 6, the hyperbola
its
are
to as
may
be
is called
conies can, in many particulars, be conveniently studied together, and in this chapter we shall use the equation 2 2 1 to represent a central conic ; for an ellipse a Oix^ + Py I/a
=
=
/3
1/6 2.
,
an hyperbola a
for
To find
the foci
and
=
directrices
a focus, and x cos directrix, then the equations If
(x'j y')
I/a-,
is
=
of a central
0+y sin
conic.
=
p
the corresponding
are identical.
Comparing
coefficients
-
we have
l-e2-cos2 2 e sin 6
Since (a)
If
e is
l-e*s ,
cos
=
not zero,
= 0,
then
#'
0,
or
is
a/
=e
= e*p cos TT.
2 jp,
'
0,
y'
=
0,
and
,
1/6'-'.
y
= e*p sin
0.
CENTRAL CONICS
304 Hence, since a
= d-a
=+
we have p
,
- and
yf
=
C
ae,
where
,-1-3. The
focus-directrix forms are
The
foci
and
ae,
(
then
and the corresponding directrices are = 0, where, for the ellipse 0), x + a/e
= (a2 + &2)/a2 then tf = 0, y' =
for the hyperbola e 2
=
(b) If
IT,
=
Hence p For the and p.
I
.
l/e
ellipse
/3/a
y'
;
=a
= 2
2
/6
For the hyperbola, since In this case
ft
(ae, 0),
&=
In both cases 2
(a
xa/c = & 2 )/a
2 ,
0,
and
e is real.
and
e ^,
e/
2
;
and
E
>
1,
is
negative, y'
.'.
e is
2
=
1
I
imaginary, and also y'
and p are imaginary.
e is real.
In any central
conic, if
e,
e'
are the
two
eccentricities,
we have
Examples. (See Figures below.)
Prove that (1)
CS.CX= CA\
(2)
AS. A' 8= CB\
(4)
SX:CX = CB = (M
(5)
CS=e*.CX.
(3)
2
3.
<2
^<
It is convenient to use definite letters to indicate the prin-
cipal points of central conies: the following should therefore be noted.
THE ELLIPSE AND THE HYPERBOLA Foci
S,
S'
;
XM, X'M'
directrices
EPR'
';
normal
vertices A, A', B. B'
305 ;
centre C;
perpendiculars from the foci and centre on the tangent, ST, S'Y', CK; ordinates of P to and Pn, the latter meeting the directrices in Jf, M'. the axes,
tangent at P,
;
at P,
PGy
;
PN
We have already proved that CA = CA' = a GB = OB' = 6 CS = OS' = ae CX = OX' = - and from the focus-directrix form ;
;
;
c
of the equation of the conic
If P(x'y')
SP = ePM, S'P = ePM
any point on the
is
;
'.
conic,
Hyperbola.
Ellipse.
SP=ePM=e(CX-CN)
SP = ePH = e(CN-CX)
= ft/
=a 7
+ CN)
S'P = ePJf
a,
= (CW+ X'C)
= co/ a. S'P-SP = 2a. -f
Hence Note.
If
P
lies
SP *=a-ex' and An ellipse can
on the left-hand branch of the hyperbola we have in which case SP-&P = 2a.
ST = -a-/,
therefore be described as the locus of points the sum of whose distances from two fixed points is constant, and an hyperbola as the locus of points the difference of whose distances from two fixed points is constant. 1207
CENTRAL CONICS
306 4.
The equation of the
I.
(Chap. VI,
is
(x',
y'\ is
4).
The
II.
chord, whose mid-point
tangent.
The equation of the tangent at the point (x', y') is axx' + fiyy' = 1. (a) To find the condition that the straight line Ix+my + n
=
should
touch the conic.
Suppose this straight line touches the conic at the point (x', y') then the equations lx + my + and ocxx' + /3yy' 1 = are 9
n~Q
Hence #' = I/not and y' = m/n/3, but since (#', i/) on the conic we have a#' 2 -f /3#' 2 = 1. Thus, substituting for a?' and #', we have the required condition
identical. lies
For the
a 2 Z 2 4-& 2 w2
ellipse this is
= n2
and
,
for the
hyperbola
from the centre on a tangent makes an angle the equation of the tangent is of the form 0. Apply the above condition that this line
If the perpendicular
6 with
the
jr-axis,
xcosO + ysmOp
=
should be a tangent, and
we have for the ellipse p* = a 2 cos 2
and
2
for the hyperbola
p
=a
2
cos
2
2
b sin
Also x cos
-f
=
y sin
2
\/a cos
-f
& 2 sin 2 0,
-f
ft
0.
=
+ y sin
Thus the
x cos straight line touches the ellipse for all values of
2
A/ a 2 cos 2
2
sin 2
0.
2
2 ?)
sin 2
touches the hyperbola.
Note. This form of the equation of a tangent is called the pedal equation the pedal of the ellipse with respect to any point (h, k) is r
the point
2
Va
2 cos B
-f
2 ft
2 sin 6
;
- h cos S - k sin 0,
&) being the pole and a line parallel to the o>axis being the
(ft,
initial line.
To find the locus of the foot of the perpendicular SY e. the pedal of the ellipse focus S on a tangent to an ellipse,
Example from
the
i.
i.
with respect
to the focus.
The equation of a tangent ia x cos 6 + y sin 6 If
of
S
Hence
cosa
SY =
r
2 a
aV = a
a
-fc 2, this
J -I- 2 acr cos & + a 2 * 2
is
the circle on
AA'
0,
a
i,^a cos3 B -f 5 8 sin 8 0, becomes
(r-f ae cos 0)
t
This
+ 6s sin 2
initial line, the polar coordinates
0.
therefore
and since
__
a
and SA as the
is taken as the pole
Y are SY and
V0
as diameter
;
it is
a8
.
called the Auxiliary Circle.
THE ELLIPSE AND THE HYPERBOLA Example to the tfiat
If p
ii.
P
tangent at
for an
=
-o
ellipse
2a --
)
P be
Let
P lies on
according as the point
y*)
(a?',
conic,
and for an hyperbola yr
1,
r
p* 1
of the perpendicular from the focus S and r is the length of SP, shotc
is the length
of a central
-,
=
l-f
p*
,
or
r
near or far branch of the curve.
the
the equation of the tangent at
;
307
=
and we have the condition aa?' a -f /3t/' a 2 Now, since e = 1 a/ft we have
P is
1.
Thus
/ 1
,(!- age*')'
hence For the
For the hyperbola, far
branch r
=
J- =
a
P^r
a
=
ellipse,
a
ft
l
r
=
and r
2
=-
ex'.
?:
ex
a
and
ex'\*
-^;
a-ex'
= JjL. _ ! ex'
(t
a-ex'. for the near
branch r
=
ex'
-a, for the
This gives the required results.
Examples. 1.
2.
Prove that SY.S'Y' = OB2 If the tangent at P meet the major axis at .
CN CT = CU
show that
Cn
2
.
,
.
Cfc
=
CJ5 2
T and
the minor axis at
t,
.
3. If T is any point of a tangent to a central conic at the point P, and TM, TN are drawn perpendicular to P and the directrix corresponding to S, show that SM = eTN. Hence prove tha.t the tangents from any point to a conic subtend equal
angles at a focus. 4. Prove that Z 5.
Show
III.
that
PSR - / PS'R' = CY = CT = CA.
The Normal.
The equation Note.
\w.
If the
of the
normal at
we have $y'(h-x hence the point
normal
f
)
(a?',
lies
(a
9
(#', ;/)
to the conic otx2
+ fiy* =
passes through a given point
(#', y')
= Otx'(k-y') i/)
at
(h, k),
or
on the rectangular hyperbola (Chap. VI,
-) a?y -
te + /3Ay U 2 Of
= 0.
1
6j
CENTRAL CONICS
308
Since, in general, two curves of the second degree intersect in four points, four normals can be drawn from any point (h, k) to a central conic and their feet lie on the above rectangular hyperbola.
Examples. Show
1.
that the pedal equation of the normal to an ellipse
x
cos 6
+ y sin
(9
=
(a
2
-
2 fe
)
2.
Prove that
SG =
PG CK = CJ5 Pg CK <= NG CN = C& CA\
3.
Prove that
4.
Show
5.
Prove that
6.
Show
7.
Prove that
that
2
.
SP:S'P= SY: &Y' - SG ',
The
P,
CG =
*
.
+ fiy'y" ~
.
S'P.
of
CK
lines. y')
(#',
with respect to the conic
1.
passes through the point
(#', i/')
The symmetry
1.
.
ffG, and hence show that
:
# lie on a circle. 6W, and a*.GN=tf.
+ ftyy' =
1 is Qiocx'
Hence the polar
PG Pg = SP
,
2
of the point
polar
=
CA*;
.
;
:
;
that S,
ax* + py* OLX'X*
SP.
.
Conjugate points and
TV. (a)
e
is
sin & cos Q^tfcotfQ-t a 8 sin 2 B.
of this result
x/
(#",
shows that
!/
if
)
if
the
X
7
passes through (x' T/' ), then the polar of (x", y") it is therefore the condition that these points (x', y'} passes through
polar of
(#',
/')
,
;
should be conjugate. (b)
conic
To find the pole of &x* + dy 2 = 1.
the line
Let the point (#', y') be the and ocxx' 4- fiyy' Ix -f my + n = tf
~
--
other) if
-n'
=
-
-f
y'
=
1
with respect to the
then evidently the equations must be -identical. Hence
pole,
The
-r--.
are conjugate
^-
+ my + n =
=
AM
7
and
lx
nn'
(i.
straight lines
Zo;
+ m^ + n
=
0,
the pole of each lies on the
e.
= 0.
If one of the straight lines
is
a diameter Ix + my
=
=
0, since
n=0
it
=
are confollows that the straight lines Ix + my 0, Vx+m'y + n' have. the all ri Hence of may poles straight jugate whatever value lie on the diameter Ix + my if lines parallel to l'x + m'y
=
=
V. Conjugate Diameters.
Two
diameters are conjugate if each bisects all chords parallel to 2. I, we the other ; referring to the condition found in Chap. VI, are conjugate see that the two diameters Ix+my 0, I'x+m'y
=
=
THE ELLIPSE AND THE HYPEEBOLA if ll'/QL -f
mm'/fi
=
This
0.
is
309
also the condition that the poles of lines
parallel to the one should lie on the other (see IV above). therefore define conjugate diameters either as in Chapter
We VI
can
or by
this polar property.
In
y = u', y = m'jc are conjugate for and are conjugate for the hyperbola if
particular, the diameter^
the ellipse
Note
i.
=
mm'
if
IP, 'a?,
Since a diametei bisects
clioids parallel to its conjugate, the
all
is its point of middle point of the choid whose equation is Lc + my + n = intersection with the diameter mO(x lfty = Note ii. A convenient general form for a pair of conjugate diameters of the ellipse or/a 2 -f t/ 2/& 2 1 is ay \bx~$, \ay + bx 0.
DD
Suppose that these diameters meet the ellipse at PP', P and P' we have
respectively
;
then for the abscissae of
#2 Hence,
if P.
2
(1-I-X
)
P' aie the points (x l1
=
/y t ),
CP - CP - x* y? = 2
and if
D,
2
diameters
Now
if
is
2 Z>
-f
)/(1
X
2 1
.
rr are the points (,r2> y 2 ), ( -//jN 2 2 2 2 2 2 2 6 /(l -f X ), X
=
2
2
2
+ X2
#1),
/)'
CD - CD' = CP
x lt
(a
-
and Hence
.
(
/2
-\
Similarly,
2 ft
* CD"
=
a2
-f
62
i.e.
,
2
(X
-I
sum
the
2 2 6 )/(l 4 X ).
of the squares of two conjugate
constant.
CP = CD, we
have
(\
a* 4 2
2
-l)(rt
-&
X2 62 2 )
=
The equations of these diameters
0;
X'-V .'.
+
2 fc ,
X
i.e.
=
+l.
are therefore
x/ayjb
$\ they are
called the equi-conjugate diameters.
Note iii. A pair of conjugate diameters of the hyperbola .r2 /a 2 may be represented by ay \b.c = 0, \ay-bx = 0. as in Note ii, we find Proceeding 2 1 -X ). CP2 - (a 2 -f X 2 ^)/(l -X 2 CD2 - - (\V- +
2
2 //
//>
=
1
2
ft
),
Hence sign,
but evidently CP CD - a either CP 01 CD is imagi-
CP
hence
2
2
-f
2
-
2
fc
2
;
Only one of two conjugate diameters meets the hyperbola in real nary.
and
),-
(
CD
2
are of diffeient
J*
points. If lengths CD 1} CD/ are taken on the diameters which meet the conic
in imaginary points so that
then CDj is often called the length of the semi-diameter conjugate to CP. In this case CP2 ^CD 1 2 = a 2 ^6 2 or, the difference of the conjugate diameters of the hyperbola is constant. ,
square-,
on
CENTRAL CONICS
310
Note on
CP
p. 309),
parallel to
1
Fis the mid-point of TT.
DD'
In particular,
Note
CD are two conjugate diameters, the polar of any point CD and to the tangent at P. Thus, in the Figure (see
If CP,
iv.
is
is
To find
v.
the polar of the
the equation
'
'
point at infinity
on CP.
of a central conic referred
to
a pair of
conjugate diameters as axes of coordinates.
PCP\ DCD' be any pair of conjugate diameters, and Q any point on the conic referred to these diameters as coordinate axes. (or, y) If QQi QQ* al*e drawn parallel to CD and CP, they are bisected by CP and CD further, if Q Q z is a chord parallel to CP, it is bisected by CD'; hence Q 1 Q2 is both parallel and equal to QQ 3 therefore Q2 Q 3 is Let
i
;
l
,
also parallel
Thus, (
#,
If
if t/),
CP =
and equal
the point (x,
',
CD
e.
\>
l
Hence #2 # 3
.
lies
(x, y)
i.
;/),
QQ
to
is
on the conic,
also bisected so
the equation of the conic
the equation of the ellipse
is
by CP'.
do the points (-#, of the form
y),
is
a*/a'* + y*/b'***l; for the
hyperbola
CD
is
imaginary, and the equation
is
a*/a'*~y*/V*=l.
we
It follows that all the results
find for the equation of a central conic
referred to its principal axes are true also for the conic referred to
of conjugate diameters except
when our
results
any pair depend upon the axes being
rectangular.
Examples. 1.
2.
3. 4.
Prove that
CF. CQ = CP2 CP CF - CZ>2 CP2
(Fig., p. 309;
Show that TF Any pair of orthogonal 2
2
:
Prove that a
ae mi-diameter
.
PG =
1>
.
2
.
:
through a focus are conjugate. CD, and b Py = a CD, where CD
lines .
conjugate to CP.
.
.
is
the
THE ELLIPSE AND THE HYPERBOLA Pair of tangents.
VI.
The equation of a x*
811
+ p,f=
1
r
pair of tangents from
P(x 2 2 (ac + /i// -l)(o(T' 2 + ^// 2 -l) =
is
y') to
>
the conic
(aa;
These are parallel to the pair of diameters f
i.e.
oc (ft
y
2
1)
If the tangents,
an angle
0,
2 a,3 x'y'xy
-f /3
(^ ,/
2
(J
^ [a^~2 / / -^^^^!)]^/ 2 --!]} *./ 2
2
~~
"~7
,
a right angle
is
Hence the locus
"-".
11
/ 1>
of point
1
./',
the tangents from which to the conic
1
+
fl
The
=
,,
1
+
/r
called the director circle.
For the ,r
i
the circle
is
,.
2
it
if
^+^ = 5 a
) ?/
~
(/
are at right angles,
is
1
and therefore the parallel diameters, include
I
which
= 0.
2
2
'hen
tan
The angle
x2
ellipse this
a
2
-6
latter
=
a2
when a >
h.
becomes x* + y*
+ 6 a and ,
for the hyperbola
2 .
is
real
only
When
the hyperbola is the asymp-
rectangular the circle reduces to a point at the centre totes are the only real orthogonal tangents.
:
Note. The equation
represents (vide
a locus passing through
p. 284)
the
points
of
infer-
^
section of the ellipse,
and
its
directrices x*
2
=
0.
=
aa
This equation reduces
t0
It follows that
the directrices.
n "
2
i.e.
a?
the
-f
y
common
a2
-f
6
2 ,
since
2 r(
e
2
chords of the ellipse and
I*.
its
director circle are
CENTRAL CONICS
312
Illustrative
The angles
(i)
Examples.
normals from
ivhich the
x*/a* + if/b
make
(/, g) to the ellipse
-l =
are given ly
tvith the x-axis
- g cos 0) 2 (a2 cos 2 6 + 6 2 sin 2 0)
(/sin
The equation of the normal
makes an angle
=
=
- 6 2 2 sin 2
2
(a
at the point (x'y')
fi(*-O anil if this
2
)
cos 2
0.
is
(y-/),
with the #-axis,
that 6 2 sin 6
a? cos S
normal passes through the point (/
If the
But since
we have
(/-*)
=
*, a/
(/-#)sin0
=
(#-y )cos0.
or
-).
on the curve, we have
(^'//O lies
g),
(ii)
(i)
1 2
62 sin
cos d
Substituting in
(ii) .
ya for x'
2
2
cos 04- 5 sin*
i.
2
(a cos
e.
(ii)
x'/aP+y
+
Ma
57/020 2
2
2
/!)
tangents are
=
A
2 (av
A
x/aa'
tangents
?/)
?5
yi]/W
formed
is
^ sin3
l)y
(a?
-
12 ) 2 cos 8
sin 2 0.
point of intersection of the
(e
x 2/a' 2 + y2 /l"2
1,
-f
-
2 -0 cos 0) =
sin 2 0) (/sin
if (,
the parallelograms
common
2 2>
2
and y\ we get
/sm 0~g cos 0= .
yV cos
2
=
=
1
their
W
1,
equations of
their
ellipses
common
and ^e product of the areas of four common point* and their four ;
Saa'W.
The common points of
S+g-i-*
+
-'-
are given by
hence we gather that the points are symmetrically placed with respect to the principal axes: let their coordinates be (g, ?), (, -rj), ( - ,-'?), (-, ?). The area of the parallelogram they form is 4 77.
THE ELLIPSE AND THE HYPERBOLA If lx
-f
my =
1
is
313
a tangent to both conies,
-l =
a /8 P + & /f w*-l
0,
=0.
Hence b'
Thus, combining
(i),
P " r-& _
which
2
m* -a' 2
_ ""
1 '
a 2 6' 2 -a' 2 fc2
is satisfied
by (,
?),
and
(ii),
1
m Hence the common tangents are
The corners
of the parallelogram formed by these tangents are
('/*, 0), (-oa'/g.O), (0,&ft'M (0, -to'M and consequently the area of the parallelogram they form is 2aa'W/rj, The product of the areas of the two parallelograms in question therefore
The tangents from any point
(iii)
to the
is
Saa'W. to
a central conic are equally inclined
focal distances of the point.
Let the point be y' (x
(#',
ae)
then the equations of the focal distances are
:
/')
y (x
ae)
=
0,
y (x -f ae)
=
0,
xy'-y (x'
- y (x
f
-f
ae)
=
;
these are parallel to xy' i.
e.
-y(x' -ae} f
to the lineb
(xy
W
or
(xy'
2
yx')
y
1
a? e*
- yxj - tf (ft - a)
-\
= -
ae)
=
0;
0,
0,
+ Ot-fyy 1 = 0. y'lxt-Z&px'y'xy + The bisectors of the angles between these lines are the same as those of the angles between the lines or
^xt
otp
which, we showed above, are parallel to the tangents from
(#', y'}
to the
conic.
Hence the bisectors of the angles between the tangents from a point to the conic and of the angles between the focal distances of the point are the same this establishes the proposition. :
Examples Villa.
A circle on a diameter PP' of an ellipse as diameter meets the tangent an end of the minor axis in Q and Q'. Show that QQ' is equal to -the difference of the distances of P from the two foci. 2. Find the equation of the pair of tangents from the point ((X, 0) to the 1.
at
hyperbola x*/a*-tj*/b* = For the case in which
1.
a
of each tangent separately.
*
5,
/3
=
3,
a
=
1,
b
-
l/v'S, find the equation
CENTRAL CONICS
314
on the ellipse /^M a^t/'^a^b2 and A, A' the k (IP a^jZaW. major axis, show that coiAPA' 4. Prove that if (.r, //) and (#', y') are any two points on the ellipse 2 2 2 Deduce a?/a* y*/b = 1, then Vx/(y -f y'} + a*y/(x -f a:') = a 6 /(;n/' -f ar'y). 3.
If (h, k)
is
extremities of
P
a point
its
-l-
that the locus of the middle points of parallel chords is a straight line. 5. Prove that the bisectors of the angles between the focal distances of
a point
P on
an
tangent and normal at P. which the ellipse subtends an angle 60
ellipse are the
6.
The
7,
Prove that the rectangle under the perpendiculars drawn to a normal centre and the pole of the normal is equal to the rectangle
at
locus of points at
is
P from the
under the focal distances of P. 8. Tangents are drawn to the
(,
ellipse
b)
from the point
show that the intercept made by them on the ordinate through the neaicr focus is equal to the major axis. 9. A rod AB of length / moves with its extremities on two fixed lines which intersect each other at right angles. If P be the point which divides AB in the ratio 2 to 3, show that the locus of P is an ellipse, and state its eccentricity.
Find the points on AB which describe ellipses whose eccentricity 10. Find the coordinates of the intersections of
x cos ft/a -y sin &/b
=
cos 2
a with
2
1
jc^/a
-\
2
-y /b
=
is
^.
1.
Find also the locus of the projection of the centre of the ellipse on the above line. 11. Find the locus of the point of intersection of tangents to an ellipse which meet at a given angle OK. Pairs of tangents to an ellipse intersect at right angles prove that their chords of contact touch a fixed concentric ellipse. 12. Show that the points in which the straight line # cos a 4 //sin a = 2 ;
= 4 subtend a right angle at the centre of the hyperbola. 13. Find the coordinates of the foci and the length of the latus rectum X 2 , where X is positive. of the conic Xo?2 4-(l + X)y2 meets the hyperbola 2#2
y
2
=
Find also the locus of the extremities of the latera recta as X varies. 14. Show that the locus of the middle points of chords of the ellipse 2 2 a- # -ffc~ 2 f/2 = 1, the tangents at the ends of which intersect on the circle
2 2 2 2 2 2 (a- ^ -f &~V) = a- (# -f ?/ )chords of an ellipse whose middle points are on the same straight line touch a parabola. 2 2 2 2 16. Tangents are drawn to the ellipse # /a + t/ /fc = 1 from any point on
<&
2
-f t,
=a
2
,
is
15. All the
the circle
x2 -f y2
= a'
2
2
-f
fc
.
Prove that (i) (ii)
the tangents are at right angles the locus of the middle points of the chord of contact
the equation
;
2 (o;
+ y2 )
-
2
(a -ft
2 )
(a* /a*
+ ?/V&2 )2
.
is
given by
THE ELLIPSE AND THE HYPEKBOLA 17.
315
Find the coordinates of the extremities of the diameter of the = 1 which is conjugate to the diameters y'x = x'y.
ellipse
#2/a 2 -f i/ftf
Two
such conjugate diameters are- inclined at angles 6 and to the a'b' are connected by 2 the relation a' 2 sin 2 6 b' sin 2 4> = 0. <
major axis of the ellipse: show that their lengths -t
18.
If S,
H are the foci of an ellipse and P is any point on the curve, show
that the locus of the centre of the inscribed circle of the triangle ellipse x*
+ (\ + e)y*/(l-e)
SPH is the
= aV.
Chords of an ellipse which subtend a right angle at the centre are 2 distant afc/VV-f 6 from the centre. 20. Find the equation to the locus of the foot of the perpendicular drawn to a tangent from one of the foci of 00? + by* = 1. 21. Express the length of the perpendicular from the centre on the 19.
normal
to
an
ellipse in
terms of the perpendicular on the corresponding
tangent.
Show
that the area of the rectangle formed by two parallel tangents and is never greater than half the square on the line
the corresponding normals joining the foci.
=* BD BE of the ellipse a? /a* + .r-axis to meet the ellipse again in D and E. 2
22. Chords to the
t/
9
2
/fc
the centre of the circle inscribed to the triangle 23. If PQ is normal to the conies
at
P and
ar ^ drawn at angles rr/3 Find the coordinates o p
BDE.
Q,
A diameter DD' of an ellipse is produced, meeting the director circle and two points P and Q are taken on the diameter produced such that the angle between the two tangents from Pis the supplement of that between the tangents from Q. Prove that PD PD' QD QD' = ODa OD' 2 25. An ellipse has its centre at 0, its axes lie on the coordinate axes OX and OF, and it passes through the points P(2, 7) and Q(4, B). Find the equation of the ellipse and give the positions of the foci. Show that the 24.
in 0,
.
.
.
length of the semi-diameter conjugate to
OP
.
is
.
y'SJl/SO, and give
its
equation. 26.
A
variable tangent
drawn and Q.
is
to the hyperbola
the circle # + y = a in P Show that the locus of the middle point of 2
2
2
P
PQ
2
2
rr
j/
=
a 2 cutting
is
and Q are extremities of two conjugate diameters of the ellipse = a 2 &2 and S is a focus. Prove that P<>2 - (SP- SQ)* = 2 b*. 28. Show that a normal to an ellipse divides the distance between the two parallel tangents most unequally when it is equally inclined to the axes. ~ I in terms 29. Find the equation of a chord of the ellipse x^/cP + y^/b* 27.
2
fc
*2 + <*y
,
CENTRAL CONICS
316 xlt yl the coordinates
of
of
its
middle point, and show that the equation of
the circle described on the chord as diameter
is
(fc'^HaV) {(*~*i)My-yiH ^ 30. The points (xl} y^ (x2 self-polar for the conic xt/cP t
Prove that the points A,
t/ 2 ),
,
(x3
,
y 3 ) are the vertices of a triangle
+ yi/tf-l
B C are }
=
on the rectangular hyperbola
=
(xl ^xJ/(a*x) + (yl toy 9 )/(V'y)
and that the
6. I.
lines 2?(7,
ABC
0.
1',
CA, AB touch the parabola
Coordinates expressed in terms of a single parameter.
The
Ellipse
Since the equation of the ellipse gives us
-
=+
and y
evident that the coordinates x and y of real points on the ellipse b respectively. in magnitude between -fa and -f & and a,
it is
lie
Now
the point whose coordinates are (acosfl, &sin0) lies on the ellipse for all values of 0, and further, since cos# and sin0 can
have any values between be so represented.
-f 1
and
1.
any point on the
ellipse
can
Geometrical interpretation when the coordinate axes are rectangular.
The coordinates of any point j? on the circle on A A' as diameter can be represented by is
nate
x*
+ y2
pCA
meet the
be acos0, is
described
and
P
where
the ordi-
if
:
ellipse
the rr-coordinate of
is
2
(acosfl, asin0)
the angle
pN
=a
at P,
must then
its
^-coordinate 6 sin P i. e. consequently I cos sin The the point (a 0, 0). ;
corresponding to any
angle
P
on the
point ellipse is called the points eccentric angle p, P are said to correspond, and its
;
the eccentric angle of P is that made with the #-axis by the radius- vector to the corresponding pointy.
the auxiliary circle.
It is evident
The
circle
ApA
r
is called
from the symmetry of the
THE ELLIPSE AND THE HYPERBOLA
317
P is the point 0, P', the other extremity of the diameter Now the ordinates of corresponding points the point (Tr-j-0). are in a fixed ratio, viz. b:a. figure that if
PC,
is
PN:pN =
Various properties of the ellipse can be deduced from properties of the circle as illustrated below.
If P, p and
(i)
ApCq A PCQ = :
a
two pairs of corresponding points, then
are
q
Q, :
b.
For if P, Q have eccentric angles pCq and PCQ are 2
^ {a cos0sin
a2 cos (/> sin 0} and
which are in the ratio a Hence it follows that sponding
9,
the areas of the triangles
,
ab cos
J {afccos0sin<
sin 0}
b.
:
if
P,
jp,
R, r are three pairs of corre-
Q, q,
points, A PQR Apqr = b a, for &PQR = &PCQ + &QCR + &RCP, :
:
Ac.
Consequently, when the triangle pqr is a maximum, the triangle is a maximum. But the maximum triangle which can be
PQR
inscribed in the circle
hence the
maximum
is
is i (3
(ii)
its vertices 0,
area
%7t
is
i(3v/3)a
2 ;
in the ellipse
+ 0, $v + 0, and
its
A/3) ab.
Conjugate diameters.
D
(a cos Suppose P (a cos a, 6 sin a), two conjugate diameters CP, CD.
The equations
of CP,
V-
CD
bx
bx
A
tana,
-^
b sin
/3,
/3)
are the extremities of
are
#:=--
TL2
whence or
its
triangle which can be inscribed
has for the eccentric angles of area
and
equilateral,
tan a tan .
/3
.
tan/3, ij2
=
-5
>
(
Fide p. 309)
= 0; a^/3 = \-n.
1-htana. tan/3
Hence the eccentric angles of the ends of conjugate diameters are of the form a, a + 77. If jp, d are the corresponding points on the auxiliary circle, it follows that the diameters cp, cd are at right angles, and are therefore conjugate diameters of the circle. If pCp', dCd' are diameters of the circle which are at right angles, ' 2 a2 is a square, and its area is 4pCd
=
.
CENTKAL CONICS
318 If POP', to these,
LCD'
are conjugate diameters of the ellipse corresponding a parallelogram, its area is
PDP'1)'
is
4POZ)
= 4-. pCtf=2a&; J
thus the parallelogram whose diagonals are a pair of conjugate diameters has a constant area. Since the tangents at P, P' are parallel to CD, and those at D, D' parallel to CP, the tangents at the extremities of a pair of conjugate
diameters form a parallelogram its area is evidently twice that of the parallelogram PDP'1/, i.e. 4a6. This is called the conjugate ;
parallelogram
.
Incidentally,
OP 2 =
a 2 cos2 a + Z> 2 sin 2 tx,
CD 2 =
a 2 sin 2 a + b 2 cos 2 a,
and, as before shown,
Since
e2
=
-
OF 2
=r
^
CP 2 -f CD 2 =
we can
,
write these values in the convenient form
a 2 (1 - e 2 sin 2 a)
CD 2 =
;
a2
(1
- e 2 cos
2
a).
The
area of a sector pCq of the auxiliary circle is ia a (0 2 the area of the sector PCQ of the ellipse is Ja&(0 2 fl x ). (iii)
0i)
;
The student should now re-read Chapter V, 5-7 by exactly similar methods to those there illustrated the following equations ;
can be found (a)
The equation
angles are
(b)
:
of the chord joining the points
The tangent
at the point,
whose
-
The point
of intersection
eccentric angles are
(0,
eccentric angle
fo sin0=
a (c)
eccentric
)
of tangents at the points
is
'
GOB i
|cos$(0-) (*
~ .
.:
r~
T *
(0-40)
tanj^-f tan
.
s
v
^
T
>
is 0, is
1.
(a cos
vl
whose
0, <>, is
v/ .
whose
THE ELLIPSE AND THE HYPERBOLA The equation
(d)
angle
is 0,
x
of the tangent at the point P, can be put in the form
qcosfl
_y ~~
a sin
where
~ V {(#q cos 0)
bainO
V >a
-^fecos
r is the distance of
point of contact
2
__
P and CD
2
any point
y)
(x,
whose
+ (y b sin 0) 2 + 6 2 cos 2 0j
}
eccentric
__r_
~"
OD*
on the tangent from the
the semi-diameter conjugate to CP.
is
Example. To ./wd M0 lengths of from any point to an ellipse. The equation of the tangent
sin
2
319
the tangents tvhich
at the point P(B)
x-acoa0
__
is
r
fesintf __ ~~
i/
-YcosJ
Tsintf
can le drawn
W
CD'
if (a?, y) is the given point from which the tangents are drawn, r is the length of the tangent if the point 6 is the point of contact. Hence the elimination of from equation (i) will give us an equation in r whose roots will be the lengths of the required tangents.
and,
X -
Now
x*
Hence
a
F
=
:'
77=:
sin 6
y*
+
-f
cos 0,
~!
*>>
f*=f.CD\
or
x2
(ii)
1
/ =-2 + ^-l.
where
Thus
= /(a
>*
or
r' (cos
2
+
8
2
sin 0)
y
sin 2 6
-f
=/(a
2
62 cos 2 0),
sin
2
-f
&a cos 2
0),
2
(^-aVHan^+^-fc /):-*).
i.e.
But the
first
of equations
(i)
(iii)
gives us 1,
a
i.e.
-f
r o
or
we
get
))'' +/ [(x +
?)"
Eliminating 6 from this equation and
fO which
is
v - 2'l<-H
/+ "'''(
''>
the required equation.
(iii)
+
+ y']
[(x
-
<)'
+ y1 ]
- 0.
CENTEAL CONICS
820 Cor.
If P,
i.
F
are the points of contact of tangents from T(x, y) to
.
CP we /
ellipse, CZ>, CD' the diameters conjugate TP8 /. CD2 and TP* =/. CD'*.
the
to
CP,
,
have from
(ii)
Tp TF^CD-.CD'. .
Cor.
The term independent
ii.
squares of the distances of
of r represents the
T from the
TP9 TP" ==/* ST8
Hence
.
.
or
product of the
foci S, S'. .
TP.TP'^-^.ST.S'T. The Normal.
(e)
The normal
at the point
being perpendicular
to the tangent, its equation is
sin
,
which
is
acos0)
,
.
=
a2
can also be put in the useful form # a cos y 6sin0
""
by cosec 6
""
a sin
b cos
6
2 .
r
CD
normals which can be drawn from any point If the
= 0,
AX
.
6sm0)
usually given in the form
ax sec It
cos 9 --(y ,
-jjr- (#
normal
at the point
through the fixed point ah sec
(h, i),
whose
to
an
ellipse.
eccentric angle
is
6 passes
then
=
We cosec
6 2,
a2
and, conversely, this equation must be satisfied by the eccentric angle of any point the normal at which passes through (A, Jc). The equation can be written in the following three forms, where c
*
= aa_ &
I.
(*
2.
cos4
0- 2
2 2 ft
*
- c4
)
cos 2
+ 2c2aA cos II.
4 c4 sin
+ 2c*Z*sin 6+ (aW + W-c4 ) sin
aW = 0.
2
III.
Each of the equations
is quartic,
can be drawn from any point to an
two
real
and two imaginary, or
all
and
follows that four normals ellipse, of which all may be real, it
imaginary.
be noted that one value of tan \0 corresponds to one definite normal, for this value gives one value only for cos (9 and It should
sin0.
THE ELLIPSE AND THE HYPERBOLA To find
the conditions that the
321
normals at four points on an
ellipse,
ichose eccentric angles arc given, should be concurrent.
If
6^ 2 @3i #4 a e the eccentric angles of the four points, then some value of h and k these must satisfy equation III above.
for
**
>
coefficient of tan 2 J
The
equal and opposite
is
and the coefficient of tan 4 \ 6 is hence we get the following
zero,
the absolute term
to
;
conditions: (a)
2 tan
(b)
tan \ O l
Ol .
.
tan
6>
\
tan
2
=
2
;
tan \
.
<9
3
.
tan
= - 1.
\
Only two conditions are necessary in order that four straight lines should be concurrent, consequently these conditions are necessary and
sufficient.
From
and
(a)
(b) it
follows immediately that
J^s
tanft^ + t^ + J^ + hence
ex
To find
+
9 = 3 + 4
2 +
the condition that the
co;
(2u + l)ir.
normals at three points on
the ellipse
should le concurrent.
Equation
(a)
above gives us
tan \ 6 1 tan $0 2 + tan
2
tan
1
3
+ tan
1
Hence, tan
J
(?!
substituting for tan
tan ^0 2 + tan
1
= But
cot 10, 21 cot
cot &
(O t
tan ^
+
2)
+
3
tan
2
3
sin
(cos (0 2
+
1 -
2
^ + tan &
2
+ tan
(b),
^ cot i0 + cot J
^ sin sin
3 -h
J
2 L tan l* 2 2 -tan **,
Hence or
which
2
tan
X
(tan J
4
from
#4
tan i
3
- tan i
-
^ -f cos
i 6! 1
cot i
2
3 4-
cot
1
a
cot
^(coB ==:
2)
6 3 ) -f sin (6 3
+
^ i)
=
0,
the required condition. This condition can also be deduced from equations I and II. is
To find the conditions that the normals at the extremities of the chords whose equations are lx + my should 1 0, i#-f J/2/1
=
=
be concurrent.
The equation l) is satisfied
I*x +
(i)
by the coordinates of the points common to lx+ my 1 = 0, = respectively, and the ellipse it therefore represents X
Myl
1267
=
:
CENTRAL CONICS
322 for different values of
A the conies which can be drawn through the
points of intersection of these chords Now we have shown in 4 (iii) that
and the ellipse. the normals at these points of
if
intersection are concurrent at the point
on the rectangular hyperbola Xy (2 _ //>) _ a *Jiy +
(A,
fc2
/?),
=
to
then these points
lie
.
( jj)
thus for some value of A the equations (i) and (ii) will be identical. Since the term independent of x and ;/ in (ii) is zero, we have at once
=
A
1.
Also equating the
coefficients of x~
LI
=
-
and
2 ,?/
Mm =
3
-
.-,
we
to zero,
get
--0
or
which are the required conditions. The equations of two chords the normals
at
whose extremities
concurrent can then be put in either of the following forms
r
f
r /
,
4-
la
~~
mb
+
=
1
are
:
,
U
6 -sec f i
j
+ fcosec + b
-=
=
d
both of which are quite general.
The equation
of the rectangular hyperbola passing
through the
ends of the chords then becomes
which
at
once reduces to
we
and comparing this with
(ii)
point of intersection of the
normals are
To find
find that the coordinates of the
the locus of the intersection of coincident normals.
The chord joining to the ellipse
:
if (a
the feet of coincident normals
cos
0,
of the normal, the chord
then a tangent
b sin 0) is the point of contact is
-cos a
is
6
+
f sin b
0-1 =0.
and the foot
THE ELLIPSE AND THE HYPERBOLA Hence the equation
of the chord the normals at
are concurrent with the
normal
-sectf
a
Hence, as in the
last
at (a cos
0,
whose extremities
b sin 6) is
=
rcosec0+l
4-
323
0.
b
paragraph,
if
(/&,
the point of intersection
k) is
of these normals, the rectangular hyperbolas
and are identical.
2
xy(a
2 b'*)--a-hy+b' kx
The former reduces &sin 0#
xy
Hence, comparing
to
a cos 3
3
=
6y
=
0.
coefficients,
'
a
For different values of
b
9 the locus of this point is
which
curve, called the evolute of the ellipse, is the locus of the intersections of consecutive normals. Incidentally, we see that
,02-62 -
( is
cos 3
^
6 2_ 6,
a2
^~
sin 3
the centre of curvature at the point (a cos
shown
0,
0) b sin
#).
from points from within it four real normals can be drawn points on it the four normals are real but two coincident; from points outside it only two real normals can be drawn.
The form
of the evolute
is
in the figure
:
to the ellipse;
x 2
CENTRAL CONICS
324
Illustrative
Example
i.
The normals
Examples.
at P,y, It,
S meet
a point, and P',
in
Q', R', S'
are the points on the auxiliary circle corresponding to P, Q, R, S respectively. If straight lines are drawn through P, Q, jR, S parallel to CP', CQ', CR' CS'j they are concurrent.
Let
P be
the point (a cos
b sin 0), then P'
0,
(a cos 0, a sin
is
The equation of CP' is #sin0 ycoetf = 0. Hence that of a line through P parallel to it is sin B) cos (x a cos B) sin B (y
= 0, = 0.
fc
y cos0
icsin0
or
Write
t
for tan 5 y' t
Hence four
this equation
,
4
-
y' cos
sin
+2
3
(#'
cos
through a given point (x
If this straight line passes a;'
6) sin
(
4-
a
cos
b) sin
(^
0).
*/'),
',
=
0.
//
=
becomes
&)
-h2(#'
a
-f
&)
0.
(i)
drawn which
intersect at the point (x y') if t 3 2 4 are the eccentric angles of P, Q, /?, S, the correfor S sponding lines of the above form are concurrent provided 0j, a 4 lines of this type can be
r
:
,
,
,
,
,
some value of x and
t/',
satisfy equation
(i).
The conditions
,
,
for this are
and these are
0n
0j
0s
04
identical with the conditions that the four normals at should be concurrent.
Example
P on
Lengths are measured off Jrom
ii.
the
normal at
in both directions equal to the semi-diameter perpendicular to the
prove that the
Let
P be
loci
of the two points thus obtained are
the point (a cos
x
0,
circles.
b sin 0), the equation of the
a cos
__
y
b sin
# sin
b cos
P
normal;
normal
is
r
_
'
C7.O
where C/> is the semi-diameter conjugate to CP and therefore perpendicular to the normal at P. We have then to find the loci of points on the normal distant + CD from the point (acos0, fcsin0). Their coordinates are therefore given by a?
a cos
&C080 i.e.
Eliminating i.e.
two
circles.
the loci are
y
6sin0_ ~
asin0
'
THE ELLIPSE AND THE HYPERBOLA The
and a
intersections of the ellipse
oircle.
825
Application
of parametric coordinates.
The equation
of
circle is of the
any
form
The eccentric angles of the points of intersection of this circle and the ellipse are given by a 2 cos 2
+ b 2 sin 2 0+ 2ga cos 0+ 2/6 sin + c
for this is the condition that the point (a cos
should
on the
lie
Writing
6 sin 0) of the ellipse
0,
circle.
= tan
t
= 0,
this equation reduces to
0,
i.e. fi
2
(a
2 -20a + c) + 4/M 3 + 1* (46 2 -2a + 2c) + 4/M + a* + 2ga + c
Hence a
circle
= 0.
intersects an ellipse in four points and, if the
eccentric angles of these points are coefficients of 3 and t are equal,
T
.
2
0s
,
#4*
we
have, since the
tf
2
=
tan |0
2 tan J^ tan J0 2 .
.
tan i0 3
=
9^\6^ + \ 3 + i 4 = 2W7T. #1 + 02 + 03+04
hence
tan (\
i.e.
)
;
0,
the condition that four points, whose eccentric one condition is sufficient angles are given, should be concyclic
Conversely, this
is
:
since any three points
Note
i.
The equations
lie
on a
of the
circle.
common
chords of the circle and ellipse
being
5 cos *
(tf t
+
- cos J
(tf s
+
the condition 20
=
*.)+! -f
4)
I
sin ( (O l
+
== cos *
sin i
+ ^4
=
2n;r shows that
(tf s
)
common
cos
(^ - *a ),
(<9 3
<9
4 ),
chords of a circle and an
ellipse are equally inclined to the axes.
The
circle of curvature.
Since the circle of curvature meets
the ellipse (p. 280) in three coincident points, if the eccentric angle of these points is 6 and that of the other point of intersection O l} we
have 30 + #x
= 27r,
i.e.
the circle of curvature at
again at the point 2^77 30. The common chord of the circle of curvature at T
is
therefore
- cos a
II
r sin b
= cos 26.
cuts the ellipse
and the
ellipse
CENTEAL CONICS
826
The equation *-
a2
of the circle of curvature
+ ys _
i
6
J- cos 6
+A
(a
Equating the
+ \ sin 6
1
e.
^ + |!
-
0.
)
#2 and y 2
coefficients of
the equation of the circle
M(a-cos 6 - %I sin 0-cos 201 =
)
we
>
get for A
Asin 2
1
i.
therefore of the form
is
287)
(p.
1
is
2 2 sin 2 e + b* cos i) (a
6}
= 0, &
(a
which reduces
to >
-
_ --
cos 3 6.
-
;
or
+
2
(a
-2 6
L sin 3 6 .
s )
y
.
cos2
Cor. The radius of curvature at the point P(a cos
or, if
CD is
the semi-diameter conjugate to CP,
Note. The work throughout
by using
(**
+ *-**),
5-;J
Z
this
2
+
its
(&
- 2 a2
)
= 0.
sin 2
0, 6 sin 0) is
length
is
r-
paragraph can be considerably shortened
(**--**) instead of acostf,
involves the use of imaginary quantities
;
we
leave
it
The method
bsintf.
as
an exercise for the
student.
Examples VIII 1.
If a circle touches
contact and the
b.
and cuts an
common
ellipse, the tangent at the point of chord are equally inclined to the axis of the
ellipse. 2.
The
inclinations to the axis of the ellipse of tangents
drawn
to it
from
the point (x y) are given by the equation t
2 (o;
3. If
-a2 jtan^-2a;t/tan^-l-y2 -62 =0.
the normals at four points whose eccentric angles are
concurrent, then
and
2 sin B l -f 2 sin t sin 2 sin 2 cos 0, 4- 2 cos 6 l cos#a cos
8 3
= 0, = 0.
0,
,
2
,
3
,
4
are
THE ELLIPSE AND THE HYPERBOLA
327
from any point four normals are drawn to an ellipse meeting one of O19
4. If
the axes in 5.
an
,
be concurrent.
ellipse to
6.
,
,
The three points Qlt Qz
are such that their three circles of curva-
<) 3
,
ture intersect on the ellipse at the point whose eccentric angle is 0. Show that Q l Qn Qs are the vertices of a triangle of maximum area inscribed in ,
the ellipse. If four concurrent
7.
normals are OP, OQ, OR, ON and T, T' the poles of make angles ex, ft with the axis, show that
PQ, RS, then, if CT, CT' tan a tan ft = 6 2 /a2 .
8.
The normal
N respectively
at
P
any point
M
of an ellipse intersects the axes at
PM
and
PN
in a constant ratio. is to prove that 2 2 9. The tangent at a a 2 Z> 2 meets the axes on the ellipse Jrur-f u // point in T, t and the normal meets them in G, g prove that the locus of the ;
=
;
2 2 2 the curve (^2 4 // 2 ) /(a 2 - ?r) j*/a + y*/b*. points on a line parallel to the axis of # normals arc drawn to show that, if a,, Of a 3 3 3 4 are the eccentric angles of the feet
intersection of Tg
From
10.
and tG
is
the ellipse of the normals drawn from a point on the both constant. ,
,
;
line,
2
(sin
3) and 2 (cosec
Show
that the equation of the chord of the ellipse ur/ a2 + & ft is joining points whose eccentric angles are tt + = sin + cos cos ft. (x (*/ (X)/a 3)/b 11.
Or)
2 !/
3
/^
are
=
1
,
If
12.
rP
and
<7()
are conjugate diameters, show that
4(CP*-C()
Two
2 )
= (SP-&P)* -
(SQ-S'Q)*.
diawn to the ellipse from the point whose coordinates are k; show that the area of the triangle TPQ al {fc 2 /a 2 + */&' - 1 } V {W 2 ^ &Y& 2 14. Prove that the normals at the points where the line 13.
tangents TP,
T$
are
//,
7] is
]
x\(a cos
0()
=
intersects the conic x*/a* + y*/l 2 2 3 1 are - c* cos 8 oc/a -f c sin CX/fc. (c ,
15.
CP
f
PP'
in Q.
ratio
is
an
is
-J-
1
I y/(b sin OC) meet at the point
=a
whose coordinates
1
b.)
a double ordinate of an ellipse, and the normal at P meets that the locus of a point which divides P, Q in a given
Show
ellipse.
A, II, C are the vertices of a triangle of maximum area inscribed a in the ellipse fc a # 2 + rc 2 // 2 = a 2 & P,
;
sponding to A, J5, C. Find the locus of the centroid of the triangle PQR. 2 17. Find the equation of the tangent to the ellipse x-fd' + if/l = 1 in terms of the angle 3. the perpendicular from the centre on it makes with the major axis. If this tangent and a perpendicular tangent-be taken as
new
coordinate axes, what will be the coordinates of the centre of the
ellipse ? 18.
The tangent
at
(#', y')
meets the auxiliary
circle at QQ'.
the lines CQ, CQ' are represented by the equations xif the centre of the ellipse.
=
y
(.c'
Show
ae)
;
that
C being
CENTRAL CONICS
328
19. Show that there is in general one conic of finite axes with given centre and direction of axes which has two given lines as normals.
The locus of the intersection of perpendicular normals to an ellipse + 2 ) (x* + */ 2 ) (a + 1" x*f = (a2 - 2) 2 (a 2 if - 2 # 2 )2 21. The normal at P meets the tangent at Q on the minor axis show that PQ touches # 2 /a 2 (a 2 -2& 2 )-7/y& 4 = l/(a 2 -& 2 ). 22. If the normals to an ellipse at ABCD are concurrent, and diameters are drawn parallel to AB and C/), their extremities are at the angular 20.
is
(a
V
2
fc
Z>
fc
.
;
points of a parallelogram whose sides are parallel to the equi-conjugate diameters. 23. If the centres of curvature of
an
ellipse at the extremities of a pair of
conjugate diameters are joined to the centre, the product of the tangents of the angles these lines make with the major axis is constant. 24. The normal at any point P meets the axis in G, a point Q is taken in the tangent so that PQ = \. PG, where X is constant prove that the locus of Q is # 2 /a 2 + f/l 2 = (a 2 + \ 2 & 2 )/a 2 ;
.
The
25.
common
line joining the centre of an ellipse to the polo of the chord and the circle of curvature at any point, arid the line
to the ellipse
joining the centre of the ellipse to the point where the circle of curvature drawn, make equal angles with the axes. 26. If
TQ
TP,
#2/--f y l j\jl lateral
\
TPCQ
=
is
is
are the tangents from the point T(f,y) to the ellipse 0, whose centre is C, prove that the area of the quadri-
(i)
V^'T^V 17^;
\
(ii)
TT /2 tan0,
where TT'
is
the
tangent from T to the director circle and 8 is the angle PTQ. Prove that the tangents at Q 27. PQ, PR are focal chords of an ellipse. and R intersect on the normal at P. 28. In an ellipse, if CPand CD are conjugate diameters, find the envelope
PD.
of
29.
on
A
PQ
chord PQ of a conic passes through a fixed point. If the circle as diameter meets the conic again in P'Q', show that P'Q' also
passes through a fixed point. 30. Find the coordinates of the intersection of the normals at the points of contact of two tangents from (, 17) to the ellipse, and show that the
normals at the points of contact of tangents from
~^/ r
P ass
ellipse parallel to the
chord
through the same point.
2 2 2 2 chord of # /a + // /& = 1 touches If 2r is the length of a diameter of the
31.
and
c
A
a
/i
first
2
2 -f
?/
/V =
*/
f)
1-
the length of the chord, (a
32.
.r
(
2
- W) &
=
4r> [a 2
-
2 fc
- (r/ 2 - ^ 2 ) + ^ (iY t
2
- &iV& 8 )]-
at a point P of the curve meets the major axis in G, and taken on the normal at P such that PQ = PG find the
The normal
a point
Q
is
;
locus of Q.
and (a*2 # 2 ) ave ^ wo points on the ellipse x^/tf + if/b* = 1, the tangents at which meet in (a?, //) and the normals in (, 77), prove that 2 = e^xx^x^ and I 2 = e^yy^jn where e is the eccentricity. 33. If
(a?j, t/j)
,
rj
THE ELLIPSE AND THE HYPERBOLA
829
= 1 are drawn parallel to the = find the locus of their middle pointfl. wy/b at P to the ellipse meets the curve again at Q; show that
Chords of the ellipse x*/a? -f i/2 /&2
34.
diameter Ix/a
-f
The normal
;
the locus of the middle point of
Normals at the ends
35.
is
PQ
given by the equation
tangent at 6 meet in
of chords parallel to the
points lying on
2(ax8\nQ + fy/cos0) (ax cos -f ly sin 0) = (a -fc ) normals to an ellipse from meet the curve 2
2 2
36. If
N
sin
20 cos2 .
28.
P19 P2 P P
in
,
3,
4
and
the major axis in 19 Ar2 JV3 JV4 then ^(OP^/^P^ is constant. 37. Show that unless the eccentricity of an ellipse be greater than l/\/2 it is impossible for the centre of curvature at any point of the ellipse to lie
on the curve 38.
The
,
,
,
itself.
locus of the poles of normal chords of an ellipse 2
V^ + //'/?/
2
2 2 -(rt -fo ) .
From any point on the normal
39.
at the
other normals are drawn to the ellipse. of intersection of corresponding tangents sin
l>x
From any
40.
drawn 41.
at is
P
is
+
point of the curve
to the ellipse
of contact
C*
.r
2
2
2
/a
-f
y^/l
//
is
2
cos
Show
ellipse
two
is
=
Oi -f .nj
0.
^ /a + if/tf = (.r2 / 1 show that the 2
2
on an
(X
point
that the locus of the point
;
2
-
2 2
2 //
/&
)
tangents are
line joining the points
the chord of curvature at one of them.
2 2 2 2 = 1. Show that the normal point on the ellipse o- / -f // /b bisects the angle between the focal distances SP and HP. If SP
P is any
produced
to Q,
Pit =
making
PQ =
P//,
and
HP
is
produced to
J?,
making
PS, show that RQ intersects the tangent at P on the major axis and find the equation of the locus of the intersection of RQ with a line
;
drawn from the centre of the ellipse, parallel to HQ. 42. If tangents TP, TQ are drawn from T(/, g) to a conic prove the difference of the angles TPQ, -i
__
4a 2 & 2 / 2yT(a
x^/
TQPis
V-
N on the ellipse meet the whose eccentric angle is a find the eccentric same point angles of L, M, N, and show that the circle circumscribing the triangle LMX passes through 0. 2 2 2 2 2 44. A point P on the ellipse & # -f a // = a b 2 is such that the centre of curvature there lies also on the ellipse. Find its coordinates, and show 43.
The
circles of curvature at the points L, 3/,
ellipse at the
;
that the radius of curvature at
P
is
{3ya
a
a
& }/{(a
2
+ &)%}.
Prove that four normals, real or imaginary, can in general be drawn from a point to an ellipse and show that the line joining the feet of any two of them is equally inclined to the axis with the diameter which bisects 45.
;
Show also that the middle the chords joining the feet of the other two. points of the diagonals of the quadrilateral formed by the tangents at the
CENTRAL CONICS
880
four feet lie on a straight line which passes through the centre and right angles to the diameter which passes jthrough the point. 46. Chorda through the point (a
-
2 fe
.
)
(V - a
2
a cos 0/(a* 4-
fc
subtend a right angle at the point
),
8 )
b sin 6/(tf
.
is
at
+ a2 )
(a cos 0, b sin 6).
are the coordinates of the centre of curvature at a point (#', y') on #2/aa + y*/W = 1, and if the centre of curvature is on the ellipse, prove 47. If
Ot, /3
48. The normals to an ellipse at P, <), R meet in a sum of the eccentric angles of these points is constant. Show that the locus of their point of intersection is
and
point,
also the
a straight line, and
that the sides of the triangle PQR touch a parabola. 49. A point P moves on the ellipse
where k
is
a constant
prove that
;
it is
a constant distance from the centroid
of the four points on the ellipse #2 /a2 -f y*/V* = 1, whose normals meet at P. 50. A tangent is drawn from the point (a cos 6, ft sin 6) of an ellipse to the circle of curvature at the other
end of the diameter through the point
show that the length of the tangent
;
2 {(a 2 -ft 2 ) cos20}i.
The Hyperbola
II.
6.
is
Several systems of parametric coordinates are possible each can be developed in a manner analogous to that used for the circle and the ;
ellipse (a)
we
:
Any
shall therefore only give a short sketch of each.
point on the hyperbola can be represented by (a cos 6,
where
i
=: v/
The
1.
ib sin 6),
results in this case can be
deduced from
those found for the ellipse by substituting throughout ib for b. objection to this system is the use of an imaginary quantity.
The
The hyperbolic trigonometrical functions can be used, and any point on the curve represented by (a cosh 0, b sinh 0). This system has the advantage of retaining the symmetry to which we have become accustomed in the case of the ellipse. The objection to the (b)
system
is
that only one branch of the curve can be so represented
for real values of
The
0.
area of a sector
The equation |cosh
PCQ
of the hyperbola is J ab (6 l
of the chord joining \
(at
+ /3) -
JJsinh
two points
\ (a
+ H)
a,
= cosh
ft
^
is
| (a-/3).
2 ).
THE ELLIPSE AND THE HYPERBOLA The equation
of the tangent at the point 6
The equation
=
smh0 f b
-cosh0
a
is
1.
of the normal at the point
+
ax sech
is
= a2 -f
by cosech 6
331
2 fr
.
This equation can also be written by tanh*| 6
+ 2 (OB 4- a 2 +
Thus the normals
2 fc
at
)
tanh 8 i 0- 2 (as- a2 - & 2 ) tanh
lf
2
2 tanh and
tanh
But
X
.
tanh
_
^ tanh
""
1
Hence,
+2 if
(c)
fl
j
.
2
.
tanh 3
+ S tanh
^
6>t
J
tanh
.
J ^4
=
l.
^ 2 tanh
tanh
^
if
=
2
tanh
tanh J Ol tanh i 6> 2 -f- tanh
tanh
i
^
fl
a
MirniTp^tanhT
the normals at these points are concurrent,
tanh
(a sec
\ 6l
^
are concurrent
^3> $4
,
J0- 6y = 0.
(J
0!+
i
2
+J
3
+4
4)
=
Again, any point on the hyperbola can be represented by The results are not analogous to those .for an b tan 0). 0,
ellipse.
The equation
of a chord joining the points a,
-cos
The equation
i
(a-/i)
-
+ /3)
=
of the tangent at the point
a
The equation
sin i (a
|
of the
normal
ax sin
=
sin | 6
+
cos
at the point 2 by =. (a
^
cos
is
(a
+ /i).
is
0.
is
+ b*) tan
0.
of coordinates is a variation (d) The most workable system we can use for any point on the hyperbola the coordinates '
The equation
of the chord joining the points
t lf
<2
is
of (b)
:
CENTRAL CONICS
332
The equation
of the tangent at the point
The equation
of the normal at the point 2
2at(l-t*)x-2bt(l + t )y
=
(
t
is
t
is
a
2
+
fc
This equation can also be written
Thus the condition
that the normals at the points
t l9
2
>
*3> ^4
should be concurrent are
2*^ = Eliminating three points
we
<4
t ly
2
,
0,
M4= - 1
-
see that the condition that the normals at the
should be concurrent
/3
is
The equation giving
the parameters of the points of intersection of the hyperbola and the circle s
Hence the condition parameters are
J
lt
2
>
that four points on the hyperbola,
^;)
should be concyclic
'4*
whose
is
i. To find the equation of the circle of curvature at the whose parameter is t, and the coordinates of the centre of
Example point
curvature.
Let the circle of curvature be
#2 + t/2 + 2gx + 2fy + c then the parameters by the equation
of its
= 0,
points of intersection with the ellipse are given
Since three of these intersections are coincident, equation be
/,
t,
/,
let
the roots of this
t'.
Then
t*t'
=
1;
.'.
*'
= -^
Also
-
A I
-PI \
--
-
= sum of roots two
=
sum
at a time
=
of roots three at a time
3
=
Q '-
+ f3
.
THE ELLIPSE AND THE HYPERBOLA
338
-J&-K)'=
2c
and
Thus the
2 ij
+
(a
circle of curvature is
and the coordinates of the centre of curvature are l
Incidentally, if
parameter
is
t,
(a;,
y)
is
the centre of curvature at the point whose
we have
Hence the equation of the evolute
Example
ii.
/SAow
is
chords which subtend a right
tliat
a fixed point on an hyperbola
all
angle at pass through a fixed point, and find
coordinates in terms of those of the given point.
its
Let the given fixed point be JPJ^aff-f-J, t
l9
1
9
* ""
(
are the parameters of the extremities of any r.hord
a right angle at P. Then the equations of the chords PQ,
-(1 a
+,)
and since these are perpendicular (1+ 1*0(1 + 0,)
^
which at once reduces
to
(l+^ But the equation
or,
^
l )(
of the chord
substituting for (^
+ *2 ),
it
QR
is
becomes
PR
are
7 )
QR
[ >
anc^ suppose
which subtends
CENTEAL CONICS
334 This can be written
(* i
+ yr ,
2
""
(a
which for different
+ &a
*
r
'values of
t
<**
a 2 -fc2
i
{a
)fj
(a
+& --*ti
(x y , , , + Mia -- 76
1
r a -6ToTa 2
6
t2
l}
*
i r
=
0,
j
represents a line passing through the
intersection of the lines a? ,
x ""
i.e.
y
_
b~
a
a*-b*'
y^
i'
*
'
through the point
Thus
if
P
the coordinates of the fixed point are (# 1T P all pass through the point
y,),
chords which
subtend a right angle at
This point
Example hyperbola
Show
z
a2
on the normal
lies
iii.
---
b2
From any
=1
We
at P.
point on the normal at a point
^/ree
normals other than that at
that the circle through their feet
is
P
oj the
P are drawn.
one of a coaxal system.
Let the parameters of the fixed point P be t, and those of the feet of the other three normals drawn from any point on the normal at t be f l9 ? a t s Also let s ^-f *a + 3 pEEf^ + fafa + Mi'
.
=
,
,
Since the normals at
t,
t lt
f
2
p + ts
Now
,
t%
=Q
are concurrent
and
^
1
^^ 8 ==~1.
suppose the circle through the points
J 1? ? 2
,
/ t
,
is
then, since the parameters of the points of intersection of the circle hyperbola are given by
three roots of this equation must be t lt f a MaMi 1 hence t+ = ~^ = constant.
=
,
f
3
;
let the fourth root
J
Thus
^^ and
~~
4
^^^^^^^ 4 ^ 4
= ^i^ + ^(i
i*i
+ '8
\
-*P
=~ \
be
and the
*4
,
then
THE ELLIPSE AND THE HYPERBOLA Hence
2//
The equation of the
=
a2
4-
/>
2
-
circle is
335
(1
+ f)
(1
- 8 t),
then
which can be written
Since
a
is the only undetermined constant, this equation represents of coaxal circles of which the radical axis is system
s
Evidently the point whose parameter it is common to all the circles.
is
lies
t)
(
on the radical axis
since
Examples VIII
c
Show
that the normal at the point whose parameter is t on the 1 a 2 meets the curve again at the point whose rectangular hyperbola rf-y' 1.
parameter 2.
8.
is
-
3
1/*
The tangents
.
at the points
t
ly fa
on
2
2
2
arya'
t/
/fc
=
*
intersect at
t 'i+V~' *!+' Find the locus of the foot of the perpendicular from the centre to
a tangent to the hyperbola tf'/a2 t/ 2 /&* = 14. The equation of the straight line joining any point on the hyperbola to the vertex is of the form x/a 4- y/b l-\ t (x/ay/b 1) = 0.
Find the locus of the mid-point of a chord of the hyperbola which subtends a right angle at the vertex. 5. Find the coordinates of the foot of the normal which meets the axis of the hyperbola at {(a 2 + b 2 )/a, 0}. 6. The tangent at any point P of the hyperbola x*/a? - y*/b* = 1 meets at the points Q, R show that CP'bisects QCR. the lines (a 2 + 6 2 ) y2 + a ? 6 2 = :
7.
The mid-points of
focal chords of
an hyperbola
lie
on an hyperbola of
equal eccentricity.
an hyperbola which passes through a which passes through the centre C of the given hyperbola and its centre is the mid-point of CP. 8.
The mid-point
fixed point
P
lies
of a chord of
on another
fixed hyperbola
CENTRAL CONICS
336 9.
Find the conditions that the chord joining two points on an hyperbola (i) pass through a focus, (ii) be a diameter when the extremities are
should
given in'each system of coordinates. 10. Show that the normals at the ends of the chords
-
(x sec B)/a (y tan ff)/b + d = 0, (x cos ff)/a + (y cot 0)/b 1/d = of the hyperbola
#2 /
a
a -#Y& -l
are concurrent.
Hence find the coordinates of the centre of curvature at 11. Find the equations of the common tangents of a:
Ya-y
6).
&'=!.
12. Find the conditions that the normals at the points where 6 is ly 3 3 4 respectively should be concurrent. ,
(a sec 0, b tan
(a sec #, b tan 6)
,
Also find the condition that those points should be concyclic. 2 1 meet at 13. If the tangents at two points on x'/cr if/b
the normals at the same points at (,
*/),
(x, y)
and
show
14. Find the locus of the mid-points of focal chords of the rectangular 1 a1 hyperbola x -]? 2 15. A circle whose centre is (/, g) cuts the hyperbola #2 -y 2 == a in four .
points 16.
and
:
A
Q.
find the coordinates of their centre of
mean
position.
2 tangent to tf/tf-f/tf = 1 cuts the ellipse x /a* + if/b* Show that the locus of the mid-point of PQ is
=
1
at
P
17. From any point on the hyperbola x^/W-if/tf^ 1 three normals other than the one at the point are drawn show that the centroid of the triangle formed by the feet of these normals lies on the hyperbola :
-sW = {('-&W +
9 {*/< 18.
& J )}'-
In the last question the locus of the circumcentre of the triangle
4(aa^-6V) = 19. If
is
(#2 , y 2 )
x*/a*-y*/b*
=
1,
a * fe4
is
-
the centre of curvature at (x lt y^ of the hyperbola
show that
x^/a^-y^/Vy^l = 0.
if (#2 y 2 ) lies on the hyperbola x z /x l 4- y 2 /^i + 1 = 0. 2 Find the locus of the intersection of a normal to j? 2 /a 2 - yY^ = 1 anc^ a chord which subtends a right angle at its foot. 2 21. Find the equation to the normal to the hyperbola .rYcr-// /^ = 1
Hence show that
,
20.
at a point
whose eccentric angle
is
6.
sum
of the eccentric angles at the points where normals from a given point meet the hyperbola is an odd multiple of two right
Show
that the
angles.
drawn from a point to the the abscissae of four points hyperbola x*l<#-\fl\? on the hyperbola, the normals at which meet in a point, prove that 22.
Show that
in general four normals can be
=1.
(XL
If
x l9 # 2 #3 #4 be ,
,
+ ffa + ^a + ^J (l/Xi+l/Xi + l/Xi + l/Xt) =
4.
THE ELLIPSE AND THE HYPERBOLA 23. Find the radius of curvature at
&v
2
337
any point of the hyperbola
-y =
tfw.
Prove that the difference of the lengths of the tangents froir any point of the hyperbola to the circles of curvature at its two vertices is onstant. 24. PN is an ordinate of an hyperbola NQ is drawn to t-mch the circle described on the major axis as diameter: show that the tangent at P i
;
intersects
NQ in
a concentric conic.
Prove that the equation of any normal to the hypeibola x^ y 1 = c 2 can be written in the form x sin 6 -f y = 2 c tan 6. Prove that the locus of the middle points of normal chords of the hyperbola is (// 2 ~ir 2 3 = 4c 2 ^y. 25.
J
The
26.
circle of curvature at
meets the curve at
(a seed,
if
=1 P of the hyperbola # /e* - y the Man be and $) Han0) (asec, 2
any point
2
2
2
/fc
coordinates of P, Q respectively, find the relation connecting 6 and
on x^-y 1
27. Prove that the chord joining the two points abscissae are a cosh 6 and a cosh <j> is the line
^
a*
whose
+ )- ysinhj(0 + $) = a cosh f (0 ). = a2 and P, Q the points whose parameters are 6 + 8 and 6-8 where d is constant. Prove that the locus of the intersection of AP and A'Q is a lectangular ajcosh J(0
A A'
are the vertices of or-?/ 2
,
hyperbola.
Determine the equation of the chord joining the two points on the 2 a & a? a 2 // 2 = a2 2 whose coordinates are
28.
hyperbola
fe
(a sec a,
Man a), (a sec ft b tan /3). A and ^4' the vertices of
the hyperbola, and foci, from any point P on the curve PS, PS' be drawn cutting the curve again in Q, Q\ then if QA, Q'A' are joined the locus of their point of intersection will be an hyperbola having double contact with the given one.
5 and
If
S' are
the
if
29.
Prove that the line joining the centre of a rectangular hyperbola to is perpendicular to the chord common to the circle
any point on the curve
of curvature at the point and the hyperbola. 30. From a point (x, y), lying on the hyperbola 2
2
=
2
2
- y- =
a--fr 2 tangents whose centre is C. If 1
x'
2
2
,
a an ellipse 6 # -f a // CO meet PQ in R, show that OP.OQ: xy 2 OR OC. 31. The common chord of an hyperbola and the circle of curvature at a point on it passes through a fixed point show that there are four such points and that they are concyclic.
OP,
OQ
drawn
are
to
:
fc
,
:
:
:
If the fixed point
2 (x 32.
1
is
(a?
+ y2 ) - xx
K
+ 6 2 )/
a
~ mi (& + &
2
2
)/&
The chords and the
-(
2
is
- &2 ) =
-
D
at four points A y B, C, of the hyperbola are is one of a fixed point show that the circle
:
33.
yj, the equation of this circle
The chords of curvature
if A is concurrent a coaxal system, and find
at
n
(a*
BCD
its
radical axis in terms of the coordinates of A.
of curvature at P, Q, R intersect on the chord of curvature PQR cuts the diameter at 0. Prove that
circle
CK.
C0(a
a
-& 3 ).
CENTRAL CONICS
338 7.
The asymptotes
The Asymptotes. 2
*>
-o 4- fo a* b*
=
9
a-
y^ cr
~+
are
1
of the ellipse
2
= 0,
are imaginary, in accordance with our original classification. The asymptotes of the hyperbola
i.e.
>
-
a2
O
b* fit
/-y ' '
written separately,
or,
'-
\
u
Ct
= 0,
-
-f-
-
=
;
u
(I
each
is
inclined to
the #-axis at an angle tan" 1 -*
Thus,
if
the angle between the asymptotes
is
co,
we have
tan 5 == ~
*
The two hyperbolas whose equations
Conjugate Hyperbolas.
referred to their principal axes are
x2
y
2
J \
a2
have the same asymptotes. The axis of x meets the points
:
b2
~
1
(i\\
()
l
first
in real
the axis of y meets the
first in
and the second in imaginary imaginary and the second in
real points.
We
note that the points (a. 0),
0),
(0,
1
6),
(0,
-tl)
and
lie
on the
lie
on the second.
first,
(-a,
(.a,0),
(-f,0).
(0, &),
(0,
-I)
THE ELLIPSE AND THE HYPEEBOLA
339
P
on the first hyperbola can be represented by point (a cosh 6) b sinh 0), and similarly any point p on the second can be represented by (a sinh 0, 6 cosh 0).
Any
Thus
for the
same value
The tangents
the equation of the diameters CP, Op
of
are
I
y
=
x - tanh
y
=
x - coth
0,
0.
a
to these hyperbolas respectively at the points -
0- f sinh (9=
cosh
a
P,p are
1,
fr
y- cosh
- sinh
a
=
1,
6
which are respectively parallel to G^> and CP. Thus (7P, Cjp are a pair of conjugate diameters of both hyperbolas. The curves are therefore called Conjugate Hyperbolas.
The following
properties should be noted
:
If CPy Cp are a pair of common conjugate diameters of the two (i) hyperbolas, each meets one hyperbola in real and the other in
imaginary points.
CP =
(ii)
a 2 cosh 2 a* sinh 2
cosh 6
hence, since
+ 1 2 sinh 2 0, + tf cosh 2
2
(9
sinh
2
=
OP 2 -Q> 2
;
0=1, a 3 -& 2
.
jf%
asymptotes.
Let POP', pCp' be a pair of conjugate diameters coordinates of their extremities are
P(acosh0, a cosh
P' ( j?
!/(
Hence the tangents
at
0,
a sinh
0,
P and
-cosh0 a
and
0),
& sinh
0,
sinh
(a
b sinh
&
0),
cosh 0), 6 cosh 0).
P' are
~fo sinh0=
1,
at p,|/ are
-sinh0 a
-
y T
b
cosh0= +1.
y 2
:
then the
CENTRAL CONICS
340
Hence the (|cosh i.
e.
they
lie
intersection of the tangents at 6
-
+
|sinh
1)
Pp
+ (^einh 9-
or P'p' satisfy
|cosh
+
9
l)
= 0,
on the line
or on the asymptote
x
~~
y
a
So the intersection of tangents cosh (M i.e.
i.e.
0-fb sinh 6 ~+
l} /
-o
'
b at Pp',
- (-sinh \
lie
P'p
a
;'
on
cosh 6
b
<-
~~
l) /
=
0,
on ^
V\
-a
I))
"~" '
'
on the other asymptote ~~
a
b
The student should prove the following (a) If PCP',pCp' are a pair two conjugate hyperbolas, then
additional properties
of conjugate diameters
The parallelogram PpP'p' is of constant area. The parallelogram formed by the tangents {ii)
:
common
to
is
of
(i)
at
PpP'p'
constant area.
The
(iii)
lines Pp,
and bisected by the
The chord
(b)
Pp,
P'p, P'p' are each parallel to one asymptote
other.
of contact of tangents
from a point on an hyperbola
to its conjugate hyperbola touches the hyperbola. (c) The polars of any point with respect to two conjugate hyperbolas are parallel and are equidistant from the common centre.
8.
"The equation of
an hyperbola referred to
its
asymptotes
as coordinate axes.
Method form
(vide
i.
The equation
Chap. VI,
6. 1)
of the hyperbola in this case is of the
xy
=
c
2 .
Now
the asymptotes are equally inclined to the axes: equation of the axis is
xy =
0.
hence the
THE ELLIPSE AND THE HYPERBOLA
341
This meets the curve where
i.
e.
at the points
(
+
c,
c).
OA 2 =
Hence
4e
or
The equation then
2
of
a2
= 4e
=a
2
=
a
2
l
+
=
a2
+
fc
sec
2
cos 2
i a>
2
|co
bin* Jw)
2 .
an hyperbola, whose semi-axes are
a,
b,
referred to its asymptote as axes, is
a0=H<* 8 + & we
shall generally use this in the
where
4c
);
form
xy 2
2
c\
=
a2
+ 62
.
Let any point P on the hyperbola have coordinates (x, y) referred to tho asymptotes as coordinate axes and (xf, y') Draw referred to its axes. perpendicular to the axis, and let
Method
ii.
PN
Plf, parallel to the
asymptote Oy, meet the asymptote Ox at M.
= x =
Then
a'
Thus af
=
ON,
y'
M
?/
,
= PiY, =
= (PJf + 0-M) cos Jco = (y + #) cos = (PJf- OJf sin |o> = (y-o:)sin )
CENTRAL CONICS
342 2
for
(#', #')
lies
y)
-(;s-y)
=
z"* (1
2
= a^ 2 seca i(.>-y
+ tan 2 1 co)-/ 2 (1 + cot 2 i
on the hyperbola
Hence the required equation
is
4xy = The reader can Chap.
Ill,
also
obtain
equation by the method of
this
6.
Note. The equation of the conjugate hyperbola
is
or a pair of conjugate hyperbolas can be represented by
xy xy
=c = - c2 2
)
J
The majority of problems dealing specially with the hyperbola are concerned with the properties of its asymptotes, which are peculiar to it in such cases it is usually convenient to use the hyperbola in the form xy = c 2 :
,
and we proceed
to develop a parametric system of
coordinates for this
form.
For the hyperbola in general the axes are then oblique, but for the rectangular hyperbola, since its asymptotes are at right angles, the coordinate axes are rectangular.
The hyperbola referred
9.
Parametric
to its asymptotes.
Notation.
The
coordinates of any point on the hyperbola xy
in the form
c\
/
-J for some value of
(ct,
whose coordinates (i)
are in this
form
lies
t,
=c
2
can be put
and further, every point
on the hyperbola.
The equation of the chord joining two points whose
parameters are
t l9 t 2 .
Let the equation of the chord be
= 0;
THE ELLIPSE AND THE HYPERBOLA then, since the given points
on
lie
348
it,
f\
87 + c = h
0,
= 0, = 0. At.? + B +
Atf + B+tt
or
and
t.
2
A
Hence
B 4 ti
/ to""
or
/
This
may
//
^1^2
^4.
Thus the equation
/
\
1 \
^"^ /
2
2
/
I/
2
/
2
1
=
=
of the chord is
also be written
x
hence,
if
(,
is
i])
the mid-point of a chord, its equation
=
is
2.
PQ meets the asymptotes at p and q, and the mid-point since the equation of the chord is x/ + y/*) = 2, the then, 17), Hence the point (, rj) is also the midpoints j, q are (0, 2 9) and (2f, 0). This property enables us to draw an hyperbola point of pq, and Pp Qq. Cor.
PQ
of
chord
If the
is (f,
when we have
for if any the asymptotes and one point P on thq curve P meets the asymptotes at p and q, we can at once ;
straight line through
construct the point
Q on
the curve.
The
figure (see p. 344) represents the same hyperbola as that shown in Chap. VI, p. 234. Any number of points, such as Q, can be found by using a ruler and a pair of dividers. Each of these gives, in the same manner,
two other points Q l} Q^ on lines parallel to the coordinate axes, which can be found conveniently when the construction is made on squared paper. (ii)
to
Conjugate Diameters.
y+mx =
0,
we have
?)
If the chord x/t + y/i]
=
2
is
parallel
wtf = 0.
Q lie on Thus the mid-points of chords parallel to y + mx = 0. The converse is evidently true the equations of a pair y mx ;
of conjugate diameters thus take the simple forms y
+ mx
=
0.
that any pair of conjugate diameters are harmonic with conjugates respect to the asymptotes.
Note.
It follows
The Tangent. The equation of the tangent at the point (iii) whose parameter is 2 is x+ t 2y = 2ct this follows from the equation of the chord by putting t l = t 2 = t. ;
CENTRAL CONICS
344
If the tangent at P meets the axis of x at T, T is the point (2c, 0), CT = 2ct. Also CS* = a + IP = 4c2 so that CS = 2c. Geometrically = CT/CS. It can be shown that has the same value in the therefore
Note.
2
so that
,
t
t
system of coordinates #
=
s
*
(
+
7 )
#
=
*
d" (
""
'
7)
\\
IN
Example.
The
bisected at the point
intercept
made by
the asymptotes
on any tangent
of contact and the triangle so formed
is
is
of constant
area.
For the tangent at the point t meets the asymptotes at the points / 2c\ (2ct, 0), f O,-T- ), and the coordinates of the point midway between these is /
c \ ct '
\
t/'
^
ie
area
^ e ^ an ^ e 80 formed
is
evidently 2c
2
,sin
.
THE ELLIPSE AND THE HYPERBOLA
345
(iv) The equation of the polar of any point (#', y') with respect to f 2 the hyperbola xy 2c 2 and if (#', y') lies outside c is xy + x'tf the hyperbola this is the equation of the chord of contact of tangents from (x'j y'} to the hyperbola.
=
,
Example. To find the point of intersection of the tangents at points whose parameters are f x / 2 .
,
Let
(x' y y')
be the point of intersection, then xy'
and
xy =
-f
=
x + t t2 y both represent the chord of contact. l
Comparing
coefficients
2 c2 c (t l
we get 2cUo ,
The equation
(v)
of the
1 -f
mi
m=
or
=
2c
at the point
(A
ct)
=
pendicular to the tangent x -\-t-y 2
/ ) 2
,
normal
Let the normal be the line
-\
(^
+ m) cos
t cos -TiT
r
r
)
=
;
this is per-
hence
2tc,
2
l
+ m(y
t.
co,
1
co
;
cos co
and the equation of the normal takes the form
2
or
a'^ -cosco)
+ yt(t*
cosco-
1)
= c(
4
-l).
a general rule questions involving the equation of the normal are more conveniently treated by referring the hyperbola to its
As
axes.
Special case of the Rectangular Hyperbola. co
=
In this case
90, and the equation of the normal becomes 1
^-.tysrcft -!), or
ct*
(a)
Concurrent Normals.
xfi-\-t
c
If the
=
0.
normal
at the point
rectangular hyperbola passes through any proposed point
have cf*-ft<
3
(A,
t
on a Jc)
we
+ j-c = 0.
Hence, conversely, this equation gives the parameters of the feet of the normals which meet at (h, Jc). Since the equation is of the fourth degree four normals can be
drawn from any point
to a rectangular hyperbola.
CENTRAL OONICS
346 (b)
Conditions that the normals at any four points on a rectangular
hyperbola should be concurrent.
whose parameters must satisfy c 4
If the normals at the point concurrent, these parameters
some values Hence or,
of h
and
are ^, hfi
f,
+ Jct
^3
2
c
^4
are
=
for
t.
2^
2
=
and
t
l
1,
t^t^t^-=^
written otherwise,
v^^v.l
=0>
which are the required conditions. If these conditions are satisfied, the coordinates of the point of
intersection are then given
h
by
=c2
Example. 2%e is
hyperbola
on
1,
C~
*
!
C-|
j
2
orthocentre of a triangle inscribed in
a rectangular
the curve.
Let the parameters of the vertices of a triangle PQR inscribed in the 1 rectangular hyperbola x\j == c be t l9 f a / 3 Suppose that the straight line P to meets the curve at the point T whose QR through perpendicular is the then lines t+ parameter ,
.
;
1. are perpendicular, hence t^t^ = that the pairs of chords PQ, RT&ndPItj
The symmetry of the
QT
result
also perpendicular,
shows
i.e.
Tis
the orthocentre of the triangle PQR. may notice that the feet of the four normals from any point to a rectangular hyperbola form a triangle and its orthocentre.
We (c)
The
of four concurrent normals
feet
lie
on another rectangular
hyperbola.
If the
we have
normal
at
any point
(ct,
-)
passes through the point
(7r,
&),
seen that
or
c t
Thus through
must at (h,
if (x, y)
7c)
in the present case
if
x
=
any normal passing
ct,
y
=
-,, t
then
(#, y)
hx + fcy # = 0, i. e. the feet of the normals meeting on the rectangular hyperbola a?2 # 2 hx + Jcy = 0, which
satisfy x lie
are the coordinates of the foot of i.e.
(h, &),
t
2
also passes through
2
(A,
A;).
THE ELLIPSE AND THE HYPERBOLA (d)
To find
the conditions that the
normals
at the ends
347
of the chords
of a rectangular hyperbola ivhose equations are
= 0, -1 = \
should be concurrent.
The points of hyperbola
lie
and the rectangular
intersection of these chords
on the conic
A(^-c2 + (te + w#-l)(^W#-l) )
='0,
(i)
and consequently, if the normals at these points are concurrent, some value of A this conic must be the rectangular hyperbola,
for
= 0.
x2 -y*-hx + ty
(ii)
2 Comparing the coefficients of x and # 2
IV =.
mm',
the chords are perpendicular, and also, since the coefficient of xy and the constant term in (ii) are zero, i.e.
lm'
-f
=
I'm
A
=
'
c
The
2
required conditions are therefore tt'
c
/
2
(/w
+ mm' + n)
= =
0,
-l.
the coefficients of x and y the coordinates of the point of intersection of the normals can be found in terms of the coefficients in the equation of either chord, thus
By comparing
7
/*
11,11 =
= +p
-- ---
K
-,
1
in
I
I
and the conditions above give
m
,
>
and w' in terms of
/'
I
and
*;z,
or
vice versa.
Cor.
If
one chord
is
the tangent at the point a?
a
+ < y-2cf
=
t,
viz.
0,
the other chord must be
2^-2ty-c(l-* = 4
)
and the point
(vi)
The
(h, k),
which
is
O
the centre of curvature at the point
intersections of the hyperbola
The equation
of re
Substituting
2
a;
circle is of
any
+y
=
2 4-
rf,
2 xy c o s o>
y
f
=
and a
t,
is
then
circle.
the form
+ 2 gx + 2/y + d
= 0.
y :
we
obtain an equation giving the
CENTRAL CONICS
348
values of the parameters of points on the hyperbola which also on the circle this equation is
lie
;
If the values of
t
given by this equation are &!
This
is
2
3
4
=
ly
2
>
^a
^4
we have
1.
the necessary and sufficient condition that any four points x 2 ^3 ^ should be concyclic.
whose parameters are Note.
r
t
,
>
>
If the points A, B, C,
D on
D
a rectangular hyperbola are concyclic, ABC are extremities of a
and the orthocentre of the triangle Vide Example, p. 346. diameter.
then
Circle of curvature.
If the circle
= is
the circle of curvature at the point (ct l9 j\ then three of the values v *v
of
t
given by
must be
t
lt
for the circle intersects the hyperbola in three coincident
points at the point of contact.
Let the values of
i.e.
t
given by the equation be
the circle of curvature at the point (ct lt
t l9
t
l9 t ly t z
) cuts the
,
then
hyperbola
*v
again at the point
The equation of curvature at
or
Further,
(-7-3,
of the
^
,
i.
e.
c^
8
).
common
chord of the hyperbola and the circle of the chord of curvature, is
THE ELLIPSE AND THE HYPERBOLA Hence the equation of the /!
+
curvature at the point
circle of
O
4
349 t
is
O
^
When the hyperthe of becomes circle of curvature the rectangular equation
Special case of the Rectangular Hyperbola. bola
is
= 0. (a)
The centre of curvature
(b)
The equation of the evolute,
curvature, can be found thus
and (c)
the point
is
viz.
the locus of the centres of
:
(x
The radius of curvature
(p)
at the point
Illustrative
Examples.
t
is
given by
i.e.
Example
^4.
i.
triangle
is
inscribed in the hyperbola
that its centroid is a fixed point on the hyperbola
touch an ellipse which touches the asymptotes
Let the vertices of the triangle be (ct^ .
fixed point
f cc?,
c
and
( ct n" \
^J
^-Kj + ^=3d,
Hence
A side Now
),
M/
\
v
and
of the triangle
is
x + t^y^c *1
+
fa
4f,
=
(^
-f fa ).
3eZ-/8 3
1
,
ocy
show that
:
=
c
2
so
its sides
the hyperbola.
,
),
**'
(
\
rt s
,
) '.I/
and the
CENTRAL CONICS
350 The equation
of the side of the triangle then becomes
t(c-dy+
.e.
For
values of
all
.e.
*3
this touches
(
i.e. 2
(3^-f3(^t/-8ccZ) -4^a?y.
i.e.
The same equation
x
=
0,
y
(i)
=
0,
(i)
evidently true for the other sides of the triangle now from its form represents a conic touching the asymptotes
result
is
:
the chord of contact being
3a?-f3d 2 t/-8cd!
=
0.
The terms of the second degree are hence the asymptotes are parallel to
i.
e.
are imaginary
Now
:
the curve
is
therefore an ellipse.
the parameters of the points of intersection of the ellipse
hyperbola are given by substituting x
T-
= 8
=
- in this equation,
and the i.
e.
*/-
-8d =
3*+
Hence
y
ct,
(i)
+ 2
~~
t
o ^2
Taking the negative sign
3
.e.
-- 6d =
1 4-
0,
t
i.e.
2
(^-^)
=0;
meets the hyperbola at two coincident points at the given fixed point and consequently touches it there. hence
it
Example
ii.
If
the circle circumscribing the triangle formed ly the
tangents at the points (x lf
passes through the centre xt
y^
(# 2 ,
+ xi + xz Xix^x^
and
the centre
of the
# 2 ),
(# 3
,
y^)
on a rectangular hyperbola
of the curve, then
circle is
+ yi
+ ys + y* y \y-2V 3
=Q
THE ELLIPSE AND THE HYPERBOLA
351
Let the parameters of the points of contact of the tangents be
t
lt * 2
f
,
3
;
=
and write SjEESJ, sa 2Va> $5=: Ma 3The coordinates of the vertices of the circumscribing triangle are
where
t
2cs s
2c
'(i-
*i-~*'
has either of the values
A
circle
If
any one of the
whose centre
is
(g,
t
/),
lt t99
or
fs
.
and which passes through the on this
vertices of the triangle lies
~ (A ~ 98* -c)t*~ gsi s3 t + cs3 2 = ft* But >
this condition is satisfied
when
then
circle, 0.
equal to
is
t
origin, is
t
or
lt /,,
f
3
hence
;
are the roots of this equation. It follows that *a> ^s
(i)
So
,.. x
(n) 8a== -0*,-*
=
.
i.e.
7
(iii) 8l
= *! -
3
^r /
CSn
--; 5i
/^,
i.e.
-
C -
flr
*3
Hence the centre of the
circle is
-,
(
\
s
+ 'i^
These
-^
s
*s
with the condition
), /
0.
results are identical
with those required, since x l
= ct
lt
t/i
=r r
&c.
i
Examples VIII d. 1.
is
The tangents
at the extremities of a chord of
#y
=c
2
whose mid-point
,
(X, Y), intersect at the point {c'/r, c*/ 2. The equation of the director circle of the hyperbola xy 2
a
# + ^ + 2ojy cos o>
=
=
2
c
is
2
4c cos o>.
What 3.
does this become for a rectangular hyperbola ? Find the locus of the intersections of perpendicular straight lines which
are tangents respectively to a rectangular hyperbola and 4. The normal to the rectangular hyperbola xy 3 c? ). meets the curve again at the point (-c/f8 5. The normals to the rectangular hyperbola xy chords whose equations are
c
1
its
conjugate.
at the point (ct,
j
,
= = ycostf
x cos 6 + y sin
c,
a?sin#
c
cos
=
c
2
at the ends of the
20
are concurrent. 2
the chords c There are four points on a rectangular hyperbola xy of curvature at which are concurrent, and these points are concyclic. 7. The sum of the'squares of the lengths of the normals which can be 6.
drawn from a point
,
P to
the rectangular hyperbola xy
=
c
2
is
3CP
2 ,
CENTRAL CONICS
352
Find the locus of the centroid of an equilateral triangle inscribed
8.
in
a rectangular hyperbola. 9. The polar of any point on an asymptote is parallel to that asymptote. 10. Pour points A, B t C, D on a rectangular hyperbola are such that the
AB CD are perpendicular. AC to BD.
straight lines to BC, and 11.
9
Show that
AD
is
perpendicular
Find the equation of a pair of conjugate hyperbolas referred to
common
a
pair of conjugate diameters. CP, Cp are conjugate diameters of two hyperbolas, P being on one, p on the other. Find the locus of the orthocentre of the triangle PCp. 13. Show that the envelope of the chords of the rectangular hyperbola 2 xy = a which subtend a given angle OC at the point (#', y') on the curve 12.
the hyperbola x*x'* + y*y'* = 2a'2 xy(l -f-2cot a)-4a 4 cosec*#. 14. A circle is described having its centre at a point P on a rectangular f hyperbola and passing through the diametrically opposite point P on 2
is
N
the hyperbola. Prove that, if L, M, are the other three points in which is equilateral. the circle cuts the hyperbola, the triangle
LMN
15.
A
normal
to a rectangular hyperbola
the transverse axis.
makes an acute angle $ with
Prove that the acute angle at which
it
cuts the curve
cot"1 (2 tan 26). again 16. If the position of a point on a rectangular hyperbola is determined ctan 0, y = ccot 0, the locus of the intersection by the variable 6 where x is
=
of tangents at the points
17.
0,
O+
Ol,
(X
being a constant angle,
is
Prove that the locus of the mid-points of chords of the rectangular = c 3 which are of constant length 2/ is
hyperbola xy
18. The sides of a triangle ABC, inscribed in a rectangular hyperbola, make angles Of, |3, y with an asymptote. Prove that the normals at A, B, C will meet in a point, if cot 2 & + cot 2 + cot 2 y = 0. 19. To a rectangular hyperbola with centre C and focus S normals are drawn from a point P. Show that, if these normals make angles 19 2 ... with one of the asymptotes 2 cosec 20 = (2 CP 2/CS 2 ). 20. The normal at P to a rectangular hyperbola whose centre is C meets 2 2 the curve again at Q\ show that PQ = 3CP + CQ\ 2 at P meets the 21. The normal to the rectangular hyperbola xy = c curve again at Q and touches the conjugate hyperbola; show that ,
a diameter of the hyperbola #Ya 2 -y 2 /& 2 == 1. A straight line is drawn through P parallel to one asymptote, and a straight line through P' parallel to the other asymptote show that the locus of the intersection 22. 'PP'
is
;
=
1. of these straight lines is the hyperbola y*/b*x*/a? and A' are the vertices of the hyperbola xt/at-if/W 23. If 1, and and the if PA' at and on meet an P-4, it, points asymptote Pany point
A
X
F,
show that
XY=
THE ELLIPSE AND THE HYPERBOLA 24.
line is
The
Show
constant.
is
AB
straight line
is
drawn, and two points P,
BQ A
bisected at
Q
C through C ;
are taken on
353
a fixed straight
such that the distance
it,
PQ
that the locus of the intersection of the straight lines
an hyperbola. and B are fixed points, and AC is a fixed straight )ine. If a line drawn through B meet AC in Q, and a point P be taken on this line, produced if necessary, so that PA = PQ, show that the locus of P is an hyperbola whose centre is the middle point of AB. 26. Show that the polar of the origin with respect to the circle of curvature at the point (a?', y') on the rectangular hyperbola xy = a 2 is
AP,
is
25.
y(x*-\ 3aY) = 6a (#' + y"). 2 27. The tangent at any point P of the hyperbola x^/a^-if/b = 1 meets and M. Find the equation of the circle OLM in terms the asymptotes in
*(3aV +
/3
t,
a
2
)-f
of the coordinates of P, and deduce the locus of the centre of this circle. Explain the result of putting a equal to b in the equation to this locus. 28. Find the equation of the normal at any point of the rectangular
2 show that from any point in its plane four noimals can hyperbola xy = c be drawn to this hyperbola, and that if x ly x2 # 3 x4 be the abscissae of the feet of these normals x x,i x3 xi c* = 0. 2 29. The normal at P to the rectangular hyperbola xy = c meets the curve :
,
,
-
1
again at Q.
If x,
y are the coordinates of
P
and
,
r;
those of Q, prove that
30. If the tangent at the point (h, k) of the hyperbola meets the asymptote bx = ay at the point 3/, and the asymptote bx + oy at the point and S is a focus of the hyperbola, show that
=
N
31. From a point P are drawn two tangents to a rectangular hyperbola. The tangents of their inclinations to an asymptote being p and g, show that, if tne ratio of 1
pq
to
(\/p~
2
v'*?)
*8
constant, the locus of
P is
another
rectangular hyperbola. 32. Show that the lengths of the tangents from P(x, y} to the rectangular are given by hyperbola f=xy-c* =
and show that the lengths of 10.
In Chapter VII,
S'P, S'
9,
P are
we
factors of the absolute term.
discussed the forms of the equations
of several loci related to the parabola: these forms clearly apply equally well to any of the conies ; thus throughout the section we is the equation of a parabola, where S may substitute S for
=
P
an
ellipse, or
Example ellipse
#
an hyperbola. i.
2
2
-f-32/
A
=4
circle is described
as diameter.
on
Find
the cJwrd
# + 3#
the straight
=
i
f
O
line joining
the the
other tivo points in which the circle cuts the ellipse.
Let the equation of the other
common chord my r 1 = 0.
Ix +
of the circle and the ellipse be
CENTRAL CONICS
354 The equation of the
form
circle is therefore of the
= The two conditions
X+
Z=3X + 3w
m = -3/
i.e.
= 0, = Z-3w = l(H,
and
and
2X
=
X
or
The equation
3/
+w
5f.
of the circle then becomes
2 2 5J(# + 3
Its
centre
is
il-l
the point
_ ~
this
lies
by hypothesis
+
1)
-
0.
3|
rr
on the diameter
x-\
3y
=
1.
J-l-9/-9 = 12J; = -4.
Hence
I
.-.
The
3/
"12
\i2~r and
0.
for a circle give
equation of the circle
is
then
2 2 5(# + 3y -4) +
(a;
=
+ 3y-l)(.r-3y-2)
2ar-f2t/ -;r-j/-G = 0, equation of the chord is x-%y =
0,
2
or
and the required
2.
ii. Find the equation of the parabola tvhich toucJtcs the 2 2 in the points in tvhich 3^ -f 2xy y + 8x+ 10^-f 14 = hyperbola
Example
it is
met by the straight
line
5x~y
2
=
0.
The equation of the parabola is in the form S == fcw 2 i.e. 3^ ? -f2a:y-t/ 2 -f-8a;-hlOy-f 14 = k(bx-y-2)*. The condition for a parabola is ab h 2 = 0, ,
(25fc-3)(fc4-l)(5fc-fl)
i.e.
25P-f22A:-3
i.e.
=
25A;
2
-f
2
or
8o;
Example
iii.
-8
Find
i.
=
;
# _ _ 2) 2
(5
*,
+ 2f/ -22a:-13y~19=
xyc ~Q 2
at the point (ct, -
of the tangent at the point f ct, ~
The equation
of the parabola 2 xy - c
is
S= (x -f t* y 2 ctf.
of the form
= lc
,
0.
of the parabola tvhich has four-point
The equation
o.
1
2
the equation
contact tvith the hyperbola
,
10^-f
The equation of the parabola is therefore 3(3** + 2#y-t/ -f 8.r + 10y + 14) 2
2
is J
ku* (Chap. VII,
9, v),
THE ELLIPSE AND THE HYPERBOLA The condition
355
for a parabola gives
=
k
Hence the required paiabola
is
4*V*/-c or
(^-^
2
2 i/)
2 )
-4c^-4d
3 f/
f
8c 2 ^ 2
=
0.
Confocal Conies. We have shown that the foci of a conic axes and are equidistant from the centre hence if conies have the same foci SS' their axes lie along the same straight lines and they have a common centre C. 11.
lie
on
its
Now CS 2
=
;
if
a2
20~ =1 -I-
~2 b
2
and consequently
,
are such that a 2
A
be any conic referred to
b2
conic, then,
=
constant
conies 2
whose equation x
2
aH~A is
all
=c
+
is
its
axes
we have
which have the same
foci
.
of the y*
form
_
&*TA~
confocal for different values of A with the conic
and this equation therefore represents a system of confocal conies. The conic is an ellipse or hyperbola according as the value of A chosen makes a 2 -f- A, & 2 -f A both positive or of opposite sign. Proposition 1. Two real confocals of a system pass through any an ellipse and one an hyperbola.
real point, one
be a given point then to find the values of A for those conies of the system which pass through this point we have
Let
(x', y'}
If A i.
e. is
=
a 2 the left-hand side of the equation becomes (a 2 ,
& 2 )#'-,
positive.
= 6 A = -f oo, 2
If A If
;
,
it
becomes
it is
2
(a
6 2 )^ /2, i.e, is negative.
positive.
a2 Hence there are always two real roots, one lying between 2 2 and & and one between 6 and +00. Thus a 2 A, 6 2 -f A are for one value of opposite sign, and for the ,
-I-
other both positive. z 2
CENTEAL CONICS
356
Hcnco there ellipse
are two real confocals passing through and one an hyperbola,
Proposition
One confocal of a system
2.
(x' 9 #'),
one an
touches every real straight
line.
Let the equation of the straight line be touches thu conic x"/(a? + \) + y 2 /(b* + \)
=
1,
lx
+ my + n =
this
;
provided that
This equation gives one value of A corresponding to the one conic of the system which touches the straight line. The equation of this conic is
x1
if
and since n 2 + (a 2 b 2 )m 2 is always positive the conic is real an ellipse or an hyperbola according as n 2 is > or < (a*& 2 ) i 2
it is
;
Proposition
Any
3.
tivo
.
confoeals cut at right angles.
Let the two confpcals be
If these conies intersect at (x l9 yj,
we have
Subtract these equations and divide through by
(A 2
^V^ + ^^ + ^+yiVl^ + A^^ + A,) = which
is
y v ) to the two confocals should be
of confocal conies point of contact.
>
y\)
Ix
b
then
0,
lie
be the equation of the straight line, and with respect to any conic of the system pole
+ my + n its
at right angles.
The poles of a straight line with respect to a system on the normal to the confocal -which it touches at the
Proposition 4.
Let
;
the condition that the tangents
at thu point (x i9
(x\
Aj)
=.
Hence te + wey + w tical, and therefore arj
Eliminating
A,
straight line, viz.
=
= we
Z
and ^,/(a2 + A) + yyi/(b* + A)
0,
2
(a
+ A)/
yx
=
1
let
are iden-
= - m (6 + A)/n. 2
get the equation of the locus of the poles of the
mnx l
lnUi
+ (a 2
6 2j
2m
=
0.
THE ELLIPSE AND THE HYPERBOLA
357
This straight line is perpendicular to the given straight line also the pole of the given straight line with respect to that conic of tho system which it touches is its point of contact hence this point of :
;
Thus
contact lies on the locus found.
given straight line
is
system which
of the
the locus of the poles of the point of contact to that conic
the normal at its touches.
it
Note. If the given straight line passes through the centre of the confocal c conies, its pole with respect to each conic is a point at infinity'. If is such a straight line, the conic of the system which it touches Ix + my
=
is (see
Prop. 2) 2
(I
i.e.
+ ma )
2
(I
x*
- mV) =
I
w
2
2
2
(a
- & 2 ),
the straight line is one of the asymptotes of the conic and is a point at infinity '.
Proposition respect to
Let
5.
(x l ,
This
The
point of
envelope of the polars of a given point with
a system of confocal conies ?/})
be the given point
conic of the system
may
a parabola touching the axea.
then
;
is
rc
its
polar with respect to any
2
2 /
(a'
+ X)-f y 2 /(b 2
-\-
X)
=
1
i*
be written 2
-
1
The equation
//')
of the envelope
(Wi + ?/'/i at
is
whose equation
\a _ A (jvrj 4 yy -
which
its
*
contact
- a* -
2 2 ?>
)
- (a*yy
is
+
*(
2 + V .^
{
- a2 6 2 = )
0.
therefore 2
Wi + &
-
2
=
0.
2
**\
W=
0,
once reduces to
+ 6 ) + 4 s y v ry (^i - lIVi The latus rectum of the parabola is
-^
2 2
{
t(<*-W)wM*+y?)\ and the equation of the axis (**i
is
+ yi/i) (*i* + 2/i 2 + (? ~ & 2 )
2 )
(y i
- ^i = 2
)
J
these results are left for the reader to prove.
Proposition draivn
to
6.
Tlie
bisectors
of
the angles
a conic from any point are the tangents
between the tangents
to the confocals
the point.
Let the point be
(a cos 0, frsin
x*/a*
and
let
6),
which
+ y*/u*=
tangents be drawn to the conic
1,
lies
on the conic
through
CENTRAL OONICS
858
If the straight line lx
and, since (acos0, &sin
+ my + n =
so that
(al
or
cos 6 +
on the line
0) lies
al cos 6
lm sin
is one of these tangents, then
= = Pa* + m
+ lm sin 2
0)
n, 2
62
+ A (Z 2 + m 2
0- ton cos 0) + X (I + w 2 = 2
2
(a? sin
)
)
0.
This equation gives the values of the ratio / m corresponding to the directions of the two tangents. Thus the equation of the :
straight lines through the origin parallel to the
two tangents
is
For all values of A these have the same bisectors of the angles between them, viz. Ix cos
+ ay sin
=
and
ax sin
fy/
= 0.
cos
The
bisectors of the angles between the tangents are the straight lines through (a cos 0, I sin 0) parallel to these, viz.
bx cos 6 + ay sin
=
ax sec
al),
l>y
cosec
=
a2
two
2 fr
,
=
1 at the the tangent and normal to the conic # 2 /a 2 + 2/ 2 /& 2 7; sin cos cut at since confocals 0, right angles, they or, 0) point (a i.
e.
;
are the tangents to the confocals
which pass through the
point.
Cor. Let PQ, PR be the tangents from P to a conic, PX, PY the tangent and normal to a con focal conic through P. We have shown (p. 308, Ex. 5)
PY
that PX, bisect the angles between SP, S'P\ it follows therefore that the tangents from any point to a conic are equally inclined to the focal distances of the point. (7)
If x +yi
confocal
Now,
= c cos( +
hyperbolas and
17 i),
then
=
constant
and
ellipses.
since
x + yi
=
ccos(-Mji)
=
c cos f cosh
v\ic sin
TJ
= constant arc
THE ELLIPSE AND THE HYPERBOLA
359
we have
=
x Eliminating
wo
T)
cosh
c cos
y
rj,
=
_
-
_ c
a
2
+ sin Again eliminating we
bolas, for c (cos
=
2
)
c
2
or
which equation, for of the same system,
/n
'
represents confocal hyper-
,
OS =
c.
find
__.
._.. 2 2 cosh
c
u
1
sin a
for different values of
2
?/.
find
c*cos*
which equation,
sinh
c sin
c
?j
2
sinh 2
different values of
for c 2 (cosh 2
represents confocal ellipses
?j,
sinh
7j
'
?j
=
2 ?j)
c
2
=
C
or
c.
Evidently then the equation x l + y l i = c cos (ft + i?i i) is the conconfocals =fi, 7 = 7?i, f/j) should lie on both the
dition that (^,
/
whose equations
are
r2 ^
a
r2
7/
.
2
-f
(8)
confocals
o;
2
correspond (a)
Two
Definition. 2
/ if
2
+ ^/ /^ xja =
P^,^)
points
=
2
and
1
a;
2
and P'(XI, y/)
2
/V + 2/
2
/&
/2
=
on two
are said
1
to
^iVa'
I/ P, P' are too corresponding points on two confocals then y
P
of the system which passes through
the second confocal
passes also
through P'.
Let the system of confocals be defined by x + yi = ccos(-M/i),
and
let
the coordinates of the given points be P(rr lt y t ), P'l*/,
Then
=
if
sponding points
t
=
,
P and P
= t and Let which pass through i]
r]
Xl
+y
l
i
=
c
TJ
are *^e
2 r
lie,
= ?]/
P and cos (^
two confocals on which the
we have by
2/iO*
corre-
definition
be the second confocals of the system thus
P' respectively
:
^/ + y / i
+ 7h i),
=
c
cos (f 2 + )//
i),
and therefore ^t
But ^/cos & confocal
rj
= ?^.
rr C
=
COS ^ t COSh
^Vcos
2
;
f] l
,
iT/
hence
7
= /i
C
=
COS 7
h'
2
COsh 7/j
anj
p
>
7 .
?'
lie
on the
CENTRAL CONICS
860
Ivory's Theorem. // P, P' and Q, Q' are two pairs of P'Qsponding points on tivo confocals, then PQ' (b)
corre-
=
Let the system of confocals be defined by
x+yi and
let
=
ccos(-M)i),
the given points be
Pfa, P,
Q
that P, P'
lie
on
Then,
?j
=
Qfa,yj,
P'(*i',0i'),
lie
if
on
=f
=
and similarly
rjj
c cos
and
t
we showed above = 2 Hence
2,
on
lie
ft Q'
rj
= c cos
#/ + ?// i
(& + >?! i),
Q'(r*',y*').
=
on
P', Q'
r?
.
(
Then
= c2 cos (f HMhO-- cos(f + %i) = 4 c sin & (f - ^ + Ty.>-i?!) sin {
2
j
2
?
2
=c Now
{
cosh
cos (^
.
(7 ?2
sin
r/ t i)
--cos (f2
.
{
sin I
Example
// a x
r/ 2
?)
(+f -
J
f- ?/ 2 )
(r/j
)
2
- cos (f + &) l
P'Q is obtained by interchanging which evidently gives us PQ' = P'ft Illustrative
-
2
?/
;
-
2
the value of
this result,
{
- ^ - i 2 - 7^) 2 - 7^) - cos (^2 - &) cosh -
2
}
^
and
2
.
\
in
Examples.
a 2 are the semi-major axes oftivo cow/cs, confocal 2 2 tvith the ellipse x*/a? + y /l 1, w/nWi can le drawn through the point x find x \ an d !/i in terms of a l and a 2 ( u yi) i.
,
=
.
If 2 (a,
fc),
Z>e
f^e aw^fZe letiveen the tangents
2 2 2 2 prove that a 1 sin (/)-ha 2 cos
^=a
from
(x t
,
^j) to thz ellipse
2 .
conic coiifocal with the given ellipse is # 2 /(a 2 + X) + and the values of X for the conies of the system which pass through 2 2 1. x*/(a? + \] +y /(6 -f-X) fo, f/ t ) are given by
Any
=
1
Let a 2 -ft 2
=
c
2 ,
and
let
Of.
stand for either 04 or a 2
a 4 - a 2 (^2 +Vi 2 + c 2 ; 4- c2 ^ 2 =
or
;
tnen we have
0.
Hence a^ + a^.^' + yj' + c', and
iV - ^ 2
c'
Thus or
x
l
= a! yi
fl
=
and
^
2 /c,
2
2
yi
-
(i)
2 .
(ii) 2
( ai
- c 2)
-c2 )(c2 ~a 3 2 )/c
=
2 (
2
- r 2 )/c 2
bc.
,
_
THE ELLIPSE AND THE HYPERBOLA We showed
From
(i)
From
(ii)
(p.
311) that
^
* a2 y
a2
2
2
-a
a 2 a^ + a 2 a 2 ?
*
t
cfxf-Vxf = a^^
2
2
o^
2 .'.
!
(1 .'
-~1
(
Note
i.
Note
ii.
We
- cos 2 0) a
.
- a2)
If Xj
,
>2
-a
4
a/
(
?
4-
>a
r/
2
;
2 2
)-(a -a 2 2
j
2 ) ,
1 4-
cos 2
cos
2
= 2a
/>)
=* a 2
2 ;
.
.
2
be the values of X for the two confocals through (x l yj, ,
we have
2 !
so that
=
2
2
+ X lt
tan 2
thus
-a 2 )-(a 2 -a 2
- a2 2 )
2
(a
r^"~5
sin 2
have taken a l
.
.
(a 1
cos 2
-f
2
(a^
u Hence
a2 62
4
361
=
2
o
^2
/H)
^
= v/~ A,/X
tan
= ~
,
x2
^-
or
2
+ X2 ;
\/-^2/^i-
Note iii. Since the angles between the tangents from any point to a conic are bisected by the tangents to the confocals through this point, the equation of the pair of tangents from P to the conic (, I) referred to the tangents at P to the confocals through 2 2 may be written ^ /A a -f ?/ /A a = 0.
Example
If
ii.
the
meet the polar of P with regard
in Q,
and ifCY
prove that Let
P
P
normal at to
be the perpendicular
from
2
y
-#
2
ellipse
tan 2
x 2 /a*
=
0,
which
+ y-/tf =
the centre
on
the tangent at
PQ.CY= A.
be the point (a cos#,
fcsin 6)
;
the polar of
s
The equation of the normal
at
P
^acpsJJ _ ~
is
y_
bco*0~
is
the
to
is
any confocdl conic
L
and r
P
equal to
PQ
if
(a?,
y) lies
on
(i).
"
I
P
with respect to
1
CENTRAL CONICS
362 Thus acosfl
r
f
gives the length P@,
rab
i.e.
{X
cos
(
2
sin
+ a 2 sin 2
+
2
But
CT^-r;
Example
2
cos 2
& sin
tf
~
-f
2
""
X a 2 sin 2
2
tf
"
2
-f
X & cos
2 ;
-X.
r.
..
2
""
=X
cos2 0}
2 fc
_a
_ ~
.-.
From a
iii.
on a conic tangents are drawn
fixed point
any confocal. Prove that the line joining the points in which they meet the conic again passes through a fixed point 0'. to
If the
conic
x* -5 a
is
2 ?/
+
72 o
=
^ie
*
^
fO'
ocus
moves
as
is
Let the tangents from to the confocal meet the conic again in P and Q that the bisectors of the angles between OP, OQ are the ;
we have shown
tangent and normal at 0. Let PQ meet this tangent and normal at G and 0' then since the pencil OG 00 '; OP, OR is harmonic, the points P, Q are ;
t
harmonic conjugates of G, 0'; hence the polar of 0' passes through Q and is the line OG. Thus PQ always passes through the fixed point 0', which is the pole of
Let
OG with
respect to the given conic. 0, b sin 0), then the equation of
be the point (a cos
a# sec 6 Hence,
We
if 0' is
the point
(a?
n
=
&y cosec
a
t^), this equation
xxja
2 -f
yy^W =
OG
is
2
2 fc
is
.
identical with
1.
have therefore 2
(a
- 62 ) #!
whence, eliminating
Example
iv.
0,
a 8 sec
we get
P/we Ma
angles can be drawn
to
9 (9,
(a
- 6 2 ) ^ = - 68 cosec
the locus of
through any ioint
form
ivith
the
is
~
-I,-
/
^T2\ 2
(&
'
(/, #)
polar of
triangle self-conjugate with respect to x*/a*
area of the triangle
<9,
0', viz.
two lines at right
(/, g)
+ y*Wl =
a right-angled 0, and that the
Where A lf A 2 are Me roote o/
+ A) - 1
=
0.
THE ELLIPSE AND THE HYPEEBOLA
363
Let P be the point (f, g), and Ijet the tangents to the conies, confocal with the given ellipse, which pass through P meet the polar of P in Q and R. We shewed in Ex. iii that E is the pole of PQ, and since QTt is, by conis struction, the polar of P, the triangle self-conjugate it is evidently at P. The of the of the triangle ere sides equations right-angled
FQR
;
)
2
)
1
=
-1 =
0, 0,
0,
where X lf
X a are the parameters of the confocals through given by the equation
P
and therefore
//(' + M + 0V( + A) =1. Tho
vertices of the triangle are the poles of the sides with respect to the
given ellipse,
viz. \
2
The area of the
triangle '
2
Now
X lf
A2
is
_ 2
(a
X2
then
_
+ Xj)
2
(a
+ >,)
2
(fc'TAj)
(fc
-f
A2)
are the roots of the equation
hence
and i.e.
the area of the triangle
is
/<;(*'
-
Examples VIII e. Find the equations of two confocal conies whose foci are and which pass through (2, 3). 1.
(1, 0),
(-1,
0),
2. Prove that the locus of points on a system of confocal ellipses, which have the same eccentric angle a, is a confocal hyperbola whose asymptotes are inclined at an angle 2 OK. 3. If two tangents to an ellipse are at right angles, the envelope of their chord of contact is a confocal ellipse. 8 a a Through a point on the director circle of the ellipse # /a -f y*/& = * two conies are drawn confocal with the ellipse: if 2a l9 2aa are their transverse axes, show that Oj'-f a a 2 = 2 a 2 5. The difference of the squares on the perpendiculars from the centre to two parallel tangents to two given confocals is fixed. 6. Find the locus of the intersection of two orthogonal tangents which are drawn one to each of two confocals. 7. If any two parallel tangents to an ellipse meet a fixed circle concentric with the ellipse in P, Q and P', Q', prove that PP', QQ' touch a confocal
4.
.
conic.
CENTRAL CONICS
364
8. If the points of intersection of two confocals lie within the circle described on the line joining the foci as .diameter, the minor axis of the ellipse will be less than the conjugate axis of the hyperbola. 9. If from P, a point on an ellipse whose semi-axes are (X, /3, two tangents are drawn to a confocal ellipse whose semi-axes are a, J>, show that if 0, are the eccentric angles of the points of contact of the tangents, and r is the distance of P from the centre, then
= a -a 2
where X
a?
2
3
/a
-
= Ot
)
2
^
2
- X >*
P
an ellipse and an hyperbola are drawn confocal find the length of the semi-diameter of the ellipse
+ ya /6 2
=
1
CP
in
terms of the parameters of the two confocals.
conjugate to
A
11
(6
.
Through a point
10.
with
a 2 V sec 2 }
2
point the normal at
;
Pis taken on the
P may
conic
# 2 /(a a + X) + ya /(6 a + X)
pass through a fixed point (h
t
=
Show
k).
such that
1,
P lies
that
on
the curve given by
P is
any point and
of # 2 /a 2
+ t/2 /6 2
=
1, show that PS. PS' equal to the difference of the squares of the major semi-axes of the two conies which can be drawn through P, confocal with the given
12. If
S, S' are the foci
is
conic. 13.
PP'
is
DP, DP' touch a confocal then DP DP' sin 2 PDP' = 4X. that
D any point
a diameter of an ellipse, and
if
ellipse
on the curve.
whose semi axes are \/a2
^>
Prove \/&
a
~ \,
.
T
are the poles of a straight line with respect to two confocal conies whose semi-major axes are a, a', and p is the perpendicular on the 2 2 Show that p. a' -a straight line from the centre of the conies. S19 S% are the 15. Pis any point on the director circle of a fixed conic S. 14.
T,
TT
.
twc conies through P confocal with S. Show (i) that the squared of the lengths of the major axes of the three conies are in arithmetical progression, (ii) that the product of the lengths of the major axes of Si and S2 varies as the distance of the point from the minor axis of the system. 16. The locus of the centres of curvature at ends of equi-conjugate diameters of ellipses of the confocal system (a 2 -f^)~ 1 a; 2 4-(fe 2 -f^)~ y 2 = 1 is = (a2 -62 ) (t/-* 2 ). the curve 17. Prove that the polars of a point (#', /') with respect to the confocals 1
8#V
#a /(aa + X)+yV(&2 + A ) 18.
A
is
=
*
drawn
tangent the coordinate axes.
to 11011
tfl e
conic
to a rectangular
Its poles
2 \/-0y'+ vV-fc
=
0.
hyperbola whose asymptotes are
are taken with respect to a series of conies
#2 /' 2 + y*/b' 2 = 1. Prove that, the polars of all these poles 2 2 = 1 meet in a point also that the to the conic # /a' 4 y^/b
confocal with
with respect locus of such points as asymptotes, 19.
"*
;
is
a rectangular hyperbola having the axes of coordinates
and conjugate
Find the condition that
to the given 2 a?
a
/a
+ i/
a
4a /a & /2 = (a 2 -fc2 ) 2 and Ax* + By* + C = may
hyperbola a
/&
1,
if
.
represent confocal ellipses. Points P, Q are taken on confocal ellipses, such that their distances from
THE ELLIPSE AND THE HYPERBOLA
365
the minor axes arc proportional to the major axes of the ellipses on which they respectively lie. .F, Q' are another such pair of points. Prove that
PQ' 20.
i"Q.
Show
that the centres of curvature at the points where y = 1 lie on the curve
= mx
meets
the conies confocal with x^/tf + if/V1
y
9
{a?
-3w3y-a*+6
8
a
}
+
e
[m
2 (f/
+ a2 -& 2 )-3;r;/} 2
=
0.
m
a 2 2 an y point Tangents are drawn to the ellipse ar/a -f f/ /& ^ 1 fr on the confocal hyperbola whose asymptotes lie along the equi-con jugate diameters; show that the centre of the circle inscribed in the triangle
21.
formed by the tangents and the chord of contact 22. Show that the two curves
*V 2 + W*> 2 =
lies
on the given
and x'/a~ - // 2 /6 2 - (a 2 - & 2 )/(a 2 -f b 2 ) + 2 X jcy
1
ellipse.
=
intersect at right angles. 23.
Find the locus of points on conies of the confocal system
the tangents at which make an angle Of. with the axis of x. Show that the tangent to the conic at such a point is a bisector of the angle between the line joining the point to the origin and the tangent to
the locus. 24. If CP, CD are conjugate semi-diameters of an ellipse, prove that 2 the parameter X of the confocal hyperbola through P is equal to If a tangent to this hyperbola cuts the ellipse at the ends of two conjugate
-CD
.
diameters, prove that the length intercepted by the ellipse on the tangent is equal to the perpendicular from the centre on the tangent at P to the ellipse.
25.
A
triangle
PQR
is
inscribed in one ellipse and circumscribed to prove that the normals at the points of
another confocal with the former
;
contact meet in a point. 26. An hyperbola cuts a concentric, but not confocal, ellipse orthogonally at four points. Prove that the tangents to the ellipse at adjacent points of intersection are perpendicular to one another. 27. Show that the curves
# 2 /a 2 -y 2 /fc 2
=
1
and * 2 /a 2 + y*/V + 2\xy
cut one another orthogonally for all values of X only
if,
= ;
2
2
2
(a
-f
fc
)/(a
-
2 fc
)
but are confocal
if,
and
= 0.
X
Miscellaneous Illustrative Examples. (i)
The
ellipse
locus
x*/a
z
of the centres of equilateral triangles described about
+ y*/b*
=
the
1 is
=
0.
The perpendiculars* from the centre of an equilateral triangle to the sides are equal if the ellipse is inscribed in the triangle, these perpendicular*; ;
CENTRAL CONICS
366
make angles
(X,
+
<X
a + Jn- with
TT,
the x-axis
the equations of the sides
;
are therefore of the form
x cos where
-f
(/>
y sin
- vV cos 3 $ -f & 2 sin 2
=
<
0,
a 4-$7r, a + Jir.
has the values
Further, the centre of the (X, the are on and the same side of each of these tangents; triangle origin hence, if p is the radius of the inscribed circle and (x y) the centre, we have t
p= for each side
2 ,/a cos 2
$
2 2 -f & sin -#cos$-f/sin
thus
:
(x cos
-f
(f>
y sin
-f jp)
by the three values of
=a
2
2
cos 2
+ 6 2 sin 2
<
given above. the ellipse is escribed to one side of the triangle, the perpendiculars from its centre to the sides make angles a, a-f JTT, (X-fJ/r with
is satisfied
Again,
<
if
of the triangle and the origin are on the same side ; the centre of two sides of the triangle and on opposite sides of the other, viz. that
the #-axis
corresponding to (OC + ^TT). For this side we have therefore
p
= x cos (a -h
TT)
+ y sin (OL + j TT) -
2
2 ir)+ 6 sin (a
+ J IT),
or, since
cos(a + JTT)
x cos (a -f $?r) +
= - cos (a +
sin (a
Thus, in this case also, (x cos
when
put
=
a?
a+
hence
+ i sin
[r (z*
cos
2
{/
2 -f * ) -f
2 -(a -6 )^} 2
4
cos 2
2|)te}
2
=a
2
cos 2
(/>
and
^
and ^
0-# = 2 (a2 -f
=
z?
=
consequently equation
a
3
*3
2
-f
2 Z>
2
)*
1/z
becomes
+
sin 0,
i
= 2 cos 20
P + (a 2 -
2 fc
) ^
2
(ii)
;
(^ -f
,3
1),
2 fe
)}
are
s4
or
Therefore
2^
8
-f
+ * sin 3(X
COB 8CX
^-a'-t2 =
/i^
Xsf
+ X/i
(since
t is
4prt = M {^-( a
X (say)
evidently not zero),
2 1
/>,
we have
;
= 0,
= 0.
4^/4 --{fV- (a Hence, eliminating jp and
=
of the form (z 3 -\) (z-p)
(ii) is
triangle, (i)
+ 4j?rte3 -f2e 2 (2^ 2 -f/-2 ~a 2 --62 )^ 3 4+ 1 2 1/-2 ^ 2 - (a 2 -
3 2;
77),
sin 2
a + i sin a,
so that
and
2 fe
(i)
cos 6
2
three of the values of z given by equation
COS
+
and equation
4^
Now
= -sin (a +
+ $77-)
a-f$7r.
?r,
=
*/* + t/z=*2
thus
or
y sin -f j>}
= rsintf,
z
sin (a
+ $ TT) +p = - v/a^cos (aT| /r) -f & 2 sin 2 ( we find that, if (x, y) is the centre of the
rcos0, y
Let
and
2
-f
has the values a,
Now
71-)
0.
(ii)
THE ELLIPSE AND THE HYPERBOLA
(ii)
Show
orthogonally
two systems of
there are
tliat
in
ttvo
points
:
circles
find the equation
system ivhich passes through thepoint
ffVa
2
which cut a conic the circle
of on the conic
(#', y')
+0 2 /& 2
367
of
either
==l-
If a circle of one system cuts orthogonally a circle of the other system, show that the line joining their centres touches the conic
If a circle cuts a conic orthogonally in
two points P, P', the tangents to and P' must meet at the centre of the circle, i.e. tangents to the conic from the centre of the circle are equal to each other and to the radius of the circle. Hence there are two such systems of circles, one having its centres on the major axis and the other having its centres on the minor axis. the conic at
P
the point (acos0, &sin0), the tangent at
if (#', y') is
Now,
a
(a?',
y)
is
b
This meets the axes at the points (asectf, 0), centres of the two circles their equations are
(0,
b cosec 0): these are the
:
- a sec 6)*
(x *
x + i.
& cosec
Q?
2
x +
# 2 -f
f=
= - 2 ax sec 6 = - 2 by cosec 6 = ~~
(y
1
e.
-i
t/
t/'-
(a sec 6
a cos
2
2
a cos B
-f
2
2
a cos 6
(b
if
b sin 0) 2
cosec 8
b* sin
-f
+ 6 2 sin 2 6,
2
6 - 2 a2
a 2 cos 2 6 -f 6 2 sin 2 B - 2 fc 2
which can at once be expressed in terras of
Now
2 <9)
and
x'
:
,
,
y'.
the circles
x 1 -f if - 2aa sec
#
2
cut orthogonally, 2
a cos
y
4-
2
2 fy/ cosec
<9
=a =a
2
2
cos cos
2
2
+ b* Bin 2 - 2 a 2
$ -f 6
2
sin 2
<
,
2 b2
we have
+ & 2 sin 2 0-2a 2 4a 2 cos 2 <-f 6 2 sin 2
2
-2
2 fc
=
0,
whence 2
(a
The
- b 2 ) cos 2
line joining their centres
a2 + 62
+ (a 2 - 6 2 ) sin 2 <.
is
-coed + ?sin=
a
From
(i)
and
(ii)
(i)
1.
(ii) v x
CENTRAL CONICS
368
Expressing the condition that this equation in the variable cos 6 should have equal roots, we get for the envelope of the line of cenires 2
52) (
(a
~
-
2
3
j
lies
-^2
a3
2
(a
=
6
-f
b2 )
on the
(a
ellipse
-
~=m
2 .
T
be'
PQ
TP
9
}
),
Show
TQ.
,
ellipse
also that
each of the hyperbolas
with one or other of the ellipses
the point (x l9 y
2
TQ
the tangents
2 -f
-
a 2 -f t 2
x contact along
Let
62 ) 2
2
=
^
Prove that two hyperbolas can be drawn touching the
and having for asymptotes
T
W) +
-
which reduces to
(iii)
(a
2
^
2
+
^=
Ms
if
double
2
j
then the equation of the tangents from T,
to the ellipse are
The equation of any hyperbola having these
for asymptotes
-1
is
'*'-*
>
where c is a constant. The eccentric angles of the points of intersection of the hyperbola (i) and the ellipse are found by substituting #= acostf, y = &sin0 in this equation: thus
x -
1
cos Q
+ Y
sin
6- 1 = ~
-f c,
This equation has equal roots and the hyperbola touches the ellipse
.
-
There are therefore two such hyperbolas.
if
THE ELLIPSE AND THE. HYPERBOLA If
T lies
on
an = ~ +
a*
fa
o
m
*>
22 +
we have ^V a
Iff
er
369
='*;
or
In this case the equations of the hyperbolas
become
or
i.e.
which
is
a conic having double contact with
x1
the chord of contact being
PQ.
i.e.
(See form
(a)
(iv)
vertices lie
(b)
The
sides touch
-I
^~
2
y
^-
0,
5=
2%c sides of a triangle touch an ellipse (a, I) and two of on an ellipse (A, B) ; find the locus of the third vertex. vertices
an
of a triangle
ellipse
the ellipse (a, b)
(A, B)
;
lie
Let the points of contact be P, Q, It let P be the point a, Q, R the
points B and . nates of A are
Then the
on an
ellipse (a, b)
and
find the envelope of the third
means x*/d* + y 2/b 2
and
1267
=
1
=
C
coordi-
A
il
si:le,
of
its
ivhere
1.
Let
A
tivo
its
ABC
be the vertices and
be the point a, the point $.
B the
point
0,
let
and
CENTRAL CONICS
370 a cos
040 -
.
,
0Sin
The equation of
040 7T-
2
x
-cos a
cos
and similarly
Now
since
for
B
B
and C. on the conic
lies
0432
and since
4
*/
j-
o
.
un
040 :
COS
'
2
this touches the conic
y'
.
1
we have
we have 2
^ cos,04^0- 4 J^ sin .
-
3
is
2
**
^
AB
B
2
2
**
,
4
^
2
T7l
COS
2
or
or a1
=
1
4
cos0-a
;
.e.
.e.
(i)
(ii)
where
where
r
-
i
.
A*
&_
, 1 '
Similarly,
Similarly, r
Lcos 3(cos04 AfBin^8in04^ Cross multiplying from
(ii)
0. (iii)
and
(iii)
Lcos3(cos043f ainO(sin04 ^=0* Cross multiplying from (i) and
Lcosa
which can therefore be written
L a
x cos a 4
M -r-
o
.
y
sin
a
= - iV,
(ii) (ii)
THE ELLIPSE AND THE HYPERBOLA and the locus of the third vertex A
which
an
is
whose axes
ellipse,
along the axes of the given
which
is
is
371
a tangent to the ellipse
at the point
lie
t
ellipses.
jya
cos
J--.i.e.
an
Of,
-
the envelope of the third side is whose axes lie along the
ellipse
axes of the given ellipses.
Miscellaneous Examples for Revision. 1. Two circles touch cuts internally at 0: any straight line through the circles in P, Q find the locus of a point R on PQ such that OR, PQ arc pairs of harmonic conjugates. :
A
system of circles touches the axis of x at the origin: find the with equation of the locus of the poles of the straight line Ix+my + n = respect to them. 2.
What
does the locus become
when the
straight line
is
(a) parallel to the
a line through the origin ? 3. Two circles touch internally, and the diameter of one is twice that of the other. Find the locus of the poles of tangents to each circle with
tf-axis
;
(b) parallel to
the
f/-axis
;
(c)
respect to the other. l 4. TP, TQ are tangents to a parabola y show that the equation of the circle TPQ is
(a) If
(b) If
is
T
= 4ax
from a point T(|,
rj)
:
the centre of this circle prove that TSO is a right angle. on the directrix, lies on a parabola whose vertex is at
lies
the focus.
T lies on the latus rectum, A BCD is a square a straight Q so that PA + AQ = 2AB: find
(c) If 5.
and
:
dicular from 6.
A
chord
lies
on the
line is
axis.
drawn cutting AB,
AD
in
P
the locus of the foot of the perpen-
C on PQ.
PQ
of a parabola passes through the point ( 3a, 0): show PQ whicb touches the parabola passes through the
that the circle through focus.
The tangents from a point T to an ellipse include an angle 0: prove 2ST H T cos ST 2 + I/I72 - 4 a 8 8. The orthocentre of a triangle P, Q% R inscribed in a parabola is at the focus if the tangents at P, Q, R make angles tf n 8 d s with the axis, show that cot0n cot0,, cot08 satisfy an equation of the form 7.
that
.
.
:
,
A
UU
+pX*
bx p
a 0.
Prove also that the centroid of the triangle formed by the tangents at P, Q, R lies on the straight line 3# + 5a - 0.
A a 2
CENTRAL CONICS
372
9. The triangle AOB has the angle orthocentre: show that the equation of
* and y and
Z/3 ,
determine the mutual ratios of
\ 19 X 3
Xj/^Zr,*
may be
=
u>) 8
~w
1
o>
and (# y ) for its OA, OB as axes of ,
referred to
x/(x + yu sec o>) -f y/(y<> -f # sec LI denotes y-f-w 1 a?~2aw 1
is
10. If
equal to
AB
1.
with similar meanings for
,
,
and
L2
X 8 so that
AjLsL^ XsLjI^ =
a circle.
the length of the tangent drawn from i9, the focus of the parabola 0, to the circle circumscribing the triangle formed by the j/ normals at P, Q, R, show that a** SP.SQ.SR. If
2
is
t
4a#
=
=
Find the locus of a point such that the line joining the points of contact of tangents drawn from it to a given conic subtends a right angle 11.
at a given point. 12. is
Through the extremities of any two
described
:
focal chords of
an
ellipse a conic
conic passes through the centre of the ellipse
if this
it
will cut
the major axis in another fixed point. 13. Find the general equation of a conic which passes through the four
given points (1, 1), (-1, 1), (2, 0), (3, -4). Show that there are two parabolas which
What
fulfil
this condition.
the nature of that conic which also passes through the origin ? 14. Find the equation of the conic which passes through the points is
(-7-5, 4), and trace it. Prove that the six points (a, 0>, (0, a), (b, 0).
(1, 1), (4, 0), (0, 1), (4, 4),
15.
(0, &),
(a,
fc),
(&,
a) lie
on an ellipse, and find its equation. Find the equation of its axes and show that their lengths are
and
A
triangle PQR is inscribed in the parabola y at the focus prove that: 16.
is
<3 2
=
lax and
its
centroid
;
(i)
The normals
intersection (ii)
is
The locus of
opposite vertex (iii)
(iv)
(v)
is
are concurrent and the locus of their point of
the intersection of a side
i
y (a
x)
=
a3
and the tangent at the
.
The poles of the sides lie on the parabola 2y2 = 2ax + 3a\ The mid-points of the sides lie on the parabola ya = 3a*-2a.r. The centroid of the triangle formed by the tangents at PQR
(- a/2,0). (vi) The line
at PQR 2x-la = 0.
4#=
circle
PQR
passes through the vertex
and
its
is
centre lies on the
lla.
of. the sides of the triangle PQR is constant (vii) The sum of the squares and equal to 81aV2. the intersection of a side and the diameter through the (viii) The locus of
2 = 0. opposite vertex is the parabola yM 2a#-3a (ix) The locus of the orthocentre of the triangle
is
2x + ba
= 0.
THE ELLIPSE AND THE HYPEKBOLA One of the tangents from Tto an ellipse subtends angles and the angles between the tangents is #: show
17. foci,
sin 6/2 a
=
sin
OJST =
sin
9
S78 lf
2
at the
/S'T.
18. The cotangent of the angle between the tangents drawn to a rectangular hyperbola from a point on one of its directrices varies as the distance
of the point from the centre. 19. Show that a circle described with
centre at any point of an ellipse
its
to touch a pair of conjugate diameters of the ellipse has an invariable radius. 20.
# 2 -y a #a -f y a
A parabola has double contact with the rectangular = a and the pole of the chord of contact lies on = 2cx. Show that its axis passes through the point (a /c, a
,
hyperbola the circle
2
0).
Chords distant d from the centre of an ellipse are divided harmonically by a coaxal conic of semi-axes or, 3 given by 21.
Find the greatest value of the angle between a diameter of an show that for the e) and the normal at its extremity this angle is earth's orbit round the sun, of which the eccentricity is ^ less than half a minute of arc. 23. A chord of length c is drawn parallel to a diameter of length 2
ellipse (eccentricity
;
,
parallel to the tangents at
these tangents 24.
A
25.
PQ
its
1
sin- [c/d
is
.
extremities.
ab/a'V
Prove that the angle between
2
-y/1
(c
/4c^)].
square is inscribed in an ellipse, whose semi-axes are a and 6, and any point on the ellipse is joined to the corners of the square. Prove that 2 2 one of the anharmonic ratios of the pencil so formed is & /a .
circle
is
a chord of an
corresponding to a
1
ellipse,
P,
Q
normal at P.
are p,
q.
r
lhe points on the auxiliary Prove that the angle yCq must
a
exceed 2tan~ 2^/l a /e where e is the eccentricity of the ellipse. 26. The circles of curvature through a point (/, g) are six in number ,
and their centres
lie
on
2 2 2 2 a a a {2(y + ^-2/#-20y)-a -& } = 12(a** + 6 y ) -3 ( -&y. 27. The circles of curvature at two points P and Q of an ellipse meet and A and the circles which respectively towb the ellipse again in the ellipse at P and pass through Q, and touch the ellipse at Q and pass through P, meet the ellipse again at R and 5; prove that the chord A/\
a
M
is
T
,
parallel to RS. 28. Triangles
& 2 #2
+
aV = a*6
formed by pairs of tangents drawn to the ellipse from points on the ellipse a*.r a + &2 y a = (a*+ & 2 ) a and
are a
their chords of contact.
,
Prove that the orthocentre of each triangle
lies
on
the former ellipse. 29. Prove that if 0, ' are the eccentric angles of the extremities of a focal chord of an ellipse, the eccentricity is equal to + cos \
(
- <j>') sec 1
CENTRAL CONICS
374 and that the
if
P
a point
^,
>//
are the inclinations to the major axis of the tangent at QR drawn so that PQ and PR pass through
and of the chord
the eccentricity is equal to (sin (^--^') cosec (^H-^')} Prove that the point on the normal at P on the ellipse
foci,
80.
chords through which subtend a right angle at P, 2
- 6 2)/(a 2 +
is
- y (a 2 - 2 )/V -f
2
2
ft ft 7> ), (x ( )}, where (a?, y) are the coordinates of P. Show that the chord IX + MIJ= 1 intersects the ellipse in points <), ft such that there are two real points on the ellipse at which QR subtends a right angle if 2 P + tf m 2 >(a 2 + ?> 2 ) 2 /(a 2 - & 2 ) 2 .
Prove that the equation to an ellipse referred to the tangent and normal at a point on the ellipse as axes i* 31.
-2
p* x*
where p
is
VV -p*J(]iF=V) xy +
2
(a
+
2 ft
- jp 2
2 )
i/
=
(2
a 2 b*/p) y,
the perpendicular from the centre on the tangent and a and b are
the semi-axes.
tangent touches an ellipse at P and meets the major axis in T. TQ parallel to the minor axis meeting AP produced in Q. Prove that the locus of Q is an hyperbola.
A
32.
is
drawn
33. Let a circle be described with a point P on an ellipse as centre, and radius a, cutting the minor axis of this ellipse at ft and let PQ intersect the is is completed, show that major axis at 1?. If the rectangle the normal to the ellipse at P.
PV
RCQV
34.
* Find the equation of the circle touching the ellipse x*/a* + # 2 /^ 2 # = acos0, t/ = &sin0 and cutting the director circle ortho-
at the point
gonally.
Prove that the cosine of the angle which the tangent at any point on ellipse makes with the line joining the point to a focus bears a constantratio to the cosine of the angle which the tangent makes with the major 35.
an
axis.
Perpendiculars SM, point
Pon
an
Prove that
PM.PN
is
ellipse,
HN are drawn to the focal distances SP, and N. and meet the tangent at P in
the eccentricity of the ellipse
if
HP
M
46 a but that otherwise ,
&V
it is
is 2
<2~i
(l-e
2
the
minimum
of
any
value of
1
)-
.
E
on a central conic chords are drawn equally 36. From a fixed point inclined to the axis and cutting the curve again at P and Q. Find the locus of the centre of gravity of the triangle PQE t
Show
that the locus of the second point of intersection of two circles 2 2 2 2 = 8 a
is the inverse of a concentric ellipse with regard to a circle whose the origin. 38. Points P, Q, one on each of the central conies
diameters centre
is
*/* 4 y*/V
=
2
1, (I/'
subtend a right angle at the circle
a?Hy
2
=
c*.
- 1/& 2
common
x> + (1/c2 - I/a
2
)
centre.
)
f = 1,
Prove that
PQ
touches the
THE ELLIPSE AND THE HYPERBOLA The normals
39.
If
an
to
ellipse at the points P, ft
F,
375
Q' are concurrent.
PQ pass through a fixed point, find the locus of the middle point of FQ\ 40. If the circles of curvature at two points intersect on the ellipse, show
is parallel to two of the chords joining the extremities of diameters conjugate to those through the given points. 41. A point moves so that the sum of the squares of its distances from
that their radical axis
an equilateral triangle is constant and equal to 2c 2 Show that the locus is an ellipse, and find the eccentricity and the position
two given
sides of
.
of the foci. 42. Find the coordinates of the point of intersection of two normals, and deduce those of the centre of curvature at P.
that the centre of curvature can only lie outside the ellipse if e 2 >|. 43. P, ft R are three points on an ellipse such that their eccentric angles
Show
by TT. Prove that each side of the triangle PQR is parallel to the tangent at the opposite vertex, and that the sum of the squares on the sides differ
is
constant.
that the lines 10y 2 7#y 6# a == coincide in direction with a pair of conjugate diameters of the ellipse 3# 2 -f 5 */ 2 - 15 0. 45. A diameter of a central conic meets one latus rectum in P; the con-
Show
44.
=
Prove that the envelope jugate diameter meets the other latus rectum in P of is a conic similar to and coaxial with the given conic. /
.
PP
46. Tangents are drawn to an ellipse at points which subtend at a focus a constant angle 20. Prove that they intersect on a conic whose eccentricity is to that of the ellipse as sec j3 1. :
D extremities
of conjugate diameters of an ellipse #y 2 -f y */b* 1 = meet the curve again in 7?, R' respectively, prove that the locus of the middle point of RR' is an ellipse. 2 2 2 2 2 48. The circle of curvature of the ellipse 6 .r -f a */ = a 2 fc at P meets the 47. If the circles of curvature at P,
GR
ellipse again at Q and the normals at P and Q meet in (7, GR, being the other two normals drawn to the ellipse from G show that the tangents at /?, intersect on the curve x*y*(b*x* + a*if) (Vx*-a*y*)*. :
R
also that the points 7, R' are imaginary if the eccentric angle of P l greater than J tan~ 2. 7 1 2 7 2 2 49. PP are points on the ellipse tfx * a // = a 6 whose eccentric angles
Show
is
a + #, ft Q' are the points y + b. If J7and Fare the poles of PP' and QQ' prove that U, P, P', F, ft Q' lie on a conic and find its equation. 50. In the ellipse a?/a* + if/b*-l = 0, if a line through a point cuts the
are
;
ellipse at is
aP
a
P and Q, show that the
external to the ellipse,
(ii)
If
If
9
a?
touches the (iii)
51.
8
>a
2
<6
a .v
6
a
a
2
(a?
and
is it
has a
/a
-f t/
>&
8
maximum
value a*(x*/a 2 + y 2 /b*-l) ar d -
2
2
8
if
variation of the rectangle OP., Oft where
such that
/6
1).
the greatest and least values are
when OP
ellipse.
Also discuss the cases in which
Prove that
if
#2
>
2
and ya
<&
2 ,
four points are taken on an ellipse such that their
OENTEAL CONICS
376
normals intersect in the same point, two of them must lie on one quadrant of the arc, and the other two, one on each of the adjacent quadrants. 52. CP, CD are conjugate semi-diameters of an ellipse, and DQ is the chord of the ellipse parallel to the minor axis. Find the locus of the intersection of the normals at P and Q. 58. Show that the normals which can be drawn from (# y ) to ,
x*/a*
+ y*b*
=
1
are given by
E
54. From a fixed point on a central conic chords are drawn equally inclined to the axis and cutting the curve again at P and Q. Find the
locus of the orthocentre of the triangle PQE. 55. Parallelograms are circumscribed to a'x*
+ Vy* = 1, the sides of any one being parallel to a pair of conjugate diameters of ax* -f 6t/2 = 1. Prove that they are inscribed in a conic similar to the latter conic, determining
its equation.
56.
Normals are drawn
to the ellipse o^/a 2
-f
y
?
2
/fc
=
1
from any point on
Show that the
locus of the centre of the circle passing through the three points of incidence is the curve 4 (a 2 # 2 -f &2 //*) 8 = rt 2 5 2 (a 2 ~fc a j 8 ^ ? y 2 57. An ellipse circumscribes a triangle and has its centre at the
its evolute.
.
ABC
centre of gravity of the triangle. Prove that the radii of curvature at A, B C are proportional to the cubes of the sides BC, CA, AB and that the product of the three radii of curvature is equal to the cube of the radius t
}
of the circle 58.
ABC.
The normal
at a point (a cos
,
2aHan0 =
When
(a*
Prove that
- &9 ) sin 2 <.
has the angle between two corresponding tangents to the ellipse
and the auxiliary
Show
on an ellipse makes an angle Q
b sin 0)
with the central radius vector of the point.
circle its greatest value ?
that that value
1
is sin"- (a
- b)/(a -f b).
T is
the intersection of tangents to an ellipse at the extremities of a chord PQ normal at P, prove that the perpendicular from T on the 59. If
diameter through
P
intercepts on
PQ
a length
PN equal
to the radius of
curvature at P.
PP
f
is a diameter of an ellipse and and PQ, P'R in Y. PR, PQ meet in diameter conjugate to PP.
60.
X
}
61.
Show
Q,
R two
points on the curve, and is parallel to the
Show that
that the centroids of the triangles
the ellipse (3a?/a-2fta/c2 ) a -f (3y/fe + 2fc6/c9 )* that the normals at PQRS meet at (h, k).
*=
XY
PQR, QRS, RSP, SPQ 1,
where
c
2
=
lie
on
a -b'l given
t
62. A circle passes through a given point and cuts off on a given line a chord which subtends at the given point a constant angle 0, Show that its centre traces out an hyperbola of eccentricity sec 8. 63. A circle is drawn to touch one side of an equilateral triangle and to make the pole of another side (with respect to it) lie on the third side.
Prove that the locus of the centre
is
an hyperbola.
THE ELLIPSE AND THE HYPERBOLA 64. x', y'
377
Prove that the equation of the hyperbola passing through the point + a2 y2 = a2 62 and confocal with it is
of the ellipse 6 2 # 3
Through a point
P
of this ellipse straight lines
PQR, PQ' R' are drawn
parallel to the asymptotes of the confocal through P, meeting the major axis in QQ' and the minor axis in RR'. Prove that QR', Q'R intersect at
a point on the normal at P, and that the locus of this point
is
Find the general equation of a rectangular hyperbola when the origin is a point on the curve and the tangent at the origin is taken as 65.
the axis of x. If a chord
a point
PQ
of a rectangular hyperbola subtend a right angle at is parallel to the normal to the curve
on the curve, show that PQ
at 0.
Prove that the axis of the second parabola which passes through the 2 is inclined to the axis of the latter points MU w a tw 8 w 4 on y -4a# = 66.
,
,
an angle cot" (m l + ma + tn, wJ/4. Deduce that, if two parabolas intersect in four points, the distances of the centroid of the four points from the axes are proportional to the latera recta. 67. An ellipse and hyperbola are concentric and coaxial, and have the 1
at
-I-
same semi-axes.
Prove that the circle circumscribing the triangle formed
by the asymptotes of the hyperbola and the tangent to intersects the director circle of the ellipse on the polar of
at
it
any point
P with
P
respect to
the ellipse. 68.
From the
drawn meet at T.
are
according as 69.
of the ellipse
focus
S
of an ellipse whose eccentricity
is e
radii
P,
SQ
P
and Q at right angles to one another, and the tangents at Show that the locus of T is an hyperbola, parabola, or ellipse, e is
>, =, or
<
l/\/2.
Prove that the eight points of contact of the four
common
= and af/af* y 2 /V - 1 ellipses x*/a* + y*/b*-l 2 2 2 2 = a 2 + a^fc 2 rr -f bf) + y (a -f a^) (ft
V
.
=
lie
tangents on the
CHAPTEK
IX
POLAR COORDINATES When the focal properties of conies are under consideration sometimes convenient to use polar coordinates, talung a focus
1.
it is
ns pole.
To find the polar equation of a conic when a focus axis is inclined at an angle y to the initial line.
is
the pole
and
the
XM
the
\
Let
SZ
directrix.
be the If
P
initial
line,
be any point
SA (r,
0)
the axis of the conic, on the conic, draw
PM perpen-
dicular to the directrix.
Now, then
if
LL'
is
the latus rectum, let
SL SX-l/e.
But using the
focus-directrix property of conies previously proved,
=
Z
0rcos0- y
;
the equation of the conic can therefore be written
Z/r=
l
+ ecos0
y.
(i)
POLAR COORDINATES N.B.
When
the axis
lies
along the
379 the equation takes
initial line,
the simpler form
l/rAgain,
1
be any point
if (r, 6)
+ ecosO. Q on
(ii)
the directrix,
SQ cos QSX = SX, rcos (0 y) = l/c,
since
we have so that
ecos(0
l/r
y)
the equation of the directrix.
is
Note. It
is
indicate the
important to remember that, since
same
(r,
0)
and
(
r, TT
+
6)
point,
= l/r =
Z/r
or
ecostf
1 1
+ e cos
y 6
y
represents the same conic as
Thus, for instance,
are
l/r
= 1 + e cos
Z/r
= l + ecos(0
y.
if
y),
L/r- l+.Ecos(0-S) conies a two with common focus, any
" i.e.
,,
'
,*
(I-
passes through points common to the two conies and, being a straight line, represents a common chord. Again,
{Z/r-1- eco*(0--y)} + {L/r+l-EcoB(0-b)} =0,
+ L)/r =
(0- y) -f .7? cos (d- 8), where we have used the alternative form for the second conic, represents the common chord through the other two points of
i.
e.
(I
e cos
intersection of the conies. 2.
conic
To find
Z/r=
l
+
equation of the chord joining two points on the tfcos(0 y) whose rectorial angles are a and ft. the
The equation
l/r=Ac<M{0-\(* + P)\+B*m{0-l(* + p)} can, by a proper choice of
straight line.
whose equation
If (r t ,a), is
A
(r2 ,
and B, be made
/3)
to represent
any
are the extremities of the chord
required, since y)
(i)
and
l/r2
= l + ecos(
y),
POLAR COORDINATES
380 these points
lie
.4 cos
on the straight J
(a- /3) +
A cos \ (a
line
B sin
/3)
(i) if
(-/) = l + ecos(a-y),
J? sin J
(a
J-
)
= 1 + e cos
(/3
y)
;
hence
and JSsin
Substituting these values of II
which
is
r- e cos (0- y) =
A
and
sec i
(<*
in
the equation becomes
(i),
-/3) cos
{
0- \ (<x 4- $ },
the equation of the chord.
It is often useful to take equations in the form shown in (i) when the conditions given are that points on the conic lie on a required locus : the following is another example.
Example. focus and
To find
the equation
also through the points a,
Let the equation of the
circle
of the circle which passes through 1 -f e cos 6. /3 on the conic l/r
the
=
be
instead of the usual form r =* a cos (6 - d). lie on it, we have Then, since the points Of,
^1 (1
and
+ e cos a) (1 + e
cos/3)
= Jsec $ (a-0)
-f e
cos|(a + 0),
J(l + ecosa)(l-f ^cos/3)
The equation of the
3.
To
circle
then becomes
^d ^e equation of the tangent to the conic l/r-
l
+ ecos(fl-y)
at the point whose rectorial angle is a.
This follows from the equation of the chord by putting hence the equation of the tangent is J/r %
ft
=a
:
= e cos (0- y) + cos (6- a).
Example. If TP, TQ are tangents to a conic, show that ST bisects the angle PSQ. Let the conic be l/r = 1 -f co80, and let P and <) be the points a and p. Then the equations of TP and TQ are //r
e cos
B + COB (d - a).
//r
e con
+ 008(0-13),
POLAR COORDINATES hence at their
common
T we have
point
cos(0-a)= therefore, since
ZST- $( +
),
cos(0-/3);
we have
not equal to 0, so that in
Ot
=
L PST =LTSQ
To find
II,
*
Tangents to a conic subtend
:
or.
0=
.
OK
is
11)
+ cos(0 a) = 0, (ecos y + cos a)/(6 sin y + sin a).
ecos(0 i.e.
the angle
through the pole parallel to the tangent at the point
Chap.
(see
i.e.
+ ecos(0-y)
1
at the point tvhose vectorial angle is
line
+ /3),
(
equation of the normal to the conic
tJte
l/r-
A
I
\ (0^0).
This property is usually referred to thus equal angles at the focus.'
4.
381
tan"
1
y)
The normal, being perpendicular
to this, is therefore parallel to
= 77/2 tan~ cos y cos &)/(e sin y -f sin tan 6= (siny-f sin a) /(ecos y-f cos = 0. e sin (0 sin (0 y) *
-f-
(e
i.e. i.
OK),
e.
4-
The equation
of the normal
or)
therefore of the form
is
&/r= esin(0 and, since the point k
1 -f e
{
e sin (a 1
1
Example.
the
y)-f sin(0-- a),
we have = k sin (a y)}
it,
cos (a ft,
-
IA
.
y)
P
normal at
+ sm(^
a conic cuts
to
,
.
.
"
r
y)
y).
the equation of the normal Z
y)'
+ ecos (OK If
on
lies
ex
Hence, substituting for
SG =
OK),
is
.
a).
the axis at G,
then
eSP.
Let the conic be l/r
=
equation of the normal at
1
4 e cos
P
e sin OK 1 -f a
The normal
cos
6,
and
P
let
be the point a.
The
is I
-
a
r
=
.
e sin
A
sm (6A .
6 4-
,
OK).
cuts the axis on that side of the focus remote from the corre-
sponding vertex,
i.e.
at the point on the '
oS"5 56?
=
le/(l
normal where 6
__
.
ss~ 8
+ e cos (X)
;
-
e
SP.
n.
Thus
POLAR COORDINATES
382
We conclude this section Example the focuS to
To find
i.
a tangent
the locus
a
to
with some illustrative examples
:
of the foot of the perpendicular from
conic.
Let the equation of the conic be l/r= 1 + ecos0 the equation of the = 6 at is e at cos cos the + and the equation of the tangent l/r (0-a), point = is e sin 6 line the to this sin (6 a). straight through pole perpendicular The locus of the point of intersection of these lines is therefore ;
-I-
(7/r-ecos0) r
i.e.
which
is
a
2
2
(l-
circle.
For an ellipse
a (I -6*), and the equation can be written = a2 r3 + 2aer cos 6 -f
I
aV
the centre of the circle
point is
(ae,
and
TT),
;
therefore at the centre of the ellipse, the radius of the circle is a. is
For a parabola e = 1, so that the equation reduces to r cos 6 = the tangent at the vertex.
Example intersect
angle
ii.
at 1\
TSR
\l>
the
which
If the tangents at the points P and Q on a conic and the clwrd PQ meets tJie directrix at R, then the
a right
is
viz.
Let the conic be
angle,
= l/r
1 -f ecostf,
The equation of the chord PQ
and
let
Pand Q
be the points
a and
0.
is
J/r-ecos0 = sec !(<*-) cos {0-1(8 + 0)}* and the equation of the directrix is At the point
JR,
J/r-ecos0 = 0. where these lines meet, we have therefore cos {tf-
which
We Z.
K
6 = + $n
hence is
the equation of SR.
showed
#ST is
+ 3)} =0, + J (a 4- /3),
in
3 that the equation of
ST
is
6
=
!(a + 0); therefore
a right angle.
Example
iii.
Prove that any chord of the conic
l/rl + c cos 0, which
is
normal at a point where l/r (e
will
+ l/e)
=
the conic is
sin
+ (e
2
met by
the straight lines
- 1) cos 0,
subtend a right angle at the pole.
The vpctorial angles of the points of and the conic are given by
lines
intersection of the given straight
POLAR COORDINATES +
e
i.e.
l/e
-f
(e*
+
2
if
=
2
either of these angles a (1 -f e cos a)
The equation
sin B +
a
(e
- 1) cos 0,
+
(l-fecos0)
Hence,
= +
cos 6
1)
1-f e
i.e.
383
+
is Of,
=
+
we have 3 - e 2 sin 2 a.
e sin
(i)
of the normal at the point whose vectorial angle
esin(X
I =
------
.
1
-f
This meets the line 8
cos
e
=
\n
a
sin 6
-f
at the point
-f tx
-esintt
- =
a
cos
i -f e
-
.
.
c sin
r
e
r
cos
^
a,
is
is
given by
a.
whose radius vector (X
is
-
+
1,
i
r or, substituting
from
(i),
l/r
=
Hh
1-
i.e.
l/r= Thus the normal meets the line 0= |7r-t-(X at the point on the conic whose vectorial angle is (J TT -f a) in other words the normal chord subtends ;
a right angle at the pole.
Example {
tihow that the equation of the pair of tangents ivhich 7 1 -f e cos hyperbola l/r from the point (r 6') is
iv.
can be drawn
-1 (l/r- e cos 6)*
and
=
to the }
,
- e cos 0') 2 - 1 (l/r'
{
}
that the equation of the asymptotes is l/r
The tangent
at
OK
= (e-e~
l )
sin 6 v/i-tr*.
cos 6
to the conic l/r
=
1
-I-
e
cos
viz.
l/r
passes through
(;', ^')
_
0,
= c cos 6 + cos 0-a,
(i)
if
l/r'
e
cos
tf'
-f
cos ('
-*
a).
(ii)
Hence, any point on a tangent from (r', 6') to the hyperbola satisfies equation (i), if & has a value given by equation (ii). Thus, if we eliminate (a) from equations (i) and (ii), we get an equation .
by the coordinates of any point
satisfied
The equations can be
oil
these tangents.
written sin 6
cos Q
cos
ft
cos 6
-f
cos
a
cos
+ sin a sin 6' - (l/r' - e cos 0')
0'
sin
0(.
(l/r
e
)
= 0,
= 0.
Hence cos a
__ (l/r'
___
_
-ecosfl') ciwtf- (l/r"-"coBtf)co8 0'
POLAR COORDINATES
884 [(//r- e cos 6) sin
- (l/r' - e cos ff) sin 0]
ff
.c
(///- e cos 0;
which
is
2
4 (J/r' - e cos 0') 2 - 2 (f/r- e cos 0)
(l/r
- * cos 0') cos 6^W
= sin* (0-0'),
the same as
the point {Je/( 2 -l), 0), and the asymptotes being the tangents from the centre to the hyperbola are given by putting r' = te/(e 2 - 1), 0' = in the above equation. In this case 7/r'-ecos0' = (e*-I)/e-e = -l/e.
The centre
of the conic
is
Hence the equation becomes .e.
or i.
(7r (1/r-e cos
9
d)
4-
2
(0
e
- 1 cos
cos 0)
=
1 l/r- eco&0 + e- cos 6
e.
+ e~* -v/1
cos
- e~
=
2
2 .
sin
(1
-
^?.
-4 v. hyperbola and a parabola have a common focus and one touch another, awl the line joining their other common points passes
Example
through the focus.
e=
Shoiv that
parabola and 2 c
Take the
.
{5
where 21
4 \/Z/c}*,
tlw length
of
the latus rectum
is
of
the
common chord
the
and
axis of the parabola for initial line
let
the equation of the
conies be
(parabola)
l/r
L/r
(hyperbola)
A
pair of
common
= 1 cos 6, = 1 + e cos (6 - y). 4-
chords are
(L
- l)/r
*= e
cos
0-y-cos
0,
(i)
and --y
The first will pass through the pole The second then becomes 2 //r
=
e
+ cos0.
(focus) if
L
cos 8^y -f cos
(ii)
=
/.
0,
and is by hypothesis a tangent to the parabola and the hyperbola. For some value of a it is therefore identical with
Comparing
coefficients,
p
e
cos y 4-
1 4 cos .*.
2 cos
a
1
isiny
a
ecosy-1, 2
sin
'
ex
shift
=
esin-y;
(iii)
POLAE COORDINATES Now
the
first
common chord e
cos (6
386
(i) is
y)
- cos 6 .
tan
or
A
*=
0,
=
1-ecosy :
ismy Hence,
if
B satisfies this equation, $
of the ends of the
chord, say
ir
+
are the vectorial angles
.
~ 1 ~ cos 6 2 c/l - SP/l + SF/l - 2/sin
l/SP =
Thus
and
PSP
f
common
1
-f
cos
6,
l/SP*
;
a
.-.
;
2
^
v
*
i
>
i
sin ---
a
yY ^n
(1-tfCOSyf
l-2coi y + * _ (l-*cosy)
Thus
__>
.
8
a
(From v
iii.)
= cosy =
1-ecosy 2
/.
Examples IX. The
1.
&enu-latu8 rectum of a conic
segments of any focal
is
the harmonic
mean between
the
cjiord.
of the tangents to a conic from a point on its directrix subtends a right angle at the focus. 3. In any conic the projection of the normal PG on the focal distance 2.
SP
Each
equal to the semi-latus rectum. that the circumcircle of a triangle formed by three tangents to a parabola passes through the focus. 5. Tangents are drawn to a parabola from any point on its latus rectum show that the harmonic mean of the focal distances of the points of contact is
4.
Show
:
is
equal to the semi-latus rectum. Trace the following conies
6.
:
= 1 -f- cos 1 -f 2 costf; (c) 2/r (b) 2/r = 1 -f >/3 cos (0 ir)/2. J 3/r (d) Show how to 7. The axes of an ellipse are 8 and 12 inches respectively. How in 9 the inches many possible ellipse. place a focal chord of length positions are there ? 8. If PSF is any focal chord of a conic section, show that (a)
where
2/r=
e is
1-f Jcostf;
;
the eccentricity, 2
1
the latus rectum, and B the inclination of the
chord to the axis of the conic. A focal chord of a parabola is twice the of this chord from the parallel length of the latus rectum find the distance :
tangent. conies have a common focus, equal latera recta, and four real Rrove that one of them is an hyperbola, and that if the of intersection. points 9.
Two
1397
B b
POLAR COORDINATES
886 other
is
an
ellipse the
points from the
sum
common
of the reciprocals of the distances of the
focus
is
y
~-~
-75
^
,
where 2J
of the latus rectum, e is the eccentricity of the ellipse, and y is the angle between the axes.
e'
is
common
the length
of the hyperbola,
10. Two parabolas have a common focus and axes inclined at an angle ft. Prove that the locus of the intersection of two perpendicular tangents, one to each of the parabolas, is a conic. 11. From the polar equation of a parabola deduce a quadratic equation for the lengths of the latera recta of the two parabolas, each of which has two focal radii of lengths r and r2 making an angle Of. with one another. 12.
Conies with latus rectum of given length are described with a fixed Prove that the locus of
point as focus and touching a given straight line. their centres is a conic. 13.
Two
conies have a
at right angles. section, fj
where 14.
e, e'
If
r^
,
common
focus
S and have
their corresponding axes S of the points of inter-
ra r3 r4 are the distances from ,
,
being the greatest and r4 the
are the eccentricities, and
I,
least,
I'
show that
the latera recta of the conies.
two points on an ellipse whose vectorial angles referred to one Prove that if PH and QS meet (Si H) are (CX + j3) and (a-0).
PQ are
of the foci
on the curve then the eccentricity e is given by sin a/sin j3 = ^ (e -f l/). 15. Find the equation of the normal at the point Q = a of the conic I = r(\ + e cos0), and prove that the part intercepted by the curve subtends at the origin an angle 2 tan- 1 (1 e* -f 2 e cos QL)/e sin (X. 16. If the ellipses whose latera recta are J, V and eccentricities e, e' have a common focus and touch one another, show that the cosine of the angle between their axes is {(J-O f -( a ^ 2 + /ap )}/2 * /flr 17. From the focus 5 of an ellipse whose eccentricity is e, radii SP, SQ are drawn at right angles to one another, and the tangents at P and Q meet at T. Show that the locus of T is an hyperbola, parabola, or ellipse, -I-
-
according as
e is
>, =, or
<
l/\/2.
SP is drawn
through a focus of an hyperbola parallel to one asymptote and meeting the curve at P, prove that the tangent at P meets the other asymptote on the latus rectum produced. 19. Trace the curve rcos'a^ acos(0-3a), and show that for all values 2 of a it touches the parabola r = a Bee \ 6. 18. If
20. If a normal is drawn at one extremity of the latus rectum, prove that the distance from the focus of the other point in which it meets the .
.
curve
is
i+
e *-e*
P
of any conic meets the transverse axis at a fixed point that points on the curve which are equidistant from G are such that the arithmetic mean of their focal radii is equal to the focal 21.
at G.
The normal
Show
distance of P.
POLAR COORDINATES
387
22. If T is the pole of a chord PQ of I = r(l 4 *cos0) which subtends a constant angle 2ot at the focus, 1/SP+ l/SQ-2co*(X/ST is constant. 23. If (X, ^, y are vectorial angles of three points on 2a/r = 1 -f cos 0, the normals at which are concurrent, 2 tan \ Oi = 0. Two points P, ^ are taken on a parabola whose vectorial angles are supplementary. Show that the locus of the intersection of normals at P
and Q
is
a parabola.
A parabola is drawn with its focus at the origin to touch
24.
///
=
1 -f e
cos 6
Show
that the equation of the locua of its vertex r = a (I -e cos0), where 2 a is the major axis of the given conic. 25. If t 2 are extremities of conjugate diameters, prove
at
any
point.
is
,
(i) (e 4-
and
(ii)
26.
+
cos 0,)
^l - e*
cos
(e 4.
2 ) 4-
sin I (0,
Show that the two
-
(1 a)
conies
Z
sin sin 8 = = cos $ (0, - A,) cos $ (6 + = r (1 4- cos 6), = r (1 e' cos 6)
- e*)
t
;
-f
2 ).
l
I
-I-
have two
tangents if e *+>e' < 2. 27. A chord of a circle through a fixed point O on its circumference meets the circle in P, and a fixed straight line in Q and (07?, PQ) is harmonic. real
common
Prove that the locus of
R is
a conic.
A
right-angled triangle has its right angle at a focus of a conic section while the hypotenuse envelopes the curve and one other vertex moves on a given line prove the remaining vertex describes a conic. 28.
:
A
PQ of a conic subtends a right angle at a focus 8. Show that the locus of the intersection of the circles on SP and SQ as diameter is 29.
a
chord
circle.
30.
The tangent
parallel to the tangent at 6
=&
is
31. In the parabola r= a sec* \6, p, q, r are the perpendiculars from the focus on any three tangents, and R is the radius of the circle circumscribing 2i?a a the triangle formed by the tangents. Show that pqr
=
32.
The conic
= l/r
pole in the points (r l9 (i)
2l/r=2/J;
1
-f
ecostf
0,), (r, f
a ),
(1-f *)2
(ii)
is
(r3
,
*y,
tan|0
(>-4
,
4)
:
prove that
Stan^tan^tan J0 a = rsin whose 'vectorial
=* (l-
Three points are taken on the parabola Prove that the radius of the angles are Oi y. 33.
t
.
cut by a circle which passes through the
,
4.
a
circle circumscribed to the
triangle of the tangents at the points is ^a cosec Ja cosec^ coseciy. 34. circle is drawn through the focus of the parabola 21 r(l -f costf) (X. to touch it at the point where Q Obtain its equation in the form
A
=
Show that no circle of curvature at any point on a parabola can pass through the focus. 35. Show that the equation of the polar of the point (rlt 0j) with respect to the conic lr~~ l
=
1
4-
e
cos 6
is
l
(lr~
e
cos 6) (7t\~ 1
e
cos # n )
cos (6
6J.
If pairs of points collinear with the pole are conjugates for each of the conies lr~ l 1 -f e cos0, lr~ l 1 -f esin 0, prove that they all lie on the conic (J/r-* sin 0) + l 0, or on the line }*r.
=
0=
B b 2
POLAR COORDINATES
388 Normals
36.
angles are
,
A, B, C)
D
On SA, SB
t
to the ellipse l/r *= l
+ fco*6 at three points whose vectovial
y meet at a point.
0,
Prove that
are the feet of four concurrent normals to an ellipse, focus S. SD as diameters circles are described. If two of these
SC,
on a fixed circle r = ccos0, prove that the other two will on another fixed circle r = c' cos 0, and determine the relation between c and c'. 37. Tangents of lengths p and q are drawn to the parabola 2a/r = 1 -f cos B from a point on the latus rectum produced. Prove that p q satisfy the relation p 4 q* a a (jp a + g2 ) (i? 2 -g 2 ) 2 and that the angle subtended by the chord of contact at the focus is 20, where 2pqt&&6 =jp 2 -g8 38. Prove that two equal conies which have a common focus and whose axes are inclined at an angle 2 a intersect at an angle circles intersect
intersect
9
,
.
-i
r
L
PSQ
39.
is
2
i cos2a-f20cosa-f 1 J 1
a focal chord of a conic.
Find the locus of the intersection of
the tangent at Q with the perpendicular from S on the tangent at P. 40. Chords of the conic J/r = 1 a cos are drawn through the origin
and on these chords
as diameters circles are described.
envelope consists of the 41. Prove,
two
circles l/r (l/r
+ e cos 0)
=
l
Prove that their
+ e.
from the general polar equation of a conic
r2 (a cos8
+ 2ft sin
cos
4-
b sin 3 0)
-f
2r(g cos
-f-/sin 0)
-f
c *= 0,
that the locus of the extremity of a radius from a fixed point equal to the harmonic mean of the two radii of the conic in the same direction is
a straight line which, when the point is external, passes through the points of contact of tangents from the point. Prove also that, if the mean is arithmetic or geometric, the loci are
and that the centre of the conic in the case of the arithmetic mean midway between the centres of the original conic and of the geometric
conies, is
mean
conic.
Find the equation to the chord joining the points 0==a~|3 and 0=sa+j3 of the hyperbola Z/* 1 + * COB 0, and hence show that the equations to its asymptotes are e/r = cos0/asin0/5. A line drawn through a focus S perpendicular to an asymptote cuts the hyperbola again in P, Q. Prove that SP.SQ = b*P/(b* P). 3 = i* drawn is 2 a the conic the cos of At 43. parabola having any point .4 point contact prove that the locus of the foot of the perpendicular from 42.
:
the centre of the hyperbola to the tangent at the vertex of the parabola r1 cos 20
a 8 (cos 4
4
-f
sin 0)
is
9 .
Pis the pole of a chord which subtends a constant angle at the focus a conic, and SPis cut internally bfc.the conic in Q. Find the locus of the harmonic conjugate of 5 with respect to Pand Q. 44.
S of
The parabola I** r(l-cos0) angle cos-'-J. If the new parabola 45.
turned about the origin through an cuts the initial line at Pand Q, show
is
POLAR COORDINATES that the intercept points P, Q to the 46. line
made on the old parabola by the tangent at one of new parabola subtends a right angle at the origin.
389 the
Two parallel straight lines and a circle,
PAB is
centre F, being given, a straight drawn from any point of the circle to meet the lines in A and B
and through A a straight line is drawn parallel to BF to meet Q: show that the locus of Q is a conic, and determine whether it is
respectively,
FP in
a parabola, ellipse, or hyperbola. 2 47. Prove that the equation r (1 -f e cos (X) = I cos (0 - a) + le cos (6 - 2 (X) represents a circle which passes through the origin and touches the conic l/r
1 -f e
cos
6.
Find the equation of the polar reciprocal of the conic l/r~\-\e cos 0, with respect to a circle, whose centre is at the focus and diameter equal to 48.
the latus rectum of the conic. 49.
Find the equation of the polar of the point (r', ff) with respect to the = c, and obtain the envelope of such polars when (r', 6') lies on the
circle r
conic c/r
=
1
-f
e
cos
0.
CHAPTER X LINE COORDINATES AND TANGENTIAL EQUATIONS IN the previous chapters the position of a point has been indicated by the coordinates x and y, and the locus of a point, under 1.
various circumstances, has been represented by an equation involving the variables x and y.
The
position of a line can also be indicated by two coordinates for example, these coordinates might be the lengths of the intercepts ;
made by
the straight line on the coordinate axes. culled line coordinates.
Such coordinates are
It is possible to of line a coordinates quite independently ; we shall develop system Line this. illustrate coordinates, however, are most valuable when
used in conjunction with point coordinates
;
we have
that system of line coordinates which works previous work on point coordinates.
therefore adopted
most
easily with our
straight line, including the line at infinity, can be 0. If we represented by an equation of the form lx + my + nz exclude the line at infinity, we can use the simpler form of this
Any
2.
equation, Ix
=
-f
my
+n
= 0,
as the general equation of a straight line
;
this equation involves three coefficients, ?, m, and n. If, further, we exclude straight lines which pass through the origin, so that n is not zero,
we can
lx + my+ 1 =
mutual
If the Ix
take the general equation of a straight line in the form 0; this equation involves only two coefficients, I and m.
+ my + nz
=
is
m
n are known, the position of the line and known, similarly, if the values of I and m are
ratios
I
:
:
known, the position of the (/,
(?,
my +
1
=
is
known.
Thus
be regarded as the coordinates of the line Ix + my + nz0; as the coordinates of the line lx + my+ 1 =0. m)
m, n)
and
line lx +
may
When using point coordinates we found two coordinates (x, if) except when we had to deal with points at infinity, in which case used the homogeneous coordinates (a?, y, z).
Note.
sufficient
we
When using line coordinates we shall find two coordinates (I, m) sufficient except when we have to deal with lines passing through the origin, in wh ich case we shall use the homogeneous coordinates (7, m, n).
LINE COORDINATES Let OL,
391
OM
be the coordinate axes, rectangular or oblique, and let AB The Cartesian equation of the line AB is (/, in). Thus the straight line 0, hence OA -1/J, OB = -I/at.
be the straight line 1
=
M
\ A
1
the coordinate axes. So also the m) makes intercepts I//, -1/w on makes (I, m, n) n/tn on the axes. intercepts n/l, The usual convention of signs can be followed thus in the figure the straight line A' B' is the line (-Z, -in). (/,
line
;
Note
The Cartesian equation of any straight line through the origin = is remembering that (?, in, n) is the straight line Ix + my -f n 0, we see that the homogeneous coordinates of this straight line are (/, m, 0). Noteii. The coordinates of the axes OL and OM are (0,1,0) and Ix -f
i.
my =
;
(1,0,0).
OL
Note ill. The coordinates of (e.g. y + a0) are (0, l/o). Note iv. The equation of the zero the equation Ix + my -f nz =
if I and are line at infinity is z ; reduces to z = 0. Thus the coordinates
of the 'line at infinity' are
or in
(0, 0),
a straight line parallel to the axis
=
homogeneous coordinates
Examples 1. 2.
(0, 0, 1).
X a.
Show
that the straight lines ( + a, +6) form a parallelogram. Prove that the coordinates (Xa, \b) } represent, for different values of
a system of parallel straight 3.
m
The
straight lines
X,
lines.
(a, &),
(6,
a) cut the axes of coordinates in four con-
cyclic points. 4.
Find the condition that the straight lines
(l lt
M,),
(Ja
,
wa
)
should be at
right angles. 5.
(*1
l)
7.
angle does the straight line (^ in, make with the axis of x ? the axes are rectangular, find the angle between the. straight lines
What
6. If
,
)
(*!Wl)-
What
does
(ft,
k, 0)
represent (a) in point coordinates, (b) in line
coordinates ? 8.
The perpendicular from
and makes angles
OK
and
the origin on a straight line is of length p, what are the coorwith the coordinate axes
dinates of the straight line ?
:
LINE COOKDINATES
892 9. If
the axes are rectangular the straight lines
right angles. 10. If the axes are oblique (/,, wij,
0) should
(
(7,
m,
0),
(w, -7, 0) are at
the condition that the lines
(J x
,
m, 0), ,
be harmonic conjugates with respect to the straight lines
(6,1,0), (1,0, Q\
=
0, involving the coordinates /, m is equation, (?, m) of the coordinates by any one of a group of straight lines same that an is satisfied by the coordiin the way equation f(xy) =
3.
Any
satisfied
nates of any one of a group of points. If any value be given to / in such an equation, corresponding values of in can be found, giving the coordinates of lines of the group. coordinates are (?, m) and (l + h, m
Any two lines of + k) intersect in a
the group whose
point as A, and therefore k, are indefinitely diminished, this point takes a definite limiting position, which is usually referred to as the point of intersection of
The
;
lines of the group defined by ((/, m) = 0. determined trace out a curve, and the equation
two consecutive
series of points so
of this curve in line coordinates
=
0. Every straight line m), whose coordinates satisfy this equation, touches this curve for such a straight line meets the curve in two coincident points, viz. the limiting positions of its intersections with the lines
is
(Z,
m)
(I,
(l
;
+ h, m + k),
h',
(I
m
k')
when h and
//
and therefore k and
k'
are indefinitely diminished. The curve is therefore the envelope of the line (Z, m), when its coordinates are subject to the condition
Thus the equation
/(#, y)
=
represents the locus of a point
whose coordinates satisfy this equation, and the straight line joining two points on the curve is, in the limiting position when these two points become coincident, a tangent to the curve. (Xj
y)
The equation
=
represents the envelope of a line (J, m) whose coordinates satisfy this equation, and the point of intersection of two tangents to the curve, in the limiting position when these two tangents become coincident, is a point on the curve. = represents the same curve as /(#, y) = 0, it is called If (i, w) (i,
m)
=
the tangential equation of this curve is evidently ({, m) the condition that the straight line (Z, m) or lx + my+ 1 should ;
=
=
touch the curve f(x, y) 0. Cartesian coordinates are useful for the investigation of loci and line coordinates are useful for the investigation of their properties ;
envelopes.
We
have learned
the object of this chapter of tangential equations.
how
is to
to interpret the equations of loci
interpret
;
and discover the properties
4.
AND TANGENTIAL EQUATIONS equation of the first degree Al + Bm + C =
The
The Cartesian equation of the line and Al + Bm + C = is the condition
sents a point. lx +
=
my + 1
0,
393 repre(?,
straight line should pass through the point (A/C, B/C). Hence, +C straight line whose coordinates satisfy the equation Al +
Bm
passes through the point (A/C, B/C). tangential equation of this point.
We lx
now
have
= and m
+ my + 1 If
This equation
is
is
m)
that
this
any
=
then the
the following reciprocal property of the equation
:
are constants, it is the Cartesian equation of the straight line whose line coordinates are (Z, m). If x and y are constants, it is the Tangential equation of the point I
whose point coordinates are
y).
(.r,
So also in homogeneous coordinates Ix -f my -f nz = is the the of line and line the point equation (/, m, it) equation of straight
Note
i.
;
the point (x
Note
t
y, *).
The origin
ii.
equation of the origin It
lines
is
the point
is
n
=
(0,
0,
1)
;
therefore the tangential
0.
should be carefully noted that when we are dealing with the origin or through the origin in line coordinates we must use homogeneous
coordinates.
Note iii. The homogeneous coordinates of a point at infinity are of the form (x, y, 0) therefore the tangential equation of a point at infinity is of the form Jo?-f my 0. ;
6.
(i)
To fin
lines (li,mi\
the equation of the yoint of intersection of the straight
I
w*J-
(J 2 >
If the point is al
then, since
it lies
+ Im +
on the given al Y
and
al 2
lines,
=
1
:
(i)
we have
+ bnii + 1=0, + fcw2 + 1 = 0:
(ii) (iii)
these equations give the values of a and b or, in the determinant the unknown constants a, b from (i), (ii), and notation, eliminating ;
(iii),
we
find that the equation of the point is I
m
1 t
= Cor.
i.
the form section
is
0.
If the straight lines are parallel, their coordinates are of the equation of their point of inter(Z lf Wj), (W,, ftrnj, and
mj
l^m
=
0,
i.e.
a point at infinity.
LINE COORDINATES
394 Cor.
The
ii.
lines
tnj,
(Zx ,
(12
h Cor.
(ii)
w3
(Z 3 ,
),
AM
7
are concurrent if
)
f\
1
1
3
The coordinates
iii.
section of
wa
,
(?!, nij),
To find
(?2 ,
m2
of a straight line through the interare of the form
)
the coordinates
aZ
line joining the points
of the straight
+ fcw+1 =
a'Z
0,
= 0.
+ 6'w + l
These are evidently found by regarding the equations of the points I and w.
as simultaneous equations giving
= 0,
If u
(iii)
u -f A t;
the equation
= = 0.
are
Me
tivo points,
equations of
to
interpret
=
= =
coordinates of the line which joins the points u 0, v 0. w At; each of these and + consequently satisfy satisfy equations, Hence this equation for different values of A represents points on
The
= 0, v = 0.
the straight line joining the points u
To find
(iv)
points al
of the point which divides the join of the in the ratio p:q. 6'w + +l
the equation
+ bm+ = 0, 1
The Cartesian
=
a'Z
coordinates of the points are
(a, b),
(a', {/),
and
those of the required points are
(qa+pa'
gb+pb'l '
p+q
\
p+q
m
y
hence the equation of the point in line coordinates
I
q+p q+p = lm + + q (al 1) +p (a'l + b'm + 1)
e.
Cor.
i.
is
0.
The equation of the point midway between the points al 4- 6m + 1 =0, s
Cor.
ii.
al + bm
+1
If
=
u
0,
=
0,
v
=
are the equations of two points in the form are harmonic con0, pu-qv
then the points pu + qv
jugates with respect to u =
Example. that
The
sides of
BP^p.PC; CQ =
P, Q,
R should
0, v
q
be collinear.
.
=
=
= 0.
ABC are = AE r KB QA a triangle ;
.
;
divided at P, Q,
M
so
find the condition that
(Menelaus* Theorem.)
AND TANGENTIAL EQUATIONS
395
Let the vertices of the triangle be
u =E al + bm + v M>
a'l
(?!
W|)
,
if
t\
+pWi
i.e. if
l-
-
(>,
satisfied
0,
u>!
by the coordinates of a straight line
+ gMj =
0,
pqr
-f
M!
-3,
ri^
-
-
1
MJ
or is
E are
^=
= -,
/#i
which
0,
1
then the equations of the points P,
and these are simultaneously
=
1
= + b'm + = 0, = a*J + &*m-l-l = 0,
=
0,
r.
t?j
=
1,
the required condition.
The reader should compare the
solutions in point and line co-
ordinates of the following elementary problems :
To tvfiose
the origin is
To find the envelope of a line whose distance from the origin is
c.
constant and equal
find the locus of a point distance
from and equal to
constant
(Line coordinates.)
(Point coordinates.)
Let any such point be
which
to c.
(#, y),
Let any such line be
then
m), then
(I,
the equation of the locus.
is
or
which
is
(Point coordinates.)
(Line coordinates.)
Let xl
+ ym + \
point, then
=0
x* 4- y*
Now on any line of the locus will are given
m') two points whose x and y
1
Ix +
a
c (f
I
m
and Ix'
2
and
,
ratio x/y' is given
any such
8
-f
4-
are given by
my + 1 =
and the
by
= be m ) = 1.
1
2
Now through any point (a?', y') two tangents to the envelope will pass,
= 0,
=c
my 4
then
whose
x2 4 y1
and the
Let line,
.
(/',
lie,
by xV 4 y m' 4
and
be any such
= c2
the equation of the envelope.
2
m = 9
4
)
ratio 7/m
is
0, 1
,
given by
and the
These two tangents are coincident, and the point (x' y) lies on the en-
if
velope if
These two points are coincident, line (/', m') touches the locus
or i.e.
s
c (r*
+m
*)
=
or
1,
the equation of the locus
t
is
i.e.
the equation of the envelope
is
LINE COORDINATES
396
Examples Xb. (The axes are rectangular except in questions marked with an
Show
*1.
that the distance between the points 1 = 0, a'l + b'm + 1 -0 y {(a a')* + (6 - b')f + 2 (a - a') (6 - V) cos *}
is
line
(7,
,
1
=
from the
m,).
*3.
Find the angle between the
*4.
Prove
lines
(l lt
mj,
the straight lines (J^Wi),
'that
+ !! wa -f (/x m.2 -f 7a W| i) cos o> = 0. *5. The points al + bm + l = 0, a'l + b'm + 1 the origin if aa 4 -f (aV -f a'&) cos o> = 0.
/!
.
Find the perpendicular distance of the point a/-f &w-f
2.
asterisk.)
(J 2
(Ja
,
wi a ).
,
are perpendicular if
*w 2 )
72
=
subtend a right angle at
W
Find the equation of the mid-point of the line joining the points = 0, Ix9 4- my^ -f n^2 = 0. Plot the following points at+ 1 = 0, 6m -1=0, J-f w = 2, l-m = 2,
6.
+ myl -f nzv
?xl
7.
:
Find the general equations of two points equidistant from the axis of /. Show that for all values of X the point al + bm + 1 -f X (a'l + &' w 4- 1) =
*8. *9. lies
on a
fixed line,
and
find its coordinates.
Show
X reprethat for different values of X the equation a/-f bm sents points lying on a straight line through the origin. *11. Find the coordinates of the line joining the points 3/-f4m= 12,
*10.
5J-6w=2. What
12.
a8 Z
a w + et
-I- fc
is
the area of the triangle whose vertices are a l l+b l m + cl a 8 7 -f 6 8 m -f ct = 0.
*13. Find the coordinates of a line parallel to
(llt
mj
the point al -f bm + 1 = 0. 14. Find the coordinates of a line perpendicular to al
= 0,
= 0,
+ bm +
1
=
and passing through (Jlf
m^
and passing
0.
through 15. Find the condition that the line joining the points al -f bm + 1 = 0, a'l + b'm + l s should be perpendicular to (llt m l ). 16. Find the condition that the point ^-hmt/-fl=0 should lie on a circle whose diameter is the join of the points
bm + l =
al + 17. is
a point on the circle
;
0,
a'J
is
A straight
line
(I,
= 0.
is
through this point. *18. Find the condition that the points a, + &,m + l*=0 should be collinear. 19.
+ fc'wH-1
the point al -f bm + 1 = 0, and Al 4- Bm -f 1 find the equation of the other end of the diameter
The centre of a circle
m)
aj -f
cuts the axes at
fc
x
w+1
^ and B
=0,
a a J -f 62
so that
m + 1 = 0,
I/O A* + 1/0 B
2
Find the equation of its envelope. Find the coordinates of the straight lines bisecting the angles between
constant. 20.
the straight lines (3, 4) and (5, 12). *21. Find the coordinates of the straight line joining the intersection of 1 0. Ci i)> (*if **) to the point af-f bm+
m
AND TANGENTIAL EQUATIONS 22.
What
897
are the line equations of the curves enveloped by the lines
2 (a) (a cos 0, a sin 0), (b) (a cos 0, b sin 0), (c) (a<9 , a6), where 6 is variable ? *28. A straight line cuts the axes at the points P, Q find the equation of its envelope in the following cases: ;
(i)
1/OP+ l/OQ =
c
(iv)
I/OP- 1/00 = c; (iii) a/OP 6/00 * OP+ 00 - c; (v) OP- 00 = c.
c;
(ii)
;
Show that the line at infinity touches the last two curves. *24. Find the equation of the centroid of the triangle whose vertices are a 1 l + 2 1 + 1 =0, a a Z + &.2 + l 0. +l 0, a s Z-f& 3
m
=
w
=
w
Find the equation of the envelope of a line which moves so that the
25.
=
foot of the perpendicular on it from al+ 1 lies on the axis of ni. 26. Find the envelope of a line whose distance from the point (a, & t ) ,
double
6.
u
is
,
The general equation
= aZ
is
distance from the point (a 2 &2 ).
its
2
+ 2w + &w
2
of the second degree in line coordinates 0. If u has factors, i.e. if
+ 20J + 2/w + c
=
=
A=
The co0, the equation u represents a pair of points. ordinates of the line joining them are (6r/C, F/C). (Cf. p. 109).
The equation
al*
+ 2hlm + bm* =
=
equations are of the form pl + qm Definition.
Two
two points whose two points at infinity.
represents 0,
i.e.
points at infinity which subtend a right angle
at the origin are called orthogonal points.
To find aZ
2
the condition that the points at
4-2Ww +
Let
a? 2 4-
fcw 2
=
infinity wJwse equation is
should be orthogonal
2hlm + bm 2
Ha
(Z -f
Jem)
equations of the points are coordinates are (1, ft, 0), (1, &',
I
(1 4-
k'm\ then, since the tangential
+ km = 0, + Ic'm = 0, I
0).
These points
lie
through the origin whose point equations are k'ory = 0, or, in one equation, lines
their
point
on the straight
=
y
A\r
0,
x
.e. i.e.
Thus the Cartesian equation 2 infinity, aZ
+ 2Mm + Zrtn 2
These are perpendicular
= 0, if
of the lines joining the points at
to the origin is kr 2
a + & + 27*cosa>
Note i. The lines ax* + 2 hxy -f by2 = The points
ii.
are orthogonal
orthogonal
The Cartesian equation of the
if
if
# 4y
2 -I
0.
a + b - 2 h cos o> =0.
a4 b4
2/t cos
o>
=
0.
lines joining the circular points
at infinity to the origin (called the circular lines through the origin) 2
=
2hxy + ay*
= 0.
2xy cos w
0.
is
LINE COORDINATES
398
This equation can be written
= 0. (x + e^y) (x + e-tvy) Hence the line coordinates of these circular lines are (1,
0,
e~ i
;
i.e.
Z
3
or, in
one equation,
-2?mcosa)-ft
8
=
0;
this is therefore the tangential equation of the circular points at infinity. If the axes are rectangular, the circular lines are (1, i, 0), (1, -/, 0), and
the circular points P +
Note at to
ma =
0.
The Cartesian equations of the lines joining the pairs of points a whose infinity, equations are aP + 2hlm -f 6m = 0, P-2lm cos o> -f w? = 0, a the origin are bx*-2hxy + ay* = 0, # -t-2arycos6>-f a = 0. These lines iii.
i/
=
are harmonic conjugates if a-f fc + 2ftcoso> hence, orthogonal points ; at infinity are harmonic conjugates with respect to the circular points at infinity; and, orthogonal lines through the origin are harmonic conjugates
with respect to the circular lines through the origin.
Example
If
i.
aZ 2
+ 2Wm + bm* + 2gl + 2fm + c =
represents
two points, find the equation of the point midway 'between them. Let ai* + 2Um + bm* + 20J + 2fm + c^c(pl + qm + l) (p'l + q'm + the equation of the point midway between the points pl + qm+l^Q,
p'l
+ q'm + l
then
l),
=0
is
or
Example points P,
where
ii.
To find
the angle between the straight lines joining the
Q, whose equation
A=
Method
0,
1.
to the point
is
E whose equation is pl + qm + 1 = 0.
The equation
aP + 2hlm + 6wa - 2
(gl
+fm) (pi + qm) + c (pi + gm)
2
(i)
by the coordinates of lines which pass through one of the points and also the point R.
is satisfied
P,
Q It
can be written
or
This equation therefore gives the ratios of the coordinates l/m of the PR and QR. If the roots of equation (ii) be ^/w^ and J 2/w a since
lines
,
the Cartesian equations of the lines are l^x +
we have
tan
PRQ
t
1
y , + ^?2/W l tWj ,
/
m^ + l = 0, M
-f f/;
ItX + rnrf
+l
=
0,
AND TANGENTIAL EQUATIONS Method
399
The equation t/P-4 2t?Jw-H0w = represents two points at the intersections of the lines PR, QR with the line at infinity. a
2.
infinity, viz.
Hence the joins of the
origin to these points are parallel to PR, QR the Cartesian equation of these straight lines is wx? 2vxy + wy a = 0, hence ;
W+U The
7.
Let the centre of the
oirole.
be the point (x, y\ m) be any tangent to
circle
whose equation is Ix + my + l = 0, and let (?, the circle. The perpendicular from the centre on
7^_^r The
and this
,
tangential equation of the circle
is
r.
therefore
this equation with the general equation
Compare and
equal to the radius
is
aP -f 2 him + 6m2 + 2gl + 2 fin + c
= 0,
_o_
_
we have
hence
=
fg
__ ~" _J>__
c 2 rcy,
_
"~"
h_ __ """
g
__/
c
U
= -cV
2
&c-
also
ac-
and
2
=-
Hence the conditions that the general equation
= JB
degree should represent a circle are -4
When
I.
becomes o;
2
r2
+y 2 = r The
form
and
-ff
of the second
= 0.
is at the origin, the equation this is the tangential equation of the circle
the centre of the circle 2 (I
-f
m = 2
)
1
;
2 .
coordinates of any tangent to the circle can be put in the
the Cartesian equation of this tangent
(-cos0, -sin0);
+ y sin + r =
x cos are
this tangent is
rcostf,
(
contact
is
then
so that the coordinates of its point of contact rsiji0); the tangential equation of the point of
rl
0,
cos
+
m
sin
0=1.
(a^, yj with respect to the hence the polar of the point
Again, the polar of the point 2
= r2
is
is
xx l + yy l
with respect to the
= r2
circle r2
;
2 (Z
Similarly, the pole of the line
+ m2) (1 19
=
Wj)
1
is
is
the line
(
circle
--1> --!
the point rtj +r^n l
m=
1
LINE COORDINATES
400
These results can be found independently as illustrated in the following example :
To find
the equation
of intersection of the tangents whose
the point
of
~sin0\
coordinates are (-cos#,
(-cos<J>, -sinV
If the equation of the point of intei section is
since the given tangents pass
through
p cos
p cos and by
_ ~"
p i.e.
6
-f
q sin
-f
q sin
-f <
r
=
+r
(0 +
= <,
(-COS0,
-sintfj
is r/
)
0,
by sin J(0-<), we have
____
_j?
r ^
is
cos
+ rwi
sin
=
1
;
this
is
therefore the general
circle.
of the circle in homogeneous coordinates
by the coordinates of the circular i, 0), (1, 0), and therefore the
this is satisfied
origin, viz. (1, circular lines through the centre. Q or J + wi contact are r(lim) ?',
=
infinity. II.
(Vide Chap. V,
The equations
have,
the equation of the point of contact of the tangent
equation of a point on the
The equation
= 0, we
8i
the equation of the point
Hence, putting
___
1
0,
cross- multiplication, dividing the result
cos
pl+ grw+
it,
of
is
lines through the circle
touches the
The equations
= 0,
i.e.
of the points of the circular points at
16).
any two
circles are (i)
(ii)
and the coordinates of any line which (i) and (ii) will also satisfy the equation which
is
satisfy
both the equations
equivalent to the two equations
+ qm + 1) -f r (pj, + qjn + 1) - r (pj + q^m + 1) fi ( pi + qm + 1)
rj (pi
= 0, = 0.
These equations represent two points dividing the line joining the centres of the circles in the ratio of the radii, i.e. the centres of similitude. Hence, the common tangents of two circles pass through
the centres of similitude.
Example
i.
its
401 lines in points
a right angle at a fixed point midway between the
ivhich subtend
find
AND TANGENTIAL EQUATIONS A straight line cuts tivo parallel straight
lines
:
envelope.
Let the fixed point be the origin and let # + a = 0, tf-a be the Cartesian equations of the given parallel lines. Let (/, m) or Ix 4- my + 1 = be the straight line then the lines joining its intersections with x + a = and #-a = to the origin are Ix + tny + x/a = and Ix -f my - x/a = 0.
=
:
These are at right angles, hence f envelope of the line is
m)
(I,
is
the origin and radius
a.
Example
point
A, and
PQ
Any
ii.
is
a
a (P
P
w = 2
^ -f
+w
2 )
=
1,
on a fixed
drawn perpendicular
to
0, i.e.
which
is
the equation of the
a circle whose centre
a fixed point
circle is joined to
AP
:
find
its envelope.
Let the straight line PQ be Ix + my + 1 = 0, and the fixed point (a, 0) and let the circle be #2 -f y* = r2 The equation of the straight line through (a, "0) perpendicular to Ix + my -f 1 = is mxly = am, and these lines intersect on the circle. x and y from these equations and a:2 + y2 = >*, we have Eliminating ;
.
(Ix i.
e.
-f
=
+ (mx - ly)* 8 a (P + m ) (x* + y
my)*
)
r*(P + m*)
i.e.
which
is
1
=1 =
+ a* w% -f
a'm',
the equation of the envelope required.
Examples Xo. 1.
Find the coordinatesof the chordjoining the points
=
rm sin Show that
rl cos
-f
rl
cos 6
-f
rm sin 6 = 1
,
1.
2. the polars of all points on the line (?', m') with respect to the circle r1 (F -f a ) = 1 are concurrent. 3. The locus of the intersection of tangents to the circle ^(P + m') = 1, which include a right angle, is 2r* (J3 + a ) = 1.
m
m
4.
A
chord of a
the equation of
5.
point
:
and radius. Find the condition that the point
circle r8
(/
-f
m
a )
1
aZ
+ &m +
l0
should
lie
on the
.
Find the coordinates of the tangents from the point 2aJ circle 2a*(P-f m 8 ) 1. 1*67 C C 7.
show that
:
find its centre 6.
fixed,
The sum of the perpendiculars from n fixed points on a straight line is show that its envelope is a circle. the points are aj + \m + 1 = 0, a a Z + 6 2 wi + l = 0, a 3 f-f& 8 w + l = 0,
constant If
its
subtends a right angle at a envelope can be put in the form
circle
=
1 to
the
LINE COORDINATES
402
Show
8.
and
3w -4//n-4J-2wt 8
that
1
=
a
is
circle,
and
find its centre
radius.
9. Find the conation that two circles whose line equations are given should be orthogonal. 10. Find the equation of the circle of similitude of the circles
(pi -f I)
Show
11.
=r
2
2
(I
+ w2 ),
(ql
+ 1) 2
= s2 (P m -f
2
).
that the Cartesian equation of the circle
r
2
2
(Z
= (to-fm& + l) points A and A',
+ m 2)
Through two
12.
2
2
(x-a)* + (y-6)
is
AP,
AA', and A'P, AP' are perpendicular. the mid-points of AP and A'P'.
2
=r
2 .
AP are drawn perpendicular to
Find the envelope of the join of
=
13. Show that the equation c J -p(/i -f c )w -2/iJ-l represents for different values of n a system of circles passing through two fixed points. 14. Show that the tangents from any point on the axis of to the circles 2 2
2
2
2
m
2
a
+ (c 2 -X 2 )w 2 + 2X/-fl
= 0,
2
2
+ (c 2 -/i2 )w 2 + 2 M J+l=0 are
equal, and for X different values of equation represents a system of coaxal circles whose limiting points are cl 1 =0.
c
/
hence that the
Show
15.
in Question
c J
first
that the coordinates of the 1
4 are given
m = 27.
common
by the equations
2
2
tangents of the two circles + 2 (c 2 4- X/i) I = 0,
(c Z -f 1) (X -f p)
2
(X-f
/i)
Deduce that they intersect on the line of centres of the from the radical axis whose product is c 2 Two circles of a coaxal system can be drawn to touch any given
circles at distances 16.
.
straight line. 17. Pairs of circles of a coaxal system are taken such that the sum of the distances of their centre from the radical axis is constant. Show that their 2 tangents envelope the curve m == 2ld. 18. For different values of c the equation c 2 w 2 -2cZ-l circles touching each other at a fixed point.
common
=0
represents
A
system of circles touch one another at a fixed point, and a line Show that the tangents to the parallel to their common tangent cuts them. circles at the points of intersection with the line all touch another fixed 19.
circle.
The general equation
8. I.
To find the
of the second degree.
Cartesian equation
To find
the tangential equation
of
of the envelope
the locus
In other words, to find the equation of the envelope of the straight line
In other words, to find the equation of the locus of the point
lx + my + I
when
/
and
m
=
0,
satisfy the equation
xl + ym +
when x and y
l
=
0,
satisfy the equation
AND TANGENTIAL EQUATIONS Through any point
(a?,
On any
two
y)
straight line (/, m) points of the system defined by /7 for x values of such the and for lie, y
straight lines of the
system defined by 2 the for values of I and pass, for such lines are given by
=
m
lx + mi/+
=
1
points are given by
ax 1 + 2hxy-{ by 9 - 2
-f
2
-f
(ft
- 2 (jx -f ex *) (a 1
- 2fy + cy 2 (b
=
(ft
-f
which at once reduces 11
-f
2 7te/
-f
By*
-f
)
,
0,
-f
cJw) xy
- 2fm -f cw 2 )y2
=
0.
if
~ 2/m + cm 2 ) (b = (ft /J -
>
which at once reduces to
to
u4P
the
=
2
2
m) touches the locus
2
2 Gar
therefore
is
(/,
)
=
-f2Ft/-f
which
(&
my)
(fc? -f
+ wy) 2
These two points are coincident, and the straight line through them
- 2gl -f c/ 2 (a
-fx -gy -f c.rt/;
c?
+/y)
(lx
)* 2 (ft -// - gm
0.
These two straight lines are coincident, and their point of intersection (x, y) lies on the envelope if
Ax
(a~2^-f
i.e.
-/# - #i/ 4- c#y ) / w + (b - 2/# -f ey) ma =
for* -f c
2 (a-2gx + cx*)l
i.e.
0,
and
+ 2 him -f lm* - 2 (gl +/w) (lx + my] 2 = -f c (te + my) 0,
2
ym+l =
xl +
0,
and al
408
4-
2 #/m + J5m 2 + 2
<7 /
+2Fm-f
0,
which
Cartesian
equation of the envelope.
is
therefore the
=
0,
tangential
equation of the locus.
These propositions are very important, and another proof will be given
later.
Since the envelope a^-f 2ft/m-f few 2 -f 2gl + 2fm + c = is the same as the locus Ax* + 2 xy -f By 2 -f 2 6?# -f 2 Fy C = 0, the envelope is a oonio.
Note
i.
H
-I-
Note ii. The Cartesian equation of the conic 2 = and the tangential are formed from these equations by the same equation of the conic S = process, viz. replacing a, 6, c,/, g, ft by A, B, C, F, G, //and interchanging ly
m
and
x, y.
S = is ^IP-f 2Jftw-f #m + 2GZ + 2FtH + C
Since the tangential equation of
a
=
0,
the Cartesian equation of the latter must be S = 0. This fact indicates the reciprocal relations which exist between the letters a, b, c, /, g, ft and
A, B, C, F, G, H,
viz.
a
6c-/ fg-ch
^, BC-f* = = H, FG-CH~
A
a,
Aft,
&c.
;
these can be easily verified algebraically.
Note
iii.
The conic Ax* -f 2
//a?y
+ By*
-f-
2 Ga? + 2
Fy + C
=
is
an
ellipse,
parabola, or hyperbola according as AB-H*, i.e. Ac, is positive, zero, or a 2 is an negative: hence the conic 2 =EaJ + 2ft/m-f fcw + 20rJ-f 2/m-f c = or ellipse, parabola, hyperbola according as Ac is positive, zero, or negative.
The condition at infinity
(0,
for a parabola, c
0,
0) touches the conic 2
r
corresponds to the fact that the line
= 0. p.
2
LINE COORDINATES
404
we can show
Similarly, (a)
2
=
(b)
2
=
that
a rectangular hyperbola if A -f JB-2#coso> passes through the origin if C = 0.
is
=
;
The latter can bo shown independently thus since we are considering the origin we must use homogeneous coordinates the equation 2 = can then be written aZ 2 + bni? + en* + 2fmn + 2 gnl 2 him = 0. :
;
-f-
The coordinates therefore of the lines which pass through the origin and and at* -f 2 him + km* = 0. These tangents ti* = 0. are coincident, i.e. the origin n = lies on the curve if afc touch the curve are given by n
Note
iv.
(a)
When A =
the equation
S
represents a pair of straight
lines.
Now
AC-G* = &b, C-F*=:Aa, FG-CH=*h, G = AC, F = BC, FG = CH. tangential equation corresponding to S = 0, viz. Al* + 2 Him Bw 2 GJ 2Fw C = 0, 2
hence
The
2
2
-f
4-
4-
-f
can therefore be written
.e.
which
is the square of the line equation of the point (G/C, F/C), i.e. of the of intersection of the lines 5=0. This is geometrically evident since point all lines which meet S = in coincident points pass through their point of
intersection.
In the same way if A = 0, 2 = represents a pair of points, and the a Cartesian reduces to (Gx + Fy+
square of the Cartesian equation of the line (G/C, F/C), joining the two points.
The geometrical two points two points. to the
To find
II. line
(Zj
,
The
significance is that all pairs of points, the joins of which are coincident, must lie on the line joining tue
2=0
v
n (Z
u
y
and
the
2
=
/7 (Z 2
(l^,
. ,
tw 2 ) are
+ $i
pnti *
j
+ qm^
r.
.
.
This
if
2
This equation lines, and
such
x
w^),
+ 2ft (pli -f ^ 2 (jPWj + 2>n 2 + b (pm -f qm,)* 2g (pi, -f ql2 (p + q) + 2/(jpm + qm 2 (p + q) + c (p +
+
=
coordinates of any straight line through the point of inter-
line touches the conic
-f
of the points of intersection of 2
the equation
section of the lines
(pli
of the line
m^.
.,,
..
i.e.
)
)
)
is
if
1
l
)
2
=
0.
(i)
quadratic in tfee ratio p/q, so that there are two they are coincident the point of intersection of e
on
2=0.
AND TANGENTIAL EQUATIONS
We adopt a
405
notation similar to that used for the general Cartesian
equation of the second degree, thus
L=
al
M=
+ hm + g,
hi -f
N = yl+fm +
bm +/;
c,
and so LI
=
mi + g
ajj 4. j
The equation and
(i)
M
+ 6m! +/,
Mi
l
9
JVi
=
reduces to
this has equal roots if
2 1 .S a ={Z a which
therefore the condition that the point of intersection of the lines (li,mi), (Z 2 2 0. Hence the coordinates a ) should lie on is
,
=
w
of any line through either point of intersection of conic 2 satisfy
(i x
=
2.2 1 which Cor. (ii)
= (LL + w2lf + tf 1
1
If (I l9
m
is
v)
and the
x)
2 1)
(ii)
,
a tangent to the conic, then 2 t
+ wJ/i + N^ =
point of contact of the tangent
The equation
is
;
(l lt
To find
Is (l' y
=
m
mM^^
0.
of intersection of tangents whose
m).
Let the tangents be
(l it
w^), (l^
ma
then their points of contact
),
mM
=
0, IL 2 + 2 + JV 2 of hence both these points, through
1^ +
and the equation
algebraically equivalent to
the equation of the point
chord of contact
=
this is therefore the equation of the to the conic 2 0. t)
=
are
m
therefore the equation of the points of intersection.
is
reduces to IL V
III.
,
and these are the conditions that the
=
lines
each pass through tlie point l'L + m'M+ quently the equation required. It is algebraically equivalent to IT/
0.
(7 lf
N=
But
(Z',
Wj),
(Z 2
IV. To find the equation of the pole of
m
which
0,
+ mM.' + N'
,
m
=
7
)
passes
2)
should
is
conse-
0.
(V, w').
Let (^ be any chord such that the intersection of the tangents at its extremities lies on (i', m'). ,
m^
The equation
of this point of intersection is IL 1
and since by hypothesis
it lies
on
(I',
m')
we have
4-
mM
l -f
j^
=
0,
LINE COORDINATES
406
But fixed
this is the condition that ft,
l'L +
point
m'H+N
tangents from any point on is
therefore the pole of
The equation
is
which
(V, m').
mM! + JV' =
algebraically equivalent to IL' +
N = 0,
(0, 0) is
therefore the centre of the conic 2
is
contact of
of
this point,
m') passes through
(J',
Cor. The pole of the line at infinity
which
should pass through the
m^
Thus the chord
0.
= 0.
i.e.
0.
gl+fm + c
=
0,
Its point coordinates
are (g/c,f/c).
V. To find Let
the coordinates
of the asymptotes of
be an asymptote
ft, Wj)
this point
must
its
on the line at
lie
infinity (0, 0)
l
l
is
0,
N = gl +fa + c = Now since
= 0.
point of contact
+ mMi + Ni =
ILi
and
;
2
hence
;
0.
(i)
we have ft, mj 2 al^ + 2 Ml m l + 6m! + 2^ + 2fa + c = 0. is
The equations
and
(i)
a tangent to the conic,
also (ii)
therefore give the required coordinates
(ii)
;
these equations reduce to
=
gl+fm + c The
0;
aP + 2ftZw + &m 2
ratio of the coordinates is given
If the asymptotes are perpendicular,
l^ 2
is
of
we have
the condition
tfiat
(l ly
Wj),
(/2 ,
m
2)
should be conjugate
the conic.
The pole of the and
by
= 0, A+B 2Tcosa> = 0, 2 = should be a rectangular hyperbola.
the condition that
To find
VI. lines
c.
+ mjmj 4- (?iW 2 + l^mj cos co
or
which
=
this lies
on
(Z 2
,
line
(1 19
w
if
<2
)
12
L
m^
is
}
which gives the condition, symmetrical with respect and Z 2 wi2 ,
Z
^2) "h c ==
0.
lf
m
l
,
alil%
If ft,
to
+ &frWi*w2 4" A ftwi 2 + ^2^1)
m
A)
and
(Z2
,
W2
)
*4" fl'
ft
+
? ) 2
"f"t/(^i
~i~
are diameters of the conic, each passes
AND TANGENTIAL EQUATIONS = and N2 = through the centre gl +fm c = 0, so that = 4-
407
JVji
;
hence
c,
or
acl
which reduces
to
this is the condition that the diameters
m^,
(Z lf
(Z 2
,
w
2)
should be
conjugate.
VII. To find
the director circle
= 0.
of the conic 2
=
Let ?#-fw#+l be a point on the director circle, then the coordinates of the tangents from it to the conic satisfy the equations lx
The
+ my+l =
= 0.
2
and
coordinates of these tangents therefore satisfy
+ 2 him + lm2 - 2 (gl +fm) (lx + my) + c(lx + my)* = 0, 2 (w*-2gx + a)P + 2(cxy-gy-fx + h)lm + (cy*-2fy + b)m = 0. al
i.e.
2
This equation therefore represents two points at infinity on the if the tangents are at to the conic tangents from te-f tny-f 1 = right angles these points are orthogonal, hence ;
c (x
i.e.
2
+ y2 4- 2xy cos co)
2x (g -f/cos a>)
2y (f+g cos a>) -f
which
is
a + 6 + 2Acosa>
=
0,
therefore the Cartesian equation of the director circle.
If the axes are rectangular the equation of the director circle is
and
its
tangential equation
is
2 2 (A + S) (I + m
If the conic is a parabola director circle reduces to
which
is
)
4- (gl
+fm + c) 2 =
we have
the directrix of the parabola.
the foci
of the conic
=
The
2g/(a +
directrix of the parabola are then [
VIIL To find
c
2
0.
the equation of the
;
line coordinates of the Z>),
2f/(a +
6)].
= 0.
We have just shown that the tangents from
the point
(#,
y) to the
conic pass through the two points at infinity whose equation
is
LINE COORDINATES
408
Consequently, if (x, y) is a focus, these points must coincide with the circular points at infinity whose equation is Z
if (x, y) is
Hence, (xr
When
=
0.
a focus,
= c#
2#r + a
2
-2Zwcos
2
2
2/y +
=
Z>
seca>(ary
gy + h).
fx
the axes are rectangular these equations become
c(^-y2)-20tf-f2/# + a-b = cxyfx-gy + h =
and
0,
(cx-g)*-(cy-f)* = 0*-/2- c a -&) = 4-5 =H (cz 0) /) = ^
i.e.
(
and
eft
(ct/
.
Note. The tangential equation of the conic s
The
=
foci of the conic
are therefore given by
(Cx-G)(Cy-F) - FG-CII=Ah.
When
the conic
2
=
is
tangents from the point
(a?,
a parabola
we have
(See p. 245.) c
=
0,
and the
through the two points
y) to it pass
at infinity
and
infinity.
the focus, these points must be the circular points at Comparing this equation with that of the circular points
we
two
if (r, #) is
obtain
linear equations in x
and y giving the coordinates of
the focus.
Example of its
A
i.
come
touches three given straight lines,
director circle is given
(r)
:
show that
its
and
the radius
centre lies on a fixed circle.
Let the conic be of + 2 A/m + &m* + 2gl + 2ftn + l = (since the conic has a director circle c^O), and let the coordinates of the fixed lines be (/!,
m^
(72 , wf 2 ) f (/ 3
,
*
3 ).
The centre of the conic we have x=g, y=f.
the point
is
(g,
Since the radius of the director circle
hence
a?
Since the lines (7^
and
a
ni l ) 9 (/ 2
,
2
is r,
+ y1 - a - b
w
2 ), (1 3>
-ft
m
/), so that for the required locus
s)
we have
=r
2 .
touch the conic,
- (x* + //2 - r2 )
AND TANGENTIAL EQUATIONS Eliminating
a, h,
and
409
we have
b,
1 !
2
o,
:
which
is
evidently a circle.
9. The tangential equations of the conies referred to their principal axes.
These equations can be found from the known Cartesian equations of 1, or by comparing the general equation of a to the conic with the equation lx + my + 1 = 0. We shall tangent
by the method
illustrate the latter (1)
The Parabola. ax =
2 2/
method
x
is
ty
in this section.
The equation
+ at
= 0,
2
lx
we have
I
=
m/t
=
I/at
and
if
+ my+1 =
2
am = The
Ellipse.
The equation
is (x cos 6) fa + (y sin 6)/b = = we have la = cos 0,
1
irib
(3)
0, t
Eliminating
.
2
(2)
any tangent to the parabola this is identical with
of
;
we
get
1.
+y 2/l 2 = 1 lx + my+1 = 0,
of a tangent to x 2 /a 2
identifying this with sin 6 ;
In the same manner we
The Hyperbola.
2 2 tangential equation of the hyperbola x /a
a2 !2 -b 2
m = 2
2
y /b
2
=
1
find
that the
is
1.
If the hyperbola is referred to its asymptotes as coordinate axes, its
equation
is
xy
= ca any tangent is A 2 a;+y 2c\ = 0, = -A/2c and m = -l/2c\, 4c lm = 1. ;
whence
Z
2
and
Illustrative Examples.
Example point
lies
i.
Find
the envelope
on a fixed straight
of a chord of an
ellipse
whose mid-
line.
=0
the chord, then Let #'/* + yY& 9 = 1 be the ellipse and te + my + 1 9 the diameter conjugate to this chord is 6 twa?-a*fy 0, and this meets the The coordinates of the mid-point are therefore chord at its mid-point.
{
f
/(a
P + &2 m*), - ft*m/(af P + Vat)}.
This point
lies
on a fixed ntraight
line, say
(?',
m'), so that
LINE COORDINATES
410
This is the tangential equation of the envelope of the line The (/, m). coordinates of the straight line at infinity (0, 0) satisfy it the envelope is therefore a parabola. It also touches the given straight line (V, m') and the tangents to the ellipse at its points of intersection with this line. ;
Example are
ii.
drawn parallel
Through points on a given straight line straight lines to their polars ivith respect to a fixed ellipse : find the
envelope of these lines.
be a2 J* + 6 8 m 2 = 1 and (h, k) the given straight the equation of the pole of any straight line (X, p) is a 2 ZX+6 2 w/i this point lies on the line (h, k) we have
Let the
ellipse
Now
line.
=
1,
and
a*\h + b*nk=l.
(i)
The coordinates
of a line parallel to (X, p) are (cX, 8 the a JX-f b*mp = 1 we have through point
a'cX 2 +
Hence, using condition
aVX Consequently, the line
2
2
6V
+&
2
2
2
/(a*-f X) -f y /(6 the fixed point. (h, k)
2
(cX, cp)
+ X) =
(ii)
envelopes the parabola
The coordinates of this
(h, k).
the polars
of a fixed point with
be one of a system of confocal conies, and
1
The equation of the polar of
I
passes
a*h\ + b*kp,
iii. Find the envelope of a system ofconfocal conies.
a?
if this
cV = <**hc\ + b*kcn.
2
Example Let
and
=l.
This parabola touches the given straight line
respect to
c/u),
(i),
a2 cX 2 4 or
&V
if
(ft,
k) is
straight line are therefore
= - V(
8
m=
+ *),
Eliminating the variable X we have 2
or
(a
This
is
a parabola, since
at infinity 2
(a
it is
- V ) lm = kl- hm. 1
satisfied
by the coordinates of the line
Writing the equation in homogeneous coordinates, kln~hmn, we see that it is satisfied by the coordinates of the
(0, 0).
-62 ) Jw =
axes ; the parabola therefore touches the coordinate axes.
Example
iv.
To find
vertex of the parabolcf (a
2
the focus, directrix,
l
2
)lm = Td
is
another method.
the tangent at the
hm.
The coordinates of the focus can be found following
and
as above
(
8,
VIII).
The
AND TANGENTIAL EQUATIONS
411
If the point (xl9 y t ) is the focus, then the line
which
is
the join of the focus to a circular point 'at infinity, touches the
parabola.
*==t + fy19
Let
we
then, since the line (
,
-J
touches the parabola,
have, by substitution in the equation of the parabola,
k~ih
(a*-b*)h equation of the directrix of
5=
(when
2Gx + 2Fy-A-B =
it is
directrix
This
is
is
B = 0, hy =
kx
H= o
2
-fc
2 ,
G=
-A?,
is
F**h
i.e.
t
the equation of the
0.
a straight line through the origin,
the coordinated of a line parallel to it are the parabola, it is the tangent at the vertex.
The condition
is
Q,
and since the tangential equation of the parabola
we have A
a parabola)
its
coordinates are
(Afc,
(fr,
ft,
0);
-Afc), and, if this touches
for this is
-
i "" A
or
The coordinates
and
its
equation
Example intersect
Let
C
v.
of the
tangent at the vertex are
Points
(7,
D
of
AC, A'D
i.e.
intersect
D
-62 ) hk.
are such that :
AC, A'D of CD.
find the envelope
the point (acos<, frsin^).
are
= cos $ 0, = - sin | )/b
cos $ 6)/ a H- (y sin \ 6)/b
(# sin | )/a
and these
2
(a
ellipse
be the point (acostf, &sin#) and
(a:
4)
on an
on one of the eyukonjugate diameters
The equations
sin 1
+ h*) (kx-hy) =
therefore (k*
is
- (y cos
cf ,
on the line through the origin
+ (y sin $ <9)/fe} + COB $ {(# sin $)/a - (y cos \ (f>)/b} cos J ^/a - y cos J (0 + )/& = 0. 2a? sin J
{(x cos 1 6)1 a
<^>
= 0,
LINE COORDINATES
412
If this is the equiconjugate diameter x/a y/b 2 sin \ $ cos \6 cos \ (6 -f
=
we have
0, ),
or
8inJ(0-f$)-sin(0-0) = cos | (0 <). = cosJ(d-<), a? cos \ (6 + $)/a + y sin i (0 -f 0)/& -i-
Now CD that
the line
is
(i)
so
coordinates are given by
its
af= -cosj(0 + <)sec$(0-<); &w = -sin$(0 + 0) sec \(6 -). Substituting in
we have
(i)
tan(0-) a 2 /3 -f 6 2 w 8 - 1
and therefore
=
2aWw =
or
al-bm, (a/
- bm)*,
1.
This is the tangential equation of the envelope of CD hyperbola whose asymptotes are the axes of the ellipse.
it is
;
Interpretation of some special forms equations.
of tangential
10.
Let u 5,
-4,
(7,
degree
2
= 0,
;
t'
and k a constant.
uv
1.
= 0, v = 0, w = 0, z = be the equations of four points D 2 = the general tangential equation of the second = 0, = .0 the equations of two points T, Z" on the conic
t
;
a rectangular
=k
.
wz.
This equation
is satisfied
by the coordinates of lines which pass through the pairs of points w = 0,
w t;
= = 0,
u
;
#
=
0,
= 0,
=
;
t;
= 0, w =
;
by the coordinates of It is of -AD, BC, SD.
i.
the lines AC,
& e.
it therefore reprethe second degree sents a conic inscribed in the quadri;
lateral
AGED.
Example i. Find the locus of the centres of conies inscribed in a given quadrilateral.
ABCD
Let
be the quadrilateral, and take AC, BD as coordinate axes. points A, B, C, D are of the form al+ 1 = 0, bm -f 1 = 0,
The equations of the cl+l
=
0,
dw-f
1
=
0.
The equation of a conic inscribed (al ,
and
..
its
,
centre
-f
1) (cl + 1)
.
,,
is
the point
.
-f-
in this quadrilateral
k (bm +
This
is
(dm
-f 1 )
=
is
0,
,
The Cartesian equation of its locus
{0,
1)
is
therefore, eliminating k>
a straight line which passes through the points {J( + c), 0},
i(&4 d)},
i.e.
the mid-points of the diagonals.
AND TANGENTIAL EQUATIONS Example fixed points
On
ii.
418
two given intersecting straight lines tico sets of Points P, P' on time A', B'< C' are taken.
C and
A, B,
lines are chosen so tliat
=
{AB, OP]
Find
M'J3', C'P'\.
the envelope
of PP'. Let
k A'C'/C'B', then we are given ihat AC AP A'C' CB PB C'B'
AC/CB
.
AF
Hence -4P
PB Take the given points
~
*
A'P
.
'
(l)
PP''
and let A, B be the Then from (i) we have - OP).
straight lines for coordinate axes,
(a, 0); (&, 0)
A', B' the points
;
(0, a') (0, b').
(OP- a) (V - OF) = k (OP - a') (6 The coordinates of the line PP' are I = - I/OP, m = - I/OP (am)(&'m4a) = fc(a'w + l)(W-fl),
7
which
;
hence (ii)
the equation of the envelope of PP'. This is a conic inscribed in the quadrilateral ABA'B' it touches the lines AA', BB' and also the axes AB, A'B'. is
;
If the equation
(ii) is
(al
-f
made homogeneous, n)
(Vm
4-
n)
=
becomes
it
k (a'm +
)
(U 4- n)
;
can be readily shown that the points of contact of the conic with the axes = {A'B*, (TO,} and (0, 1, 0), (1, 0, 0) are points 0,, 2 such that {AB, CO]
it
{AB, CO,} 2.
uv
=
{A'B' C'O} where 9
=k w .
is
the origin.
2 .
In this case the points C, D coincide, and the pairs of tangents The AO, AD] BC, BD also coincide. conic therefore touches the lines C4,
CB
A and B. Hence uv = kw represents a conic touchand ing the lines joining u = 0, w = v = 0, w = at the points u = 0, v = 0. at the points
2
Show
Example.
that the parabola
lm + al + &w touches
the
=
coordinate axes,
and find
the
points of contact.
The equation
of the parabola in homogeneous coordinates lm 4- aln -f btnn *=
is
;
this can be written
and n 0-4,
(I
+ bri) (m 4- a n) =
abn*.
the point B(l/b, 0), + an = is the point ^i(0, I/a), is the origin. Hence the parabola touches the coordinate axes at the points A and #,
Now 1 + bn
is
m
LINE COORDINATES
414
2
3.
=
This represents a conic touching the tangents to 2
points u conies 2
= 0, v = in = 0, 2 kuv = ;
other words the
common
intersect at the points
=
from the
tangents of the
u =
0, v
= 0.
Note i. If u = 0, v = are the circular points O, Q' at infinity, the tangents from them to a conic intersect at the foci of the conic. In this P + w2 case uv~P-2lm cos o> -f 2 , or in rectangular coordinates wt> 3 a = k (Z -f m j represents a conic having the same foci as 2 = 0. Hence 2
=
w
.
=
are the equations of the foci S, S* of a conic, since 0, v = Again, if u the conic is inscribed in the quadrilateral SS'QQ' its equation must be of the
k (I2 + w 2 ). u = 0, v = are the foci of the conic 2 = 0, the equations Thus, uv-k (P + w 2 ) ax and 2 = are identical it follows that
form
ttt?
if
;
2 + k(P + m*)~uv. 2 m a * represents a pair of Hence, for values of k for which 2 + k (Z = 0. points, these points are a pair of foci of the conic 2 -I-
;
The condition that
should represent a pair of points
is
h
g
h
b+k
f
9
f
-f&
*
If c 96 0, this gives two values of k corresponding to the two pairs of foci, one real and one imaginary, of a central conic.
a = the equation 2 4- k (Z*40, for ) this value has two factors, one of which represents the finite focus, the other the point at infinity on the axis.
If c
= 0,
it
gives only one value of k
U. If /! + myl -f equation of the conic is
Note
(fcr,
4-
1
m^
= 0, -f
lx^
+ my^ -f
1) (te2 -f
w
;
1
wy3 -f 1)
=
are the foci of a conic, the
k (?
-I-
m9
).
AND TANGENTIAL EQUATIONS
415
This corresponds to the geometrical property that the product of the perpendiculars from the foci on any tangent to a central conic is constant and equal to the square on the semi-minor axis. of an ellipse, a 2 J 2
Example. The equation or
(1
The
foci are (ae, 0),
tangent Ix -f my +
Note
iii.
(~ae,
=
1
- lae)
are
(1
.---
2 I
=b
2
&2
w = 2
may
1,
be written
+ w2 ).
2
(I
and the perpendiculars from them on any
0),
-
V
+ lae)
-f
+ w2
and
:
v ** + w
,
2
J their product being 6 .
The general equation of a conic having a a (J2 -fm 2 + 2^ + 2/m + c ===<),
focus at the origin
is
)
for this equation in
homogeneous coordinates becomes a (I2 -f w 2 ) -f n (2gl -f 2/m -f en) = 0,
so that the foci are
n
Example a fixed point
i. :
=
Two
and 2p7 + 2/m + cw
points P,
show that
PQ
Q on a
= 0.
circle
subtend a right angle at
envelopes a conic, and find
its foci.
Let the fixed point be the origin and the centre of the circle the point a (, 0). The equation of the circle is then x* + t/ - 2 ax + c = 0. Let to 4 tny + I = be the chord P, Q the equation of the lines OP, OQ is ;
2
a? -f
these are orthogonal i.
y*
4-
2 a* (to 4- wiy)
if
1 -f
e.
-f
-f
m
2
(/a? 2
2 a/ -f
c (P
c c/
) -f
-f
+ my )* 1
-f
cm
=
;
2
2 (a/ -f 1)
0,
=
0,
which
is the tangential equation of the envelope. The envelope is therefore a conic with its foci at the origin and the point (a, 0), i.e. the centre of the
circle.
Example
ii.
Find
the equation of the auxiliary circle
of the conic
The focus of the conic is the origin, the auxiliary circle is therefore the locus of the foot of the perpendicular from the origin to a tangent to the conic. Let
be a point on the auxiliary circle, the line joining it to the xyl -yxl =X) a perpendicular to this through the point (xlt yj is origin xx\ + y^i ^ x \ + yi** This is a tangent to ftie conic, its coordinates are (a?!, i/,)
is
;
Hence, substituting these in the given equation of the conic, we have
cK
9
+ yia )~2^ ~2/y1 -fa0,
or the equation of the auxiliary circle
1
is
LINE COORDINATES
416
A S+kD + = 0, where i$ the equation of the director circle of the conic S = 0, represents
Example
D=
Show
iii.
that
a system of conies .confocal with
the
A;
equation
8=0.
2
Rectangular Coordinates.
The tangential equation of the conic 2
s
Hence the tangential equation of any conic confocal with this S + fcCP+m 1 ) 0, 2Hlm i.e. + (B + k)m*+2Gl+2Fm + C~Q. (A+k)P + The coefficients of the corresponding point equation are
is
(A + k)C-
FH-( i.e.
the equation
& (ax
1
-h
is
2 hxy + by* +
2gx + 2fy -f c)
or
4.
2
= ku
2
In this case
.
A
and
B
coincide
contact, the pole of their
cpmmon
8 Example. The equation am
parabolas confocal with
am
;
8 /.
2
=
chord
= leu have being A (u = 0).
/-ffc(a/ + l)
and 2
a
=
2
double
represents a system of
AND TANGENTIAL EQUATIONS 5.
T =
2
2 is
=
kut.
T
lies
and 2 = common tangents meet at u
Thus 2 2
= 0; the two tangents from T to on both the curves, and they touch
2
a point on the curve
coincide, hence each other at T.
6.
= kut
=k
touch at
=
=
0.
7.
2
at
t
= 0,
and their two other
tt'.
.
2
=
at the points
t
=
0,
T
=
kut where u
^-~z^
=
point on the tangent to
=
t
0.
This curve has double contact with *'
417
is
2
a
=
0.
In this case one tangent from A coincides with the tangent at T thus the curve 2 = kut has three coincident tangents in common with 2 = at T, and the fourth common tangent meets ;
them at ,1. Thus 2 =
0,
2
= kut
have con-
tact of the second order at
u
=
their 8.
t
= 0,
being the intersection of
common 2
=
kt 2
tangents.
.
In this case the two points A and B become coincident at T on the conic 2 = 0; hence, the conies 2 = 0, 2 = kt 2 have four coincident
common
of the third order.
tangents at the point
t
= 0,
i.e.
have contact
LINE COOKDINATES
418
=k S = 0, 2
9.
If
.
S'.
2'
touching the
=
two
are
common
conies, then
fc2' represents
2
=
0,
2'
a conic
= 0.
Chords of a conic subtend a right angle at a fixed point
Example.
shotv that tlicy envelope anotlier conic,
a fixed quadrangle
scribed to
=
S
tangents of the conies
and if
:
the first conic is circum-
tlw second conic is inscribed in
a
fixed
quadrilateral.
Let the fixed point be the origin and the fixed quadrangle that formed by the common chords of
The
conic has then an equation of the form
first
The equation of the straight + 1 =3 to the origin is
lines joining its intersections
with the line
Ix + my
ax*
+ 2hxi/ + by* -2 (gx +fy) (Ix+my) + c (Ix -f my)* + k[aa? + 2h'xy + b'y-2(g'x+f'y) (Ix-t '
These are perpendicular
which
is
'y)
+ c'(lx-\ wy) 8 J
= 0.
if
therefore the tangential equation of the envelope of Ix
+ my -f
1
=
0.
But this ib the equation of a conic inscribed by the common tangents of the conies
c'
(P 1 m*)
-2g'l- 2f m -f
in the quadrilateral
'
-f //
=
formed
0.
When the solution of a problem in point coordinates has been obtained, the solution of the corresponding reciprocal problem in line coordinates follows by interchanging points and lines. The following examples worked out side by side illustrate this. 11.
Example!. Point Coordinates. The vertices A, B, are joined to these joins
of
t'lvo
C of a triangle
points P, P'j and
meet the opposite sides
the triangle in the points
The
sides a,
Show
that
D, D' ;E,E'
conic.
;
F, F'
&,
are met by two
c of
a triangle
lines p,
%>',
and
these intersections arc joined to the
opposite vertices by the lines
D,D'; E,E'-, F,f. on a
Line Coordinates.
Example!.
d, d' lie
Show
;
that d, d'
a conic.
e,
;
c'
;
e, c'
/,/'. ;
f,
f touch
AND TANGENTIAL EQUATIONS Let the equations
(in abridged notation) of the sides of the triangle be n = 0, v = 0, w 0, and let the
=
values of w,
v,
w,
when the coordinates
of P, P' respectively are substituted in them, be w lf t> n t* a , va , t0 a . l ;
w
Then the equation of the join A,
P is
--
--
i
= Of
Let
equations (in abridged the vertices of the
notation) of triangle be u
=
w,
p respectively are substituted in them, be u l1 v l9 w^;
Then the equation and p
tion of a
P
=
is
of the intersec-
=
is
0,
and
the equation of the intersection of b
0.
,
w =0. w
u
.
is "i
Since the equation
i
Since the equation .
Mt
can be written
u
w
/ r 4-
(
ri
\
"i
~ + flL - w .\ _
\ )
=
respectively,
i.
e*
represents
it
is
the
equation of the join of D and E. Similarly, the equation of the join of Z>', E' is
U
V
-
a.
or
/ w -f
1?
+ /i?
i
~
o
M>,
t?!
t^j t?
0, it
+
can be written
M 'i/
a straight line through the points of intersection of the lines AP, BC and
AC
w = 0, and when the
0,
v,
coordinates of p,
and p
J2P,
u,
i
of the join of B>
or
=
0, v
the values of
let
of
and the equation
the
410
\ I
1
(
\ TJ
= 0, it
=
ifj/
represents
a point on the join of the points ap, be
and
bp,
ac respectively,
i.e. it is
the equation of the point of intersection of d and
e.
Similarly, the equation of the point of intersection of d', e' is
W 0.
ua
t>
9
u>a
Hence the equation of any conic
D d 2
Hence the equation of any conic
LINE COORDINATES
420
through the points D, D'; E, E'
is
of
the form
w\ ----
fu +
[
\
touching the lines
d, d'\
e
e,
of
is
the form i>
x
X HV
I
W1
Vl
=
/
it
\
tt>j/
w\ v ----
+ ,
(
)
t>
/!
ttV
t
(i)
This touches the join of the points
This passes through the intersection of the
lines
=0,
CP,
and AB, w
=
0, i.e.
=
cp
*i
i
the point Fif X
Ui
and
0,
w=
ab,
i.
0,
e.
V,
the line /if X has the value given by
has the value given by X Hi Vi t?
2
i.e.
i.e.
The symmetry of the that this
is
result
shows
also the value of X for
The symmetry of the that this
is
result
shows
also the value of X for
which the conic (i) passes through f*. Hence D, J9'; E, E'\ F F* lie on a
which the conic (i) touches/'. Hence d, d''; e, e'; /, /' touch a
conic.
conic.
y
Example
The polars of a
2.
given point with respect to a series of conies circumscribed to tJie same quadrilateral are concurrent.
Let
^ EE ajic-f&jy-f = 0, 1
1 1
=
The
2.
Let
= 0, = 0,
quadrilateral.
The equation of the conic
is
poles
of a
with respect to a series of conies inscribed in the quadrilateral are collinear.
given straight line
MI
4 1
= =
0,
+ 64 w+l
=
0,
a^-f ^1/1+
0,
be the equations of the sides of the
U^lfi- &U 3 W 4 =
Example
u4
=a
4
l
1
0,
be the equations of the vertices of the quadrilateral.
The equation of the conic
is
0.
The equation of the polar of the fixed point (#', y) is therefore ' HI u* -f w a / + k (w s u4 -f w 4 w 8 ') 0,
The equation of the pole of the fixed line
(l' t
m')
is
therefore
t^t^'-f Wjtt/4- &(w s tf 4 '-f
=a
w4 tV)
~
0>
for all values of & passes through the point of intersection of the lines
V + \ m + 1 &c., which t for all values olf k lies on the line joining the points whose equations
whose equations are
are
where
and i.e.
ti/SEojaj'-f ^y'-fl, &c., which
wa !
'
+
w a w4
/
4M
a fixed point.
,/ /
4
3
where u^
0,
= 0,
MjWj'-f
t/
a
M/= '
and i.e.
,
3 u/-f te 4 t/ 8
a fixed
line.
0,
= 0,
AND TANGENTIAL EQUATIONS Example
Find
3.
Example
the envelope
of the polar s of points on an ellipse with respect to a coaxal ellipse.
Any
point on
tvith
and the coaxal
has coordinates
and the polar of
.
.
(i)
.
A
this
sin0\ -7
to
b
a
\
,
,,
writing
/costf
1,
==
.,.
this
(ii) is
.
J
t
,
and the pole of
J
line with respect to
or,
has coordinates
(i) ,
-
[
is
yb
8
of
ellipse
Any tangent /cos 6 >
,..,.
point with respect to (n)
xa C A*
the locus
Let the ellipse be
ellipse
(a cos 0,' b sin 0), '
Find
3.
of tangents to an ellipse respect to a coaxal ellipse.
the poles
Let the ellipse be
and the coaxal
421
-f -r-
t
writing
or,
1,
b
a t
\
i.e.
i.e.
-
1
The envelope
of this line
The
there-
is
locus of this point
0.
is
therefore
fore /
o
-
\
r t
'
or
which
is
another coaxal
which
ellipse.
is
another coaxal
ellipse.
We
8 how the tangential equation of have shown in a conic can be obtained from its Cartesian equation, and conversely in this section we give another proof of this property, and some 12.
;
immediate deductions.
In order
to obtain
equations are given in homogeneous I.
then
+ my + nz = touches the conic S=ax* + bf + C2* + 2fy2 + 2gtx + 2hxy = 2 ^J + m2 -hCn2 4-2Fwm + 2Gi2Z-f2HZm=:0. If the
line Ix
If the line Ix +
then
complete symmetry the
coordinates.
its
my + nz
=
0,
the tangent at the point (.r', equation must be identical with that of the tangent f
)
is
x + (hx + by' +/*') y + (gtf +fy' 4- <*>
=
0.
j/',
/),
LINE COORDINATES
422
Thus
I
=
aaf + hy'+ge',
JiA
+ lH+fG- =
Hence
Al + Hm + Gn but
=
aA + hH+gG
A,
^ +fH+cG ^
0,
;
= A x'. HI + Bm + Fn = A #',
therefore
.4 Z -f
Similarly,
-Hw + $w
Hence
7
But
since
+ my + nz
(n?
= 0,
r
/
,
by hypothesis, a point on the
is,
^')
,
we have
/
Lc
+ wiy
/
+ w/ =
;
line
thus
=
^P4--Bm2 -hCw2 4-2FmM-|-26?^-f 27/Zm
0.
Conversely, II.
I/
+ my + n# = Z/es <w f7ie cowic aP 4- ?>w 2 + m 2 + 2/wm -h 2/jfwZ -f 2hlm =
the point lx
If the point
Z.r-f
m', w'), then of contact
(r,
its
my + nz
=
0,
the point of contact of the tangent equation must be identical with that of the point
(aV + 1m' + tfO
Z
is
+ (W + ton' + /n r) m + (gV +fm' + cn')n
Thus
x
=
=
0.
a
Hence
= (aA+hH+gG) + (hA + bH+fS)m' + {gA +fH+cG)n' Ax + Hy -h Gz = A T. l'
therefore
;
Hx + By + F# = A m', Gx + Fy + Cz = An'.
Similarly,
Hence
+ (ff0 + l?y + J But lx
since
x
(Z',
+ my + nz =
0,
m',
w)
is,
by hypothesis, a
we have Vx + w^ + its
= 0,
line
through the point
thus
=
0.
AND TANGENTIAL EQUATIONS Note i. If (I, m, n) are the line coordinates S = ax* + by 2 + cz, 2 -f 2/i/s + 2gzx + 2 /*## 0, and nates of
its
point of contact,
we have the
I
__
and
_
Al + Hm + Gn
If
y,
(a:,
2-)
is
n
__ ~~
__
hx -f by +fz
gx +fy -f cz
_ ___ = Hl + Bm + Fn
__
.
Gl + Fm + Cn
a point on the conic al*
and
of a tangent to the conic (x, y, 2:) the point coordi-
reciprocal relations
m
__
ax + hy + gz
423
+ 6m2 + en* + 2/wn -f 20nJ + 2hlnt
=
0,
m, n) the coordinates of the tangent at this point, similar relations hold, the small and capital letters being interchanged. Note ii. The above formulae connecting the line coordinates of a tangent (J,
and the point coordinates of its point of contact were obtained from the equations of the tangent and the point of contact respectively.
When
(r, y, z) is not on the conic, or when (l m, n) does not touch the the conic, equation of the polar of fa:, y, z) or of the pole of (?, m, n) are of same forms as those of the tangent and point of contact. the exactly = 0, and (/, wi, n) are the if Hence, (x, y, z) is a point not on the conic S y
then
line coordinates of its polar,
__
and (a?,
'
hx + by +/S
ax + hy + gz
<jx
+fy -f cz
m, n) is a line not touching the conic are the coordinates of its pole, then
reciprocally, if
y, z)
(I,
5=
0,
and
y Al +
Any
III.
7 fii
Hm
-f
Git
III
-f
Bm + ^t
<7J -f
Fw
-f
Cn
conic being
being the value of S when #', ?/' are substituted for .r, y and being the equation of the polar of (x', y'\ the equation of the
P=
pair of tangents
from
(x', y')
to the conic is (Chap. VI,
SS'
So
also
2
=P
= 4P + 2///w + J?w
tangential equation of the (r, M') are substituted for ?,
same M and
2
4
(v))
2 .
+ 2GZ + 2JFm + C = conic,
being the
S' the value of
S when
II =
being the equation of the of the the equation (I', m'), points of intersection of (Z', w') and the conic is (Chap. X, p. 404) pole of
vv _
These can be put in the same form.
Making the equations homogeneous,
we have S
= nx*
4- fy/
2
2'
-f cz' -f
2
for convenience of notation,
LINE COORDINATES
424
Now i.e.
the equation of the polar of (x\
x (ax' + liy' + gJ) the coordinates (I,
=
I
ax' + hy' + ggf,
thus
y', z'}
with respect to
=
+ y (hx' + ly' +//) + e (gxf +./// + of) in, n)
of this polar are hx' + +//,
V
m= 8' =
w
=
^ +/y +
*
=
is
0,
'
as'
;
te' + my' + w/.
Also
+ Bm + Fn = A/, Gl + Fm + Cn = A/;
So
7/7
S
i.e.
= A,.
Also the polar of (a/, T/ /) being to + my + fl# = 0, we have P = Ix + my + 7?^ hence the equation of the pairs of tangents from = 0, viz. SS X = P 2 can be written 8 2 = A (to + my -f r^) 2 (#', y', /) to S Similarly, the equation of the pole of (?', m', M ) with respect to 7
,
;
.
.
,
7
2
=
is
= 0,
^^r + #w'+G<) + m(tfr+#w'4^^ i.
coordinates
e. its
x
=
Al' + Urn' -f
are
(x, y, e)
Gn\
y
= X
e=
//^ + i?m' -f Fn',
=
Thus
2
Also
a^c -f 7iy 4-
xl' -f y m'
Fm' 4-
Cn'.
+ zri.
=A
gz
Gl' +
Z',
cz)
Also the pole of
(V, m', n')
is
IT == Ix
'v*
^^
1
+ nz = + my -f nz
the point lx + my
0,
i.e.
;
hence the equation of the points of
inter-
(Z', m', n') and the conic = n 2 becomes being 22'
section of the line
2
=
2.5= A (te-f my + M#) 2
.Q
which
is
the
same form
,
as the equation of
a pair of tangents.
Thus,
if
lj
m, n are
fixed, viz. the coordi-
nates of PQ,
2.5= A (Ix -f my is
z -f-
nz)
the equation of the tangents at the point of intersection of the
AND TANGENTIAL EQUATIONS conic and is
and
425
are fixed, viz. the coordinates of 1\ the tangential equation of the points of contact of tangents from
PQ
;
if (#, y, z)
it
T
to the conic.
Illustrative
For an
T
Let
Example.
ellipse
be the point /a:"
4
if
v
(a?,
/)
- ,\/ l
)(
Z
;
2
v
then the equation of the points P,
- w'
Q
is
1
1
7
or (i)
Let the coordinates of
P
and Q be
(a? lf
i/^,
(a?2
are (/j?! -f
The equations in a factor to
(i)
and
make the
t/ 2 )
;
then their equations
= 0.
(ii)
are therefore identical,
and we have, putting
w// t
(ii)
,
+
1
)
(to a
+ w?f/2 +
1)
absolute terms the same,
or 5f
2
and
2 /
(a
+
9 fc
wi
-
1)
~(1x -f my -f
2
I)
= - (S+ 1)
this is identically true for all values of
(a) v
Now
put L
I
==
--
,
wi
=
---- r -
(toj -f wiy, 4- 1) (/^2 -f /
and m. ;
then
wy2 + 1),
LINE COORDINATES
426
_
____ (x +
i.e.
iyy-tfet^^-g-fa-XL Of
-I
,
-
e.
Now
make
TP, TV, TS, TS'
let
then
,
_ .^ __ -f
{ar-ara
*.f/-
..
-
^
angles
<9
0,,
2
,
<
with the axis of
a
a;
;
#-#!
x-ae
=
x + ae Substituting and using
ST.S'T (cos ^ +
i 2 -f
.
De
sin
STco8
y
.S'Tcos^j,
//
^S
Moivre's Theorem, -f-
t
^2} 1
=
-
.TP.TQ
(cos
LSTP=
W^TQ
VH+
=
and imaginary parts we have ^1 + ^2 tS'TQ, and further
real
Equating that
__
+ i.y-fr}
l
8in
^+^}
^4^2, which shows
J3
ST.S'T (b) v
Put
J
= (a?
So that
if
m=- ~
--
- ae -ft//) 2
SP, S@ make
ST* (cos 2 0,
+
1
J
= (+ 1)
angles
sin 20,)
=
then (xl -ae
+ (?/!) (ir2 - ae -f
i
;/ 2 ).
\^ 1? \^ 2
with the axis of x>
(S+l) SP. 5^ .
(cos
^+^3 +
1
sin
^7
LTSP^tTSQ,
i.e.
and
(c)
Put
/
=
--
9
m=
--
and we get 2
2
\~ a ** + (aj-a?, + )
whence, as above, SP,
t.v
~yi)
5'P are equally inclined to
=
TP
y
0,
and
Similarly,
Examples 1.
Find the envelope of chords of an
X
d.
ellipse
which subtend a right angle
at the centre.
Find the envelope of chords of a parabola which subtend an angle a at What does the envelope become when a = 90 ? 3. Find the condition that the point pl + qm + l =0 should lie on the parabola am* -f 2 1 = 0. 2.
the vertex.
AND TANGENTIAL EQUATIONS Two circles common
4.
of their
touch externally, the centres are
427
Find the envelope
fixed.
tangents.
N
Three points L, M, on the sides of a triangle are collinear and = h k. Show that LN touches a fixed parabola. 6. Find the envelope of the polars of a fixed point with respect to a series of circles which touch two given straight lines. 5.
LM MN :
A
7.
:
variable tangent meets a fixed tangent of a parabola
their point of intersection a straight line variable tangent find its envelope.
is
and through drawn perpendicular to the
:
Find the equation of a parabola whose focus is coordinates of the tangent at the vertex being (p, q). 8.
al
-\
bin
+1
=
0,
the
Also find the coordinates of its directrix. 9. Circles
are
drawn with given centres
tangents of pairs of these touch a fixed parabola.
circles,
A
and
B
show that the common
:
the difference of whose radii
is
constant,
Find the coordinates of the normal through the point of contact of 2 ( /a, t/a) of the parabola am* = I Hence show that three normals can be drawn through a given point pl + qm+ 1 = 0, and the parameters of the corresponding tangents are such 10.
the tangent
that 2 (\/t) 11.
=
0.
Interpret this geometrically.
Find the envelope of a line on which two fixed circles intercept equal
lengths. 12. Two finite lines are each divided into n equal parts find the envelope of the lines joining corresponding points of division. 13. Find the envelope of the polar of any point on the circle :
with respect
to
the
3a 2 J 2 + 4a2 m2 -2aZ-l = 2 2 circle 3a Z + 4a mH2aJ-l = 2
0.
=
=
A
1 0. 0, al system of conies have their foci at the points al + 1 Find the envelope of the tangents to them at their points of intersection
14.
with the straight line is
(X,
/ot).
Two
points are taken on an ellipse so that the sum of their ordinates constant show that their join envelopes a parabola.
15.
:
X, Y are fixed points on two lines OX, OF, and points P, on the lines such that 16.
(i)
XP.r
2
c
;
(ii)
*P+r
(iii)
Q
A'P-r
are taken
=
c;
find in each case the envelope of PQ. 17. Circles are drawn with their centres
on the tangent at the vertex of a parabola and touching the parabola. Show that the envelope of the common chords of the circles and the parabola is another parabola, and find its focus. 18.
The envelope of the polars of points on the auxiliary
ellipse with respect to the ellipse
Find
is
another
circle of
an
ellipse.
equation and its foci. 19. The bisector of the angle between the tangents from a point to an 2 w 2 = 1 is parallel to the line (l }9 wj): find the envelope of ellipse a*P + the chord of contact and the coordinates of its asymptotes. its
J>
LINE COORDINATES
428
20. A variable circle of radius r passes through a fixed point (/*, 0) show that the envelope of the common chord of this circle and the circle x* + y* i* is a conic. Find the foci of the envelope. :
=
21.
that
PR is a diameter of a circle, QR passes PQ envelopes an ellipse or an hyperbola
lies inside
through a fixed point show according as the fixed point :
or outside the circle.
A
chord of a parabola which subtends a right angle at the focus touches a conic one of whose foci is the focus of the parabola prove this 22.
:
and
find the other focus.
Find the condition that the
23.
2 conjugate diameters of the conic a/
The four
24.
lines
(Jlf
mj,
(7a
,
w
lines
(I l9 wi
=
+ fcw2
2 ),
(Z 8
,
ws
n2 ),
n
O),
(J2
>
W
2>
should be
0)
.
(J4
,
w
4)
form a quadrilateral;
find the coordinates of its three diagonals.
Three conies are drawn circumscribing the quadrilaterals formed and two given straight lines. Show that their other common chords pass each through a vertex of the triangle and are concurrent 25.
(i)
by two
sides of a triangle
Three conies are drawn inscribed in the quadrangles formed by vertices of a triangle and two given points. Show that the points of intersection of their other common tangents lie each on a side of the triangle and are collinear. (ii)
two of the
=
If a triangle circumscribes the ellipse #Ya 2 4-f/2/& 2 vertices lie on x*/A* + y*/B* 1, the locus of the third vertex 26.
(i)
=
1
and two
is
a coaxal
1
and two a coaxal
ellipse. (ii) If a triangle sides touch -4 2 /a + J9 2
is
m = 2
inscribed in the ellipse tfP + tfm* 1, the envelope of the third side
=
is
ellipse.
27.
(i)
points
Two
sides of a triangle inscribed in
:
ellipse pass
Two vertices of a triangle circumscribed to find the locus of the third vertex.
(ii)
lines
an
through fixed
find the envelope of the third side.
:
an
ellipse lie
on
fixed
28. Find the general tangential equation of a rectangular hyperbola which has contact of the third order with the parabola am 2 = I, and the
locus of its centre. 29. Find the envelope of the polar of a fixed point with respect to a conic circumscribed to a given quadrilateral. 30. A system of conies is inscribed in a given quadrilateral show that :
their director circles form a coaxal system. 31. A conic has double contact with a parabola at the ends of the latus
rectum
:
show that the envelope of the polars of points on
respect to the parabola
is
this conic with another conic having double contact with the
parabola. 32.
The sum of the
latus rectum. find its focus.
Show
abscissae of two points on a parabola is equal to the that the chord joining them envelopes a parabola and
AND TANGENTIAL EQUATIONS 33.
The normals
current
two chords of an
ellipse are con-
prove that
:
(a)
at the extremities of
429
If
parabola
one chord passes through a fixed point the other envelopes a
;
(b) If
one chord touches an hyperbola whose asymptotes are the axes
of the ellipse, the other chord envelopes a rectangular hyperbola.
the tangential equation of the conic S = 0, show that the a = equation 2 -f k (P-f represents a pair of foci of the conic if ) 34. If
2
=
is
m
35.
is
The sum of the squares of the semi-axes aP + 2hlm + bnt + 2/m
~(A+B)/<*.
of the conic
CHAPTER XI MISCELLANEOUS THEOKEMS 1. Transformation Moivre's Theorem.
I.
of
To change from one
set
Coordinates.
Application
of rectangular axes
of
Be
to another, Hie origin
being unchanged.
Let the
Let
P
axes are i.e.
ff
new
axis of
x make an angle a with the old
axis.
be any point whose coordinates referred to the two (X, Y),
(x, y),
=
and
let
OP =
sets of
r,
ex.
Thus
x + iy
=
r (cos
+ i sin 0), ':)
x + iy = (X + Y) (cos a + sin a), X + iY = (#-f ?(cosa isinoc).
Hence and
i
Equating
real
x
and II.
=
X=
and imaginary parts, = Xsina+ Xcoscx Fsina, i/ = #cosa Y a; cos a sin 2/ a, -f-
To change from one
set inclined at
1
an angle
set
of axes inclined at an angle
a/, the origin being
to
another
unchanged.
Let the new axis of x make an angle a with the old
axis,
and
let
MISCELLANEOUS THEOEEMS
P
be any point whose coordinates are
two
431
y\ (X, Y) referred to the
(#,
sets of axes.
(a)
Let the coordinates of
where Ox, Oxl Then
P referred
coincide, be (x ly t/ L ). x l = x + y cos
= y sin
yl
Thus x l -\-iy l =
+ #(costo +
a?
i
Ox ly Oy v
to rectangular axes
to,
co.
sin
to).
Again, suppose the coordinates of P referred to rectangular axes 0*2, 0#2, where 0^ 2 coincides with OX, be (x 2 y%\ then in (b)
>
the same manner
Now
(a?!,
(Xj
#j),
= X + Y (cos
+ iy 2
#2
sin a/).
are the coordinates of
2/ 2 )
9
co' -f i
P referred
of rectangular axes, the #-axes making an angle Hence, by the first part of this section,
+ ty\
*!
= (^a + ^2)
(
^+
cos
to
a with each
two
sets
other.
sin a).
i
Hence
=
r
a + sin a) [X + i (cos a/ + i sin a/)], i sin This, multiplying by (cos to to), becomes
x+y (cos + co
x (cos co
=
i
sin
i
sin
co)
(cos
+y to + isina
co)
X(cosa
co)-f
y sin
co
and
(i)
OL
of
X
writing the equations i
^(cosa + o/
whence
sin a)
Y
and (i)
+y (cos
Xsinto
x
(ii)
+ co'), 7
(co
cx
co
).
x and y can be found by in the forms
(ii)
a -f
i
sin
isina-f a/)-f # (cosco
Ysin co'
aco).
in terms of
and to
(a.
Ysiu
a-f-
co-f isinco'-f
(ii)
= X sin + ]f sin
x silicon Xsinco
The values
a
Y(cosco'+
Equating imaginary parts in
x (cos a
(i)
?'
co
a
a) r
co
= X + Y(cos a/ +
to
a/),
)
a;
to'
sin
x
+ isinco a co = Y-fX (cosco'
= y sin a sin a, = x sin a + +y sin a
i
4- to'
to.
i
sin
co'),
MISCELLANEOUS THEOREMS
432
We have seen in Chap. Invariants. change of axes the equation
+ 21u>y + by* + 2gx + 2fy + c
ax* is
Ill,
that
6,
if
by any
=:
transformed into
a change of origin only does not
(i)
terms, viz. a,
highest
ft,
&
ab expressions
h
a+b
2
--.
of
the axes does not affect the
2Acoso> .--*
>
,
------
;
sm^co
snrco
the coefficients of the
;
a change in the direction
(ii)
aft'ect
,.
and these expressions are
called invariants.
We propose here
add another important invariant, and then to
to
illustrate their use.
If
by any change
of axes ax 2
-f
21ixy
+ by 2 + 2<jx + 2fy i
c
becomes
then sin 2
in other words,
sin 2
a>
sin-4
is
'
co'
an invariant.
o>
Make both the expressions homogeneous by introducing z and Z, where in the present case z and Z will both be unity the expressions are aaP + 2hjcy + by 2 + 2gjc* + 2fg& + cz* and :
a'X 2 + 2h'XY + b'Y* + 2
We have relations between xyz, XYZ of the form x = z
This
is
for the sake of
Make
Z.
a particular case of the quite general transformation x
which
=
=
symmetry we
these substitutions in the
will first consider.
first
expression
;
then
or a'
=
I,
(al^
+ hlt+gl,) + ?2 (hi, + bl, +fl,) + 1, (yl, +// 2 + clj.
MISCELLANEOUS THEOEEMS
433
Similarly,
=m
l
(am l 4- Am 2
)
w
4-
4-
2
/* 3
HJ) -f
)
4-
Wj) 4- iw a
m Zj 4. t
4-
w 2 (A^
a 4- flfZJ 4-
An 2 4- gn 3 ) 4- L
-/w
a ) 4-
l
+ hl% + gl 3 ) +
(a/!
i
(#mj 4-/m 2 -f cm 3 ),
(^! 4-/n 2 4- en,)
Wj
J 4- ? 3 (#! +/w 2 4- cn 3 ),
(A
2
3
4-
)
=m
w
M 3 (#Wi +/w 2 4- cn 3 ),
)
1
4-
4- ? 3
m3
(^m! +/wi 2 4- cm 3 )
(r/Z l
4-/? 2 -f
cZ 3 ).
Hence, by the ordinary formulae for the multiplication of determinants, a
l>'
h'
b'
/'
9'
f
c'
?! 4-W 2 m! 4- &w 2 4-/m 3
(J
M
In the proposed we have
Now
Z
a
/i
/.'
?.'
/
/'
the factor
when we
case,
ordinates,
3
=
0,
(7
/'
Wj
h
g
A
6
/
g
f
c
a
7
= 0,
^mi 4-/m 2 4- cm 3
w2
w,
n<>
H*
are transforming Cartesian co-
n3
=
1,
hence
a
A
g
h
b
f x"
\
"
'"
c' 2
l^m^ this only affecting n x n 2 (l^m^
is
independent of the change of
to find its value then, consider origin, the particular case of the transformation of ax 2 4-^ 2 4-2A^f/4-c to ,
without change of origin, so that Then from the above
;
c
=
i.e.
by the former
invariants.
A Hence, 1207
finally,
7
ain*
/ii
,
A = -r-?
*iiii* co
E e
c'.
MISCELLANEOUS THEOREMS
434
Examples. To deduce formulae for
(a)
oblique axes
from corresponding formulae
for rectangular axes.
To find
I.
axes
the angle between the lines cu? 2
are. inclined at
an angle
4-2/w#4-&#
2
=
wJien the
co.
Suppose that when the axes are transformed to rectangular axes the 2 2 2 then expression ax* 4* 2 hxy 4- fey becomes a'X + 2h' XY+b'Y ;
,
and
.
t
a'b'-h'*
But the angle between the
a
,
=
4-
2h cos o>
b
-T-^
sura
lines
is
a
_. "~
_j
2 sin a
Cor. The lines Ix +
-f
4-
o>
V ab
\/k*
2 A cos
&
to
my 4 H = 0, I'x + my + n' = W -f (?m' 4- Z'm)xy iwmV =
V
-I-
are parallel to 0,
and hence include an angle J'w) sin
(/w'
-i
'
x
-
II. T
^nd
+ my -f n
the length
of
= 0.
(^n
-f
CD
Tmj cos w
the perpendicular
from
(#', y')
on the
line
Suppose by change of axes to rectangular axes (without changing the l'X+ w'F-f n; then the expres-
origin) tlje expression Ix + my + n becomes sion te'-f my' + n becomes rX'-H w'r' + n.
Now
i.e.
?
a:
2
(JX"
F')
,
on r-X"-fm'F4-n
Ix
f
,/ a 2 r -f
a m /a s= ^4-m a
sin i.e.
=
is
+ my becomes I'X+m'Y) (Ix 4- my/ 2 becomes (/'JT4 + 2/m-ry^w 2 y s becomes l *JP + 2l'm'XY+m'*Y*; hence
But since l
X
the perpendicular from
8
w
the required length of the perpendicular (J,x -f
my' 9
4-
n) sin
is o>
21 m cos
o>
MISCELLANEOUS THEOREMS (b)
and
of the conic ax* + 2hxy + ty* + 2gx + Zfy + v latus rectum when it is a parabola.
To find
the
the axes
If the conic is a central conic,
when the
=
coordinate axes are transformed
to the axes of the conic the equation takes the
i.e.
435
form
the process of transformation reduces the given equation to one of the
form X
OX 2
2
4- Oi
2
Y -a 2
2
2 ft )
=
0.
Note that since we choose OK, ft to be the actual lengths of the semi-axes we must introduce the arbitrary constant X, for in the above work we have assumed that the lesultant equation is obtained from the given one by substituting linear relations for x and y with no subsequent of the conic,
division or multiplication
The
invariants
now
by any
factor.
give us
+ *-2*co.._ x(ai + j9l)> sura
J3
sin 2
o)
sin
Thus
= I'YA sin 3
X
'
a>
and
Hence
Ot
2 ,
2 # are roots of the equation
When the conic is a parabola the equation, when the axes of coordinates are transferred to the axis of the parabola and the tangent at the vertex, 2 0. takes the form X(F ~4/^T)
=
Hence
~r4~ 2 <>
=
-*J 2 X S
sin
a
-f
2 h cos *>__%.
b
~
*
sin'w
A hence the latus rectum being
**
es
41.
E e 2
.
8in4 a>
,
MISCELLANEOUS THEOREMS
436
Similar and similarly situated conies.
2.
Definition
i.
one to each of
If the radii
two
drawn from two points respectively, which contain a constant
conies, in directions
angle, are in a constant ratio, the conies are similar.
Definition
If the radii in the above definition are
ii.
the same direction and their ratio
and similarly situated. Suppose that a straight
is
drawn
constant, the conies are similar
drawn from a point
line is
in
(a, ft)
to
in the points 1\ Q ; and also one from the point a conic, meeting Of (a', ft') in the same direction to meet another conic in the points it
v,
Q'-
two conies are similar and similarly
If these
OP .
,,
As
OQ
,
and
-^y,
..
.
.
.
.
situated, the ratios
OP.OQ
..
.
,
,
ls constant. are constant, hence also the ratio Xp/"7)/y
5
in Chap. VI,
we can show
(i),
that
if
the conies are
f(x, y) = ax + 2hsy + ly* + &e., /' (*, y) = o's + 2h'xy + Vy* + &c., = OP OQ /(a, -r (a cos 6 + 2 h sin cos + 1 sin 1
and then
The
8
2
.
ft)
2
6),
condition above requires that the ratio
a cos 2 a'
cos
2
0+2h sin cos 6 -f b sin 2 + 2h' sin 6 cos + b' sin 2
should be independent of h I a Hence -/= =^ ^.
0.
=
This is the required condition that the conies should be similar and similarly situated ; it is also the condition that their asymptotes should be parallel. Again,
if
directions 6
and 0' are drawn in the the straight lines through and respectively, the ratio of the rectangles OP OQ .
and O'P'.O'^' becomes a cos 2 2 a' cos
+ 2h cos -f
2 h' cos
0+ 1 sin 2 sin + / sin 2
sin (/>
(/>
If the conies are similar, this ratio is
must be constant when
constant.
If 6
= y,
the ratio becomes sin 2 (
a cos
2 -f-
2 A' sin
and must be independent of
0.
cos
+ b' sin
2
MISCELLANEOUS THEOREMS The
may be
ratio
437
written
a + 1 -f
b)
(a
cos
2r((f>
+ y) + 2 h sin 2
(
+ y)
Equate the ratios of the coefficients of cos 2, sin 2 fa and the terms independent of fa then g+ __
_ (a
6
/
Z>)cos2x
Hence, eliminating (a (a'
This
is
+ 27*sin2}' _
2/*cos2y
____,
""
7>)sin2/
(a
y,
K + bT-K-ft')
+ 6')*
_
2
-**' 2
~ <*'"&'-/?*
the condition that the conies should be similar, and
is
also
the condition that the angle between the asymptotes of each conic should be the same.
3.
The gereral equation
When
the origin
of the second degree.
= 0, and the general equation = 0. The equation of the tangent
on the curve,
is
c
becomes ax* + 2hxy 4- 6# 2 -f 2gx -f 2fy at the origin (0, 0) is
gx+fy
=
i.e. is
0,
Curvature.
represented by the terms of
the lowest degree. Cor. i. If any tangent and the corresponding normal are taken as coordia 9 nate axes, the equation of the curve takes the form na; + 2/w y -f by = 2fy >
;
this
form
Cor.
ii.
is
often useful.
Tho equation
of the circle of curvature of
ax* 4 2 Jixy at the origin
is
3
-f
fy
-f
2gx -f 2fy
-
of the form X
2 (.
f
2xy cos
a <*)
-f
y
+ 2gx + 2fy
)
=
0,
must be chosen so that one of the common chords of the circle and conic, which passes through the origin, should be gx+fy = 0. Hence ax* + 2 hxy -f by 2 - X (x* + 2xycoaa> + y*) has gx +/y for a factor, and
and
X
therefore vanishes
when gx+fy
= 0,
when gx+fy =
and unless g or
and the equation of the
or
/
is
when
x
i/
-
f
zero this
circle of curvature
is
;
thusX
is
the value of
MISCELLANEOUS THEOREMS
438
This method can be used to find the point
(#', y*)
circle of curvature at
any
of the conic
Change the
origin to the point
ax*+ 2hxy + Z#
2
(xf, */),
then the equation becomes
+ 2X'x+ 2 Y'y =
0,
where, as usual,
X'
The equation new axes is
= ax' + hy'+ g,
T = hx' + ty+f.
of the circle of curvature at the origin referred to
the
for the equation of the tangent at the origin is
or
(^
2
X'x+Y'y
=
0,
~2AX r-far + 2(X' -2X'r cosco-f F 2 (X's + Yy) = 0. 2 &X' -2feX'r + ar 2 = CS -A:=--A, since /) is on 2
4-2^cosco4-2/ )(&X
/
/2
2
)
2
)
/
But
the curve
;
.-.
/
the equation of the circle referred to the
new
axes
is
)(X .rH-r^) = 0. The equation referred to the original axes can be found by substituting (x x*)j (yy'} for x and y respectively.
A(^ + ^ -h2^cosco)-2(X -2X r cosa)^r /2
2
Example. at
tlie
To find
point (a cos
cf>,
fc
the circle
sin
/
/
of curvature of
2
ih?,
/
ellipse
-^
+
r^
=
1
r/>).
Transfer to parallel axes through the point (cos, 6sin) and the equation of the ellipse becomes
The
circle of curvature referred to the
new axes
is
=
x
where - co -f
t/
f-
sin
=
is
0,
one of the lines
(I
X i.e.
tlifi
circle of curvature
=
is
0,
MISCELLANEOUS THEOREMS
439
referred to the original axes,
or,
>)
x*
i.e.
+ i/*-2x
--
cos 8 -2y
sin
+ (a
2
=
0,
s <
- 2 62
)
cos
2
+
^- 2a
2 )
sin 2
0.
<
Equation of conic when a pair of tangents and the chord
4.
of contact are given.
Special Notation.
=
Let the equation of the tangents in abridged notation be u 0, then the equation of the conic is the chord of contact w = ;
= 0, and uv = w*. v
=
2 w; where & is a deterthe equation should be &.w minate constant ; this can, however, be included by considering Jcu to be the equation of one tangent.)
(Strictly,
=
Let
ttj,
w
v l9
l
be the values of the expressions
u,
v,
iv
when
the
coordinates of any point P are substituted in them evidently we can express these coordinates in terms of any two of the quantities MI, #i, MI, and conversely, if the values of u it i\, n\ for any point are ;
known we can determine we can use the values of position of the point
Now hence,
;
if (HiViWi) is if
%= A?r
T
,
its
coordinates.
It
should be clear that
and w at any point to indicate the two the ratios H:V:W are necessary, only 2 and a point on the curve, we have H^VI u\ M, v,
=
then v l
=
-~,
A
i.e.
,
the coordinates of a point on
=- =~ 1
the curve are connected by the relation
-J
,
and con-
versely, if X is known the ratios t^ t\ u\ are known, and the we shall refer to this point on position of the point is determined :
:
:
the curve as I.
To find
{
the point X the
'.
elation of
the chord joining two points A, p on the
curve.
Let the chord be
Au + Bv+Cw =
;
then
B
the equation of the chord
is
=
0.
MISCELLANEOUS THEOREMS
440
The equation of
II. /ut
= A,
w-f A #
viz.
Hence, have
if
the tcmgent at
2Aw =
2
A follows by putting
the point
0.
the tangent at A passes through the point (w x v l A 2 Vi _ 2 A Wl + 0, :
Wj),
:
we
%=
two tangents pass through this point, and the parameters A n A 2 Hence of their points of contact are given by this equation.
i.e.
x A l
M!
:
_2w
+ AA ,
l
:
l
_M! 2--
Al A
>
2~-^= A! A 2 w ^
:
1
(A t 4-
:
A 2 ).
Conversely, the point of intersection of the tangents at the points A n A 2 is {A^-.l-.i III.
To find
the chord
of contact of tangents from
Let the points of contact be A a A 2 ,
/.
the chord of contact
w1 .'.
,
u\).
(A x -h A 2 )i(;
=
0.
in II that in this case
=A
:v 1 :M? 1
1
A 2 :l:i(A 1
+ A 2);
is
2 wiv l
vu l
itv l 4-
By
vl
.
the equation of the chord of contact
Note.
,
is
M + A1 A2 f
But we have shown
the point (i^
= 0.
the usual argument the polar of any point (w lf v n w^) take8 the
same form. IV. If u,
Vj
and w are taken in the form x cos a + y sin a p = 0, = k ^2 in which case
the equation must be taken in the form uv any point on the conic is indicated by
u This
w
v
=
is
sometimes
i.
:
w= for
Let the conic be uv
*=*
w
1
A2 k :
to
:
=
0,
v
=
for
example that
0.
ratio
of the pencil formed "by joining any fifth point on it is constant.
and four
fixed points X,, A 2 , X 8 , A 4
X.
The equations
,
A.
we then know
between u
The anharmonic
four fixed points on a conic
point
v
useful,
bisects the angle
Example
:
.
of the rays of the pencil are 0, 0,
and any
fifth
MISCELLANEOUS THEOREMS
441
=
Now M-Xw> =
are straight lines: let us write a? 0, w~\v~ w-Xt0, then the equation of the rays of the pencil become x \^y, x = X 8 y, x = X 4 t/ and (Chap. II, 10) the anharmonic ratio of this
= w-Xv, y x
== > 8 y,
pencil depends only on A,,
Example the conic
to
OF Let OP,
X3
,
,
and
X<
PQ
let
conic
uv
is
be
=
w=
bisectors of the angle
J?S
is
POQ
the points X, /A. the line M + \/LH> (X-f /i)
lines 07?,
OS
-v =
u-\*v
are
w= 0,
O
u+v
0,
= 0,
and by hypothesis
0,
w=
is
(chap,
2
ti-fi
i;
=
=
X/i
=
0.
this
passes
1.
0.
u-v =
0, n-f-v
=
0,
which sinoe
io)
ii,
.'.
;
the anharmonic ratio of the pencil
=
u-v =
are the lines
S be
through the intersection of u
U-,A'
= #cos-f T/sin/3-^,
t>
(including any necessary constant factor), then the
R
Now
0,
w\
Let
and
are equally inclined
be the lines
The
The
OS
passing through V, prove that OJR,
OQ
X.
are tangents to a conic from a point 0, and POQ meets PQ in F, and if US is any chord
MEE tfcosa-fysina-^) = and
independent of
is
// OP, OQ
ii.
the line bisecting the angle
of
X2
I~rr
u-X 2 = t>
M=
~
=
=
the pencil is harmonic; and since M + v 0, u-t? 2 perpendicular they are the bisectors of the angles between u-X
equal to
w
p*v =
1,
i.e.
is
are t?
=
O
f
0.
Illustrative
To find
I.
0,
the locus of the poles
Examples.
of tangents
to the conic
with respect to the conic
S
= az 2 + 27m/-hfy/2 + 2^rH-2/y-f c = 0.
The tangential equation and
if
Let
Ix
of S' ==
1 is any tangent to S'= be the pole of Ix + wiy -f 1 =
+ my 4
(a?',
y')
is
equation of the polar of
0,
(Z,
m)
satisfies this
with respect to
5'
equation. the ;
=
(#', y') is
or
This equation
is
therefore identical with Ix -f my
Hence l/X' = vn/Y' (a',y') is
=
1/Z'
;
+ 1=0.
therefore (dropping the accents) the locus of
MISCELLANEOUS THEOREMS
442 II.
To find
S=0,
and
the length of the equi-conyugate diameters of the conic
show that they are inclined at an angle
to
2
sin""
1
=
a+ o
one another.
to
If 2r is the length of the equi-conjugate diameters, their extremities lie on a circle whose centre is the centre of the conic and whose radius is r hence they are common chords of S = and
:
Thus
their equation is of the
form
Transfer the axes to parallel axes through the centre of the conic equation then becomes
Now
X
is
;
their
to be determined so that this represents a pair of straight lines
through the origin
;
.-.
=
-
r
cr X
= -~
.-.
;
the equation of the equi-
conjugate diameters referred to parallel axes through the centre of the conic is
(ai*
+ -
+ 2hf*xy +
a? J
(b>*
+
|J
y
=
0.
The condition that these should be conjugate diameters of ax*
-f
a 2hxy -f % +
=
0,
.e.
(a
which gives the lengths of the equi-conjugates. The angle between the lines follows from the usual formulae. We might wa #3 = 1 the'equi-conhere usefully employ invariants: in the conic -f 5
\
jugates are
j
|
~-a,nd the Now
0,
|
*
-f
|
;
the length of the equi-conjugates
angle between them
is
sin* 1
-y --a
suppose that the equation of the conic ax*
-\
2hxy -f ^y* 4- 2 r/a? -f 2/y
-4-
c ==
is
then
MISCELLANEOUS THEOREMS by
.
448
suitable change of axes reduces to
.
,
,
- --
.-
.,
then we nave shown that
27? cos r-sa
sm
C
~
ab
a> >
2
-
sm
&>
'
ft
.
tt a
o>
A >
.
,
-7-5- are invariants.
sm'o)
~
also
What
III.
locus is represented by
= uEHlx + my + that for all vaUites
1 /o'
of a,
fo,
4
^c/^ 5wj^r
c it
awd
a, b, c
0,
are constants ?
Show
passes through a certain fxed point.
The equation is of the second degree and represents a conic it is satisfied by the coordinates of any point which also satisfy any two of the equations :
=
w2
0,
Uj
= 0, = 0, MJ
t* 3
0,
whose sides are
The
coefficient of
a
i.e. it
wa =
represents a conic circumscribing the triangle MS 0.
0,
is
say.
Now t<
a
=
0,
i
Mij (Z 3
Uj tij
=
/!
s
a
= 0. = 0, CF
/
:
a
w
=
and
s)
Hence, ^
coefficients
BE,
represents a straight line through the point of intersection of further, the coefficients of x and y in the equation are
=
0,
of b
if
us
i- e is perpendicular to Ja w s\ ^i = be the triangle whose sides BC, CA, AB are then T/j = 0, and (using similar notation for the
+ ^(fstftj
-
ABC = 0,
and
c)
of the triangle.
U^
=
0,
U9 ~ =
Hence l^
0,
represent the perpendiculars AD, T7a are concurrent at 0, C7S =
This point clearly lies on the conic since the equation of 0. ai^ Uv -f fcu 2 172 -I- cw s r/"8 = 0, U3 = i/ t/ a = 0, t/2 0, l/j = 3 Again, since the pairs of lines i*t 2 are perpendicular, the coefficients of a: and y* in each term of this equation
the orthocentre. the conic
is
=
;
;
MISCELLANEOUS THEOREMS
444
2 2 are equal and opposite, hence the sum of the coefficient of a; and y in the whole equation is zero, i.e. the conic is a rectangular hyperbola. Finally, then, the equation represents a rectangular hyperbola circum-
=
=
= 0, and all such 0, 0, x 3 2 scribing the triangle whose sides are the the orthocentre of triangle, rectangular hyperbolas pass through IV. Prove that (.r
cos
a + y sin oc
a -f y sin a
p) (x cos
j/)
_'2 ycosft
A
form of the equation of a conic ofivhich x cos a + y sin a p = 0, rr cos a -f y sin a
2
g)
=
15 the general
j/
=
are the directrices.
The tangents from the
circular points at infinity, il and fi', to a conic intersect in pairs of foci, and the directrices being the polars
of the foci
it
formed by the
follows that the conic circumscribes the quadrilateral directrices, and that the tangent to the conic at each
vertex of this quadrilateral passes through 12 or fl'. The directrices are parallel to the axes of the conic, so that one pair
is
perpendicular to the other pair.
Let the conic be (x cos
oc 4-
y sin a
p) (x cos
a + y sin a
+ &(#sina The equation tion of the lines is
p')
2/coscx
ycosa
r)(#sinai
/)
=
0.
of the tangent to this conic at the point of intersecand #sincx ycosa r #cosa-f ysmoc p
=
=
(#cosa + #sin# p)(p
p')-}-
since this passes through 12 (cos a -f i sin a) (p
(1,
p')
p
i.e.
p'
ycosa
&(#sina 0),
i,
+
a
k (sin
=
ift
r)(r
r')= 0; and,
we have
(r
i
r
cos a) (r
r')
= 0,
x
).
we put r + r* 2#, the equation of the conic can be written (# cos a -h y sin a p) (x cos a -f y sin a jp' ) + 7c(^sina-//cosa~g) 2 -p(r-r 2 = If
/
)
Now let (p-p'Y = Aft
The equation of (x cos a +y sin ex
=
A &> so that 2
/
-Aj (r-r
2 )
or
_|&(r -r')
2
= }A.
the conic then becomes
p) (xcoa ot+ysinocp')
A
y cosa
0.
MISCELLANEOUS THEOEEMS
445
Examples XI. 1.
A conic
(i)
is
circumscribed to a quadrangle
:
show that the product
of
the perpendiculars from any point on it to one pair of opposite sides is in a constant ratio to the product of the perpendiculars from it on the other pair.
show that the product of the (ii) A conic is inscribed in a quadrilateral perpendiculars from one pair of opposite vertices on any tangent to it is in a constant ratio to the product of the perpendiculars from the other pair of vertices on the tangent. 2 2. The conic #V( + ka'*)+y*/(b'* + W*) = 1/(1 + fc) is for all values of k inscribed in the same quadrilateral. :
3. Find the locus of the foci of conies which pass through the vertices of a given rhombus. 4. The director circles of all conies inscribed in a given quadrilateral are coaxal.
5. Find the equation of the chord of curvature through and the length of the radius of curvature at the origin for a# 2 -f 2hxy + fcy'-f 2gx + 2fy 0. Deduce the length of the radius of curvature at any point oil the conic
=
S=0. 6. A
variable chord subtends a constant angle at a given point of a conic
:
find its envelope. 7. Chords of a conic which subtend a right angle at a fixed point envelope a curve of which the fixed point and its polar are focus and directrix. 8. A conic is drawn touching the axis of x at the origin and having its
P
centre at the point (a, b) the axis of y intersects the conic again* at prove that the tangent at P passes through the point (a, 21) and that the normal at P touches the parabola whose equation is (y 26) a = 4a#. ;
;
9. If the four points of intersection of two parabolas lie on a circle their axes must be at right angles, and the coordinates of the centre of the circle referred to the axes of the parabolas must be equal to the serni-latera recta. 10. A straight line is drawn to be cut harmonically by two given circles.
Show
that
its
is
envelope
a conic whose foci are the centres of the two
circles.
11. is
Show
met by
that the pair of tangents to Xa: + fiy + r = are given
deduce that \x + py -f r
=
is
&=
a directrix
where the conic 2 A(X#-f M# + >*) *= 0, and
at the points
by S. 2 if (X*
- y?)/(a - b)
*
\p,/h
=
2/A.
Show
that a third pair of straight lines through the four points where the straight lines S cut the axes are cS+4Hxy 0. 13. Find the equations of the conjugate diameters common to 12.
=
14.
=
Prove that the locus of the poles of tangents to
with regard to the conic
a'a?
a
+ %h'xy -f &'y 2
=
1, is
cw?
2
+ 2hxy + by 2
=
1,
MISCELLANEOUS THEOBEMS
446
Show
that, if the pole of one common chord of a circle and a fixed on the circle, the pole of the opposite common chord lies on the circle also show that the lines joining these poles to the centre of the conic make equal angles with the major -axis of the conic, and that the rectangle under them is constant. 16. Two concentric conies are equal in every respect. Prove that the points of intersection of their common tangents always lie on the director circle, whatever the angle between the major axes. 17. Prove that the locus of the foot of the perpendicular from the origin 15.
conic
lie
;
S = 0, subtending a right angle at the origin, is the + b) (a;* + y~) + 2gx +fy + c = 0. 18. The equations of two conies touching one another at the origin are ax -f 2hxy -f b\f -f 2fy = and a'x* +2h'xy 4 &V + 2f'y 0, and they also a chord of the conic
011
circle (a
1
PQ
intersect at
;
find the equation of
PQ and
of the pair of straight lines
OP, OQ. 19. (i) If the chords of intersection of
which pass through the centre are at right angles, a 4- b *= a' + 6' (ii) If the chords which do not pass through the centre are ;
angles, 20.
a-l . a '-V
,
-^-
Show
at right
that the area of the parallelogram formed by drawing tangents
to the conic
S=
'
parallel to the axes of coordinates is
'
*
jjgv
ch*. the angle between the axes and A = abc -f 2fgh a/ bg* 21. In oblique coordinates what is the locus of a point P of which the
where
o>
2
is
coordinates are acos$, &sin
What
two of the lines joining Q,
Q, Q', then
(-VV-&
Q' to the points (\/(a
a
-& 2 ),
0)
and
2
), 0) are parallel. 22. Show that the equation of the director circle of the conic
l s
i;
Assuming that the the director circle,
Find the
a
and and two
directrices are chords of intersection of the conic
show that the equation of the four (two
imaginary) directrices 23.
=
=
foci of
is V*
- (a + b) uv + (ab - A 2) vP =
2^ + 3ary-2y9 -12a?-4t/-h8
real
0. 0.
Prove that, if at a point on a hyperbola the sum of the tangents of the angles which the normal makes with the asymptotes is 2, then the vertex of the parabola of closest contact at the point lies on a line through the point inclined at an angle tan""1 3 to the normal, and at a distance from the normal 24.
equal to three-eighths of the radius of curvature. l 25. A conic has four-point contact with tue parabola y
4a#, and the
MISCELLANEOUS THEOEEMS radius of its director circle of
centre
its
is
is
constant and equal to
Show
16aV = 0. (y*-4aaO(t/ + 4ax + that the points whose rectangular coordinates are a) lie
(c, a), (a, c), (a, ft), (fe,
of a,
and whose centre
b, c,
27.
8aV
Prove that
if b
=
6'
upon an is
ellipse
fixed if a
-f
b
whose eccentricity
+c
is
is
(b, c), (c, 6),
independent
constant.
the conies aa?
a'x*
will
Prove that the locus
the curvd 9
26.
c.
447
= x, + 2h'xy + b'y* = a?
2
-f
2hxy + by*
have three-point contact at the origin. parabolas have three-point contact at
Two
P
and intersect
that the tangents at Q, the line QP, and the line through common tangent at form a harmonic pencil.
Q
at Q.
Prove
parallel to the
P
28.
ax*
common aV-f 2h'xy + &V
Find the equation to the
+ 2hxy + by* =
1,
conjugate diameters of tho conies *
tangents inclined at the angle 45 to the positive direction of the axis of x are drawn to the conic ax* + 2hxy + by* + 2gx + 2fy + c = 0: 29. Parallel
___________
prove that their distance apart 30.
is
I -2A o v
7,
(2 ft
+ a + b)>
Prove that the conies which pass through the origin and which are #8 + 2ay-f2y2 -2#-2y + 5 = are both real and
confocal with the conic
pass both through the intersections of the circle a?*-fy'-2# conic sr'-f 2xy + 2y*-2x-2y = 0. 31.
=
and the
Show that the straight line Ix + my + n = will be a tangent to + 2 hxy + fy* = 1 if bl* - 2 him + am2 = (ab - h*) n*. Show that
conic ax
11
pair of tangents to this conic from any point on the conic (ab are equally inclined to the axis of x.
8 fc
)
the
the
xy 4- h =
32. Find the envelope of the chord common to an ellipse and its circle of curvature at any point, and show that its equation may be put in the form U* 4- Fa = 0, where 1 1 = is the equation of a similar and coaxal ellipse,
V=
that of
its
equi-conjugate diameters.
Show
that the angle subtended at the origin by the intersections with the circle ar*-f y* + 2/^ + 2 gx + c = of any tangent to the parabola 2 is bisected (0x +fy + c ) internally and externally by the axes of 4/0;ry 33.
=
coordinates. 34. Find the equation of the conic which passes through the origin and 2 confocal with the conic ax* + 2hxy + fy/ = 2x.
is
35. If e is the eccentricity and 21 the latus rectum of the conic given by the general equation in rectangular coordinates, prove that
^ ~ (a 36.
Show
to the conic 37.
Show
+ b)*
_
H(x^-y ) = (A-B)xy f ax* -f 2 hxy + 6y + 2gx + 2/y -h c = 0.
that the lines
"
(ab-h*)*~ A* l
are conjugate with regard
that the general equation of conies having the points (a, b), 6* -A) = (xy-ab)*. foci may be written (.r'-a'-A) (y*
(-a, -6) as their 38.
A
conic touches the axes of coordinates (not necessarily rectangular)
MISCELLANEOUS THEOREMS
448 and has
its
centre at the point /,
2 - 2 y rectangular hyperbola #
Show
g.
2fx
-f
2 <jy
=
that the locus of
its foci is
the
0.
39. If w a -f ! -f c = is the equation of a conic and A is its discriminant, show that the asymptotes are given by C (u 2 -f t^ 4 c) = A. Find the condition that the asymptotes should form a rhombus with the lines t< 2 = 0. 40. A series of conies being drawn having four-point contact at the origin with the conic ay? + 2 hx\j -f 2 -f 2/y = 0, prove that their director circles form a coaxal system, and that ~-/t//(a 2 -f & 2 ), + a//(a 2 -f 7i 2 ) is one of the
%
limiting points. Find the equation of the radical axis. 41. A variable circle passes through a fixed point A and also through the point of intersection of two fixed straight lines which it cuts in P and Q.
Find the locus of the centre of gravity of the triangle APQ. Show also that the envelope of the straight line PQ is a parabola touching the two fixed straight lines. 42. Interpret the equation (ax + by if it
2
2\xy
I)
=
and
0,
find the value of X
represents a parabola.
If it represents a parabola, and if a -ft is constant, prove that the locus of the focus is a circle, and find its centre and radius, the axes being either
rectangular or oblique. 43. If
u
OL
= 0, =
and y
*=
=
are the equations of a pair of tangents to the conic their chord of contact, explain the geometrical meaning of
<*'
-a = 0, w-aa' = 2
An
0,
l w-y =
0.
through the origin, touches the axis of;/, and has double contact with a fixed circle whose centre is at the origin. Show that the locus of the foci is a pair of circles. ellipse passes
LM
44. Interpret the equation
=
JV
2
where
,
=
Jlf
0,
=
0,
JV*=
each
represents a straight line. Find the equation to the tangent to this curve at the point (Z/, A/', AT '). Deduce the equation of the pair of common tangents, other than the axes,
=
4 xy, k (x/a + y/V - 1) 4 xy. to the conies k (x/a + y/l> - 1) 45. The equation of a conic referred to two tangents and. their chord of 2 JR , find the equation of the tangent at any point in the contact being 2
form
8 /i
2
=
z
2
LM = L- 2^ + ^=0.
two tangents Through the point of intersection of jK = 0, 2Z/ + iV/ = 2 are drawn to the conic LM = # touching it in PQ, Find the equation of l the conic which has double contact with LM = R at P, Q, and which has ,
=
= 0,J? = for a self-conjugate triangle. the triangle formed by L 0, lf 2 at the 46. Find the equation of the tangents to the conic ax* H- fcy -f c = in the form ends of the chord px -f qy + r = (bcp*
+ ca< + a&r2 )
(a#
2 -f
by*
-f c)
- ale (px + qy + r) 2
= 0.
equation of the family of conies inscribed in the = 0, = 0, is rectangle formed by the lines x + a 47. Prove that the
yb
Prove also that the locus of the foci intersect
on
this locus they
is
# 2 -y 2
do so at right angles.
=a
2
-J/ 2
,
and
if
two conies
MISCELLANEOUS THEOREMS 48.
Determine the magnitude of the axes and the
449
foci of
49. The coordinate axes being the lines joining the middle points of the opposite sides (2a, 26) of a parallelogram, prove that the coordinates of the point of contact with the line \x + ny-l = 0, of the conic which can
be inscribed in the parallelogram to touch this 50.
Find the equation of the director
the conic
a
is
a
-f
2 hxy
a a -/ia fc* + 1)/2X,
circle of the conic
by* + 2gx + 2fy + c = 0. show that the coordinates of its parabola
ax*
If
line, are (X
-f
focus are
{FH+j(A-B)G} 51. Prove that the four points of intersection of two conies, and the four points in which the asymptotes of one cut those of the other, all lie oh a conic, which is a rectangular hyperbola whenever the other conies are both
rectangular hyperbolas. 52. A conic circumscribes a right-angled triangle ABC, and at the right angle A it touches the circumscribing circle. It also passes through the centroid of the triangle.
Prove that
its
is
eccentricity
{2/(l
-sin
B sin C)}*.
53. Prove that
Cx,
Cy,
CiV-A
=
and
C
F,
Cf,
H,
are conjugate diameters of the conic ax* -f 2 hxy + by 9 + 2 gx -f 2/y 4- c
where A
is its
gx'+fy* +
c9
54.
Show
discriminant, A,
and
B C F G t
y
y
}
=
H are
0,
the minors of A,
N
is
any point on the curve. tangents are drawn to a series of confocal conies from
a?', y' is
that if
a fixed point on one of the axes, the locus of the points of contact is a circle, and prove that two such circles corresponding to two points, one on each axis, cut each other orthogonally. 55.
A
parabola
is
drawn touching the sides AB, BC CD, DA of the Show that its directrix passes through the y
cyclic quadrilateral ABCD. and BD. intersection of
AC
Trace the curve \/2a~2a:-f *^x + y+ */2y-2x = 0. What portions of the curve correspond to the various arrangements of the signs of the What is the length of the latus rectum ? radicals ? 57. A conic is drawn having one side of a triangle for directrix, the Prove that opposite vertex for centre, and the orthocentre for focus. 56.
the sides which meet in the centre are conjugate diameters. 58. Show that the general equation of conies, having (a, 6), (a', &')
as their foci,
may
be written
the
points
MISCELLANEOUS THEOREMS
450 59.
Find the conditions that the conies o,
a'x* + 2h'xy
+ b'y* + 2gx + 2fy
=
may have
(i) three-point contact, (ii) double contact. variable parabola S' has three-point contact with a fixed parabola S, their common chord and common tangent being equally inclined to the
A
on the curve 27aya = #(a?-9a) 3 60. A parallelogram circumscribes a given circle and one vertex moves on a fixed line prove that one of the others describes a conic. 61. Through the four points of intersection of a circle and a rectangular hyperbola two parabolas are drawn. Show that the tangents to the four
show that the focus of
axis of S:
S* lies
.
:
curves at a 62.
common
Show
may be
= ax
point form a harmonic pencil. that the equation for determining the foci of the conic
written
4-
hy + g,
rj
form aS and 4- by 4- /,
=
3
the
in
= hx
is
<
bS-q*
(hS~
i?)seca>,
where
the angle between the axes of
coordinates.
Show
that the equation of the axes of the conic
h
a 2
I
given by
b
& COS
1
is
*)*
1
o>
63. Show that the length of the diameter, conjugate to the diameter a is through the origin, of the conic ax* + 2hxy + fcy -f 2gx + 2fy + c =
ab
the axes being rectangular. 64.
-Show that the locus of a point whose coordinates are determined by 2 = l/(a s X 9 + &s \-f c3 ) is in y/(a 9 X + & 3 X-f c )
the relations x/(al \^ + b l \+cl ) general a conic section.
fl
Find the equation of the lines, through the
Can the locus be a pair of straight
origin, parallel to its asymptotes.
lines ?
Prove that the locus of the centre of a conic, with respect to which four given pairs of straight lines are conjugate lines, is a straight line. 66. The length of the diameter of the conic S = 0, which is conjugate to 65.
the diameter which passes through 2 67.
a
{A (X' +
x
(a?',
y
)>
i8
r *)/(' S' - C A)}
i.
Two straight lines OA, OBC contain an
parabolas can be drawn through B,
C
angle 0; determine how many to touch OA at A, and find their
equations.
Prove that the diameter of curvature of one of the parabolas at the point 1 -f eri) where a, 6, c are the lengths of OA, OB, OC. 7 68. Employing as coordinates '.the distances r,r of a point from two
A is a* cosec 6 (Zri
1 /* af , given points, interpret the equations r 9 1 * 1 r -*-' 2ar-f a 0, and prove that the equation
(1+ p)(r-r7~
MISCELLANEOUS THEOREMS
451
represents a pair of straight lines, the angle between which is bisected by the line joining the two poles, c being the distance between these points.
Interpret the constants p and q. 69. Show that through any four points there can in general be parabolas and one rectangular hyperbola.
drawn two
Prove also, that if ^ , /a are the semi-latera recta of the parabolas, p 1 p t the distances of the centre of the rectangular hyperbola from the axes of the ,
parabolas, then 70.
* pf + 1^ = pf + 7a
A triangle
is
.
inscribed in the conic
aa?
a
-f fy/
a
1,
Ax* + 2 Hxy -f By = 1
touch the conic 1. Show that the envelope of the third side
is
[(Ab + Ba AB + H ') + 4 (H* AB) al]
+ by* - 1)
(ax*
f
2
the conic
and two of
its sides
CHAPTER
XII
TRILINEAR AND AREAL COORDINATES THE
position of a point is fixed when its coordinates referred to two axes are known, consequently the coordinates of a point 1.
more axes (e. g. the point's distances from three straight lines) must be connected by as many identical The best relations as the number of coordinates exceeds two. known multiple coordinates are the Trilinear and the Areal. referred to three or fixed
In trilinear coordinates the position of a point is given by its perpendicular distances <x, /^, y from the sides BC, CA, AB of a triangle A BC, which is known as the triangle of reference.
P
lies within the triangle of reference, its coordinates If a point are all if lies outside the triangle, a is positive positive (a, /3, y) are on the same side or and or negative according as the points ;
P
P
on opposite
The
A
BG
sides of ; similarly for ft and y. sides of the triangle of reference divide the plane into seven
parts the signs of the coordinates of points within them are shown It may be noted that not more than two of the in the diagram. ;
coordinates can be negative. The areal coordinates (a?, y, z) of a point
&BPC &ABC
P are
ACPA
&APB
&ABC
&ABC'
the ratios
with the same sign convention as that given above for trilinear coordinates.
TRILINEAR AND AREAL COORDINATES Thus the by
areal
the relations
and
2Sx
453
trilinear coordinates of a point are connected
= aa,
the lengths of the sides, and
2S/ = 6/3, 2Sz = cy, where a, 6, c are S is the area of the triangle of reference.
i. When a figure is projected orthogonally into another the areal coordinates of a point are unaltered, since correfigure, sponding areas in the two figures are in the same ratio. The
Note
projection of the triangle of reference
taken as the
is
new
triangle of
reference.
Note
A
point is the centre of mean position of masses, to its areal coordinates, placed at the vertices of the proportional ii.
triangle of reference. Identical relations between the coordinates.
(i)
Let
P
be the point we have
(a,
y)
/?,
;
then in Fig.
(i),
where
P
is
within
the triangle,
= 2 A BPC, 1(3 = 2 A CPA, aa + M + cy = 2 &ABC.
aa Hence
c
y
= 2 A APB.
the point P is outside the triangle the coordinates point are supposed to contain their signs, so that in this case the numerical value of 3 is negative, and
In Fig.
(
a
&
(ii)
;
y) of a
aa 4-6/3 + cy=
ABPC- A CPA + A APB = 2 A ABC.
In every case we have similarly,
aa + bp-f CY
=
2S.
(i)
This result can also be written
asin4 + /3sinJ5 + ysinC= S/R where
=
4Bsin J.sinJBsin
C,
(ii)
B is the radius of the circurncircle
Now
if (x,
of the triangle of reference. are the areal coordinates of P, we have from Fig. (i), y, t)
x+y+* = AjBPC/S+ A OPA/S+ &APB/S = and, with due regard to sign,
we can show
x+y 4-2 =
1.
1,
in general that (iii)
TRILINEAR AND AREAL COORDINATES
454
Note
By means of
i.
these relations any equation can be
made homo-
geneous in the coordinates.
Note ii. Areal coordinates will be generally used rather than tvilinear because the identical relation is so simple. Throughout this chapter trilinear coordinates are called
y and areal coordinates
OK, /3,
y
x,
}
z.
Now
the position of a point is known when any two of its coordinates are known, and the third coordinate can be found from the identical relation.
Also a point OK
:
/3
y or
:
x\y\8
when
fixed
is y
the mutual ratios of
We
known.
are
shall use (a,
^3,
its
coordinates, denote a
y) to
when the actual values of the coordinates are (a /3 y) when their mutual ratios are given, In trilinears suppose OK /3 y = I m w, then we have 2S ~~ I m~~ n~~ al + bm + cn
point :
given, and
:
hence
OK
=
So also in
so that
x
=
-
2SI -
-=
areals if x
:
:
:
:
:
&c.
,
=
y z :
x
y
*
I
m
n
I
:
m
:
we have
n,
+m+n
2/({+m+i), &c.
To find the coordinates of a point points P(i, ft, Vi), Q( 2 Az* ^2)
E
dividing the distance between the
in
ti* e
Since the perpendiculars from
P
?
rai *
l
:m
-
and Q to the side
EG
of the
triangle of reference are ty and 2 we can show in exactly the same way as in Chapter I, 6, that the perpendicular from the required ,
point
~
R to BC is
(
l
-
2
The
.
required point
P
m The ,
then
+m
If the areal coordinates of the points and ^hen their trilinear coordinates are (2*2* 2/2 > ^2)9
{28x 1 /a
is
Q
are
f
,.,.,.
.
.
^-coordinate of the point
and consequently
its
RT, is .
^-coordinate
.,
-
,
therefore, as above,
is
?v
"-*
l
y-r
TRILINEAE AND AREAL COORDINATES Thus the
areal coordinates of the
between the points
y l9 zj,
(a? lt
(x 2 ,
y2
455
point dividing the distance in the ratio lim are
z.2 )
,
(mXj
+m
l
l
\
Note. It should be carefully noted that in the above work the actual values of the coordinates have been used ; if the points P, Q are given as {x l y :zi}> \x^ ya z^} dinates (viz. xl /(x l -\- Vi :
:
{
:
,
we must calculate the actual values of the coor+ ti), &c.) before using the formulae to find the
coordinates of R.
The point {fai + mx^ityyi + my^ifai line joining
the points
(x l
:
yl zj, :
(x 2
:
+ nutt)} lies on the straight y 2 z 2 f r a ^ values of the :
)
ratio l/m.
The
actual values of the ^-coordinates of the points are
hence the ^-coordinate of the point dividing the distance between (
x
\
:
Vi
:
^i)
and
(x 2
:
in the ratio
y2 zj :
m (x +y + z ):l fa +y + zj 2
2
with similar results for the other coordinates. ^
+ mx^)
:
(ly v -f
my 2
Find
Example.
)
:
(lz l 4-
tJie
mz 2 ) }
trilinear
lies
on the
2
l
is
Hence the point
line joining the points.
and areal coordinates of
the centroid
of the triangle of reference.
The
trilinear coordinates of the vertices A, B,
are (2S/a,0,0)
f
C of the triangle
(0,25/6,0), (0,0,2%). The mid-point
of reference
A of EC
is
therefore
(0,5/6, S/c).
The centroid
G
divides
AA
1
in the ratio 2:1, so that
{2 5/3 a, 25/36, 25/3c}. These are the actual values of the coordinates
;
G
is
the point
the point can be referred
to as {I/a: 1/6: 1/c}.
We
have seen that a point is the centroid of masses, proportional to its areal coordinates, placed at the vertices of the triangle of reference. Hence the areal coordinates of Q are all equal, i.e. the point Q in areals is
(1:1:1). This is also evident by elementary geometry, since the triangles
CGB, BGA are equal. The actual values of the
areal coordinates of
G
are (J, J, J).
AGO,
.
Examples XII a. 1.
What are the actual (3:-2:7)?
values of the areal coordinates of the points
(1:2:5),
2. If the sides of the triangle of reference are 3, 4, and 5, find the actual values of the trilinear coordinates of the points (2 -3 : 0), (1 6 : -8). :
:
TRILINEAR AND AREAL COORDINATES
456
Find the
3.
trilinear coordinates of the mid-point of the line joining the
points (1,8, 5), (-1, -1,3). 4. Find the areal coordinates of the point? pairs of points
midway between the following
:
(i)
-1);
(4,2, -5), (-2, 4,
(l:2:3),(5:-l:-2). The trilinear coordinates
(ii)
5.
3, 4, 1).
(
of three points are (4, -2, 2), (6, 2, -2), sides of the triangle of reference.
Find the mutual ratios of the
In what ratio does the point ( 4 15 7) divide the line joining the points (1 6 -5), ( 2 3 5) ? (Areal coordinates.) 6.
:
:
:
:
:
:
is the mid-point of BC, find the areal coordinates of a point P in = = SPA'. that (a) (b) 8. Find the coordinates of the points of trisection of the line joining the points (a? lf t/ lf ^), (#a y a *a ). 7.
If
A'
AA' such
,
9.
PA
AP
;
AP
,
Find the values and the mutual ratios of the coordinates of the follow-
ing points (a) in trilinears (b) in areals (ii) the ex-centres (i) the in-centre orthocentre of the triangle of reference.
:
;
;
;
(iii)
the circumcentre
(iv)
;
the
10. Find the areal and the trilinear coordinates of the points where the bisectors of the vertical angle A and the exterior vertical angle of the triangle of reference meet the opposite side.
2.
Transformation of trilinear and areal coordinates
to
Cartesian
coordinates.
(A)
Take the
sides
CA,
CB
and therefore or
Conversely,
Again,
o>
=
C.
(a,
/3,
y)
for axes of coordinates, so that
Let the coordinates of a point P be (X, Y) in Cartesians, Then we have in trilinears, and (x, y, z) in areals.
= 2S-aXsin C-&rsin C, y = b sin A X sin A Ysin B. Y = cosec X = a cosec x = ax/2S = X/b (7,
*= !--#=
/3
(7.
l-X/6-r/a.
TBILINEAR AND AREAL COORDINATES
457
(B) Occasionally it is convenient to change to rectangular axes. and a line through G perpendicular to it for CA as axis of
X
Take
Then
axis of Y.
X = a cosec C+fl cot G Y = f
,
ft
;
and conversely, Oi
= X sin C
f
Ycos C
ft
,
= Y
y
9
=
X sin J.
a sin U
Ycos -4.
(C) More generally, let OX, OY be any pair of rectangular axes, the origin being any point inside the triangle of reference. Now let the perpendiculars from the point to the sides of the triangle make angles a BC, CA, ^a ( a ^ measured in the positive 2
AB
,
,
A
B
The equations of the sides of the triangle of reference, referred to OX, OF as axes,. are then Xcos^+rsin^-2)! = 0, Xcos0 2 + rsin0 2 -j> 2 = 0, direction) with the axis
OX.
X cos where p {
,
j)2 ,
p3
3
+ Y sin
3
j> 3
=
0,
are the lengths of the perpendiculars from
to the
sides.
Now
if
P is
any point whose Cartesian coordinates are (X, Y) and (<x, ft, y\ we have
trilinear coordinates
ft
y
=i> 2 -Xcos0 2 - Ysin0 2 =1>3 Xcos03 rsin0 3
,
;
P
being inside the triangle, the signs of (X, /3, y, when is also inside the triangle, are the same as the signs of the perpento the sides ; the reader should verify that these diculars from the point
formulae give the correct signs for
y
ex, /3,
when
P
is
outside the
triangle in various positions.
We have,
from elementary considerations, the following relations: ^0l = TT- C; 3 -02 = IT- A 03-0! = v 2 ;
TBILINEAE AND AKEAL COORDINATES
458 3.
The
The general
straight line.
linear
and homogeneous
equation of the first degree, in trilinear or areal coordinates, when transformed to Cartesians evidently gives a linear equation ;
la+m/3 + ny
=
=
or lx + vny + nz
0, therefore,
represents a straight
line.
This can be shown independently: let ( lf y lf ^), (#2* #2> ^2) be any two points on the locus Ix + my + nz = 0, and (x, y, z) any
on this
third point
Then
locus.
Ix 4-
my + nz
x
y
= 0, = 0, ~ 0;
hence i
z
=
2/1
0.
2/2
This are not
is
the condition that constants
v
can be found, which
= 0,
Ay-f M/I + vy%
v-'y\ + v
f
of the
point
P'ZI + V'ZZ},
y
(x, y, z)
i.e.
are
the equation
of the straight
Let the equation be
Ix
+ my + nz
=
Z^ + my l + nz l Eliminating
?,
7^, w,
we
line joining (x lJ
= 0,
and
lx%
of the
the point (x,y,z)
point on the straight line joining the points (x ly y l9
To find
= 0,
= 0.
Hence the coordinates {P'XI + V'XU
jit,
such that \x + p,xl + vx2
all zero,
\Z + lJi8 l + V2.2
A,
z^
(a? 2
,
y
form is
a
^ 2)
y ly ^), (x 2 y^ ^ 2 ). ,
then we have
+ my% -f w^ 2 =
^.
obtain the equation of the straight line
in the form
y 2/i
2/2
Note i. This equation is the same whether x> y, z are the actual values of the coordinates or only their mutual ratios. Note points
ii.
The
(<*!, fl l9
Note
iii.
points (x l9 y l9
trilinear equation of the straight line joining the
yjj
(
2 , /3 2 ,
y 2)
is
exactly similar.
It follows at once that the condition that the three
e^
(x 29
y2
,
^
(a; 3
,
y a ^ 3 ) should be collinear
2/2
,
=
is
0.
2/3
Since the general linear equation in trilinears or areals transforms into a linear equation in Cartesians
and vice
versa, the following
TKILINEAE AND ABEAL COORDINATES
459
properties, proved in Chapter I for equations in Cartesian coordinates, hold also for equations in trilinear or areal coordinates. (a)
If u
= 0,
=
are the equations of two straight lines, then a represents straight line through their point of inter-
u + kv~
v
section.
Now x
=
the equations of the sides of the triazigle of reference are
= 0,
g
= 0,
=
so that, for example, my 4- nz a straight line through the point of intersection of y i. e. through the vertex A of the triangle of reference.
y
0,
represents z 0,
=
= 0,
Example. The equations of the to the vertices A, B, */*i
-#M = The
(b)
0,
straight lines joining the point (x 1 y l of the triangle of reference are y/yizfei ,
,
z^
= 0,
C
= 0.
xlx -y/ft l
= 0,
u
straight lines
v
= 0,
u + kv
= 0,
w
kv
mynz
Q
=
form
a harmonic pencil.
^
= 0, iw/ + = For example, y = 0, monic pencil whose vertex is the point A. The
Example.
triangle to a point
is
a har-
the vertices A, JB, of a of the triangle in the points
lines joining
straight
P meet
0,
the opposite sides
If B'C', C'A', A'B' meet BC, CA, AB respectively at A", B", C", these points are collinear. Show also tJiat AA', AA" are harmonic conjugates with respect to AB and AC. A',
&,
Let
(T.
ABC
be the triangle of reference and
equation of JBPis then z/z of this line and y = 0.
The equation of any */i-a?/*i + /iy = 0.
l
-x/x = l
any
at might line
The equation -rc/a^ + y/y l + z/zl fore represents B'C'. The coordinates of the point
It is
=
;
the j5oint (x lt t/ n Cj). The the point of intersection
is
straight line through B'
Similarly, the equation of
/*i
P
that B'
0, so
is
therefore of the form
through C'
is
of the form
=
is
in both of these forms
A"
are
therefore given by
evidently, then,
A"
lies
and there-
x
=
and
on the line
evident by symmetry that B*, C" also lie on this straight line. of the straight lines AC, AB, AA\ A A" are y = 0, z
The equations ylyi~ slz \
(
~
0,
y//i + zAi
^
;
To find the equation of the straight a i> ^i> 7i) I'H jf/wn direction. Let
be the point (a t
point on the straight line.
,
=
0,
these form a harmonic pencil.
/^
y^,
line
and
Suppose that
drawn through a
let P(cx,
fj,
OP makes
y)
given point
be any other
angles O l
,
2,
<9
3
TRILINEAR AND ARtiAL COORDINATES
460
with the perpendiculars from to the sides, measured in the same way as the angles between OX and these perpendiculars in 2 (C). Then,
if
ttj
OP = a
we have
r,
fap = rcos# 7i 7 = rcos0 a -^ = ^| = xqi= _ n COS
= rco&0
2
1
;
3
,
1
so that
COS
0J
COS
2
3
sometimes useful since and y) (a lf (3 19 y L ). The areal equation of a straight line, px + qy + rz = 0, can be put in this form: let (x lf y ly e^ be a fixed point on the line, (#, #, z) any other point. This equation
r
is
not homogeneous the actual distance between (a, /3, is
Then and
Hence and
p (x
since
;
it is
jw + gy + r* = 0, px l + qy l + r^ = 0. xj 4- q (yy^ + r (^ ^) (-*i) + (y-yi) + (*--*i) #+y + * = #i+yi + *i
5=S = ltl&
Hence
-fi
= 0, = 0, = 1*
.
the equation of a straight line is given in this non-homogeneous form, we can obtain the homogeneous equation Conversely,
Let the given be equation e ^
of the line.
and
let
if
each fraction equal
Then #
a^
Zfc
= 0,
=^ ^ =
m
-
1
n
,
k.
yy -.9ft= O 1
have
1=
7 I
x xl
y yl
K
I
m
n
r
y
^
nfc
= 0;
hence,
we
=
gl
for the required equation.
Examples XII b. 1.
2.
Find the equation of the line joining the points (2 - 3 5), (1 -4 -2). Find the areal equations of the straight lines joining the in-centre of :
:
:
:
the triangle of reference to the ex-centres. 3. Find the trilinear and areal equations of the medians of the triangle of reference.
Find the equation of the straight line joining A to the point of interBC and lx + my + nz= 0. 5. Find the equation of the line joining the vertex A to the point of intersection of the lines Ix + my + nz = 0, Vx + m'y + n'z = 0. 6. Find the trilinear and areal equations of the bisectors of the angles of 4.
section of
the triangle of reference. 7. Prove that a cos 6 l -f 6 cos
a -f
c cos
8
= 0,
where 6%
,
a
,
3
are
the
TRILINEAR AND AREAL COORDINATES
461
angles made by any straight line with the perpendiculars from any point to the Bides of the triangle of reference.
= m
2 (B), that the straight line lot + m$ + ny Prove, by using an angle B such that tan 6 n sin A) /(I cos C (I sin C
8.
=
AC
with
4.
Line coordinates.
=
a point, 1x + my+ nz
If
(#, y, &)
-f
makes n cos A).
are the areal coordinates of
the equation of any straight line. The fixed when we know the values of Z, m, n, or of
straight line is their mutual ratios
I
:
m
is
As we have
n.
:
seen in Chapter X,
Z,
w, n
are called the tangential coordinates of the straight line.
Let the straight line
AL, BM,
perpendiculars it
from A, B,
the
lengths
diculars PJ q
the ratios
the
by
p
:
lx
+ my + nz =
CN
We shall
C.
SO
at D,
and
lx
and draw
to
call
of
these
perpen-
r.
The
signs of
}
meet
% r are determined :
following
convention
:
q and r have the same sign
when the line cuts BC externally, and opposite signs when it cuts
BC
and similarly p internally and q have the same or opposite ;
signs according as the line cuts
AB externally or internally. the figure jp,
all
In
of the
same
sign.
D is
Now it is
q,
r are
the point of intersection of therefore the point (0 n : m).
x
=
+ my + w =
We have then ADAS _ m
:
-
~
n' for since
B and D
on the same
are on opposite sides of AC, while are side of AB, the areal coordinates of
D
are
D
C
The
C and
&DCA A ABC
(' should work out
A DAB] A ABC]
9
similarly the case where the line cuts internally, paying special attention to the signs. In all cases we have then I : : n jp q r, so that (p, q, r\ with
reader
BC
=
m
:
:
the sign convention explained above, may be regarded as the line It is evident then 0. coordinates of the straight line lx + my + nz
=
that the point equation of the straight line
Note.
In trilinear coordinates,
equation of the straight line
is
if p, q,
opa-f
fy/3-f
(jp, q, r) is
r have the same meanings, the rcy
= 0.
TRILINEAR AND AREAL COORDINATES
462
Identical relation between the coordinates
of any straight
Let the straight line make an angle from the above figure
= asin0, 3 = c sin (S
J9
a(jp
? + ccosl?sin0
from
Eliminating
Note. This or
we have (i)
6).
(ii)
+ ccosJ9( (i)
and
r)
= csinJ?cos0; = acsinJ3cos0.
(iii)
we have
(iii),
generally referred to as {op,
is
then
(ii)
p
which
J5(7;
r
g
Hence from
with
line.
=
2
&#, cr}
relation can also be written
(g
- r) a cot 4 + (r -p)* cot J?
4- (
p - g)
a
cot
C=
2
The tangential equation of a point.
The equation
in point coordinates of the straight line (jp, g, r) this passes through a particular point (x l9 y^
px + qy + rz = 0; if ##1 + #i -f ^i
So any straight
== 0.
line,
is
z^
whose coordinates
p, q, r satisfy this equation, passes through the point (x l y l jer^. Hence pxl qy l + rz v is {he tangential equation of the point ,
,
=
+
whose
areal
coordinates are
=
qy + re
Note
i.
The coordinates
reference are (1:0:0), (0:1
ylt
(x l9
equation px + or the line equation of the point is
In general, then, the (jp, q, r)
(#, y, z).
of the sides :
z^.
the point equation of the line
BC, CA,
AB
of the triangle of
(0:0:1).
0),
Note ii. The equations of the vertices A, p 0, q 0, r = 0. Note iii. The coordinates of a straight
B,
C of the triangle
of reference
are
a distance k from
it
are
pk,
qk
t
Illustrative
and at
Menelaus' Theorem.
lines joining the vertices
a triangle
q, r)
Examples.
Ceva's Theorem.
The
line parallel to (p t
rk.
A
of
ABC
straight
line
cuts the
of a triangle ABO at P, Q, JB ; show that
to a given point meet the opposite sides at P, Q; ;
B
sides
the points
show that
AR BP
SBPC
AX BP og_
_. ~~ QA
Let the given point be
i
RB'PC'QA" 1
Let the given straight line be
TRILINEAR AND AREAL COORDINATES The equation of the the given point
join of
A
The equation
to
is
i
BC and
463
of the intersection of
the given line
is
yM -*/ A
The
The coordinates
coordinates of P, the interline and BC, are
section of this therefore (0
Hence tion
:
yl
:
this point
(Orftin).
zj.
BP PC
zl
:
:
yl
;
Hence
this rela-
is
PC=-CP. Thus BP:PC = z
we viz.
viz.
:
:
;
is
PC = -CP.
Thus
:y l1
l
BP CP =
q l ^ this relatrue for all positions of P, if use the usual convention of signs,
tion
true for all positions of P, if use the usual convention of signs,
we
of AP, the join of and A, are therefore
BP:PC=q
-r lf
:
l
CQiQA^rti-p^
AR:RB = p
(i)
To find
Let (Xi, referred to
Yi),
the distance between the points (x lJ ,
^), (x 2 ,
y2 #2). ,
)
2)
1
l
l
2
2)
2)
1
l
2
1
2
1
;
l
2)
l
2)
^
= -aMyj-^fo-^-fc Thus d2 = -Sa'fo-ftH*!-*,). This expression can also be written d2 = 26ccos^l fa d2
9
2
2)
But /.
y^
= (x -x 2 -h(r -r 2 -h2(x -x )(r ~r )cos.a Now X = 'bx and YI = ay hence = 6 2 (^ - z + a (y - y + 2 (^ - x (y - y ab cos C
Then d 2 d2
-ft.
be the Cartesian coordinates of the points as axes of x and y, and d the required distance.
(X 2
CB
C4,
Y2
:
Multiplying these together we get the required result.
Multiplying these together we get the required result.
6.
l
2
o? ) 2
2 .
the distance between two points, Symmetrical expressions whose trilinear coordinates are given, can be deduced. for
TRILINEAB AND ABEAL COOBDINATES
464 Cor. line,
then
i.
~ = -- = l
-
If
the equation of a straight
where p is the distance between the points (#, y 1. Z+ w+ w = and a*mn + b*nl + cnm =
The equation homogeneous form is Cor.
ii.
of the straight line (p,
s~a?i
_ y-y\ _. 2^1
4-r and each of these
where for
is
p
y
z),
q, r)
(x l9
y l9
z^
in the non-
y
p-q'
r-jp
fractions
p is the actual distance between the points in 4 that
(a?,
y, #),
(# t
,
y x z^ ,
;
we showed
We
have taken the positive sign of the radical ; there is no loss of generality, since (i) jp, j, r can all have their signs changed, only the signs of their mutual ratios being determined by our sign convention, and to
(ii)
we can
regard p as positive or negative according
which direction we choose as the
Thus the equation
qr where
(xl9
of
other point
any (ii)
y l9 zj
To find
positive.
of the straight line (p, q, r) can be written
pq
rp
g
2/S
a fixed point on the line, and p the distance (x, y, z) on the line from it.
is
the area
of
the triangle
whose vertices are the points
Let the Cartesian coordinates of the vertices referred to (M, CB (X 19 Y^, (X 2 Y2), (X3 F3 ); then the area of the triangle
as axes be
,
,
X,
Y sinO
sinC bx3
x\
at
V\
S s
snce
1,
&c.
y*
=8 y*
TRILINEAR AND AREAL COORDINATES In
465
trilinear coordinates the result is
E
"*
ft
yi
2S
**
a
y*
3
7s
^
3
To find
(iiij
to tlte line loi
the length
+ w + ny
of the perpendicular from the point (a u
=
yj
/3 lf
0.
Transform the equation Z<x + w/3 + wy = into rectangular Carte2 the equation becomes by (C) U = (Z cos T -f w cos 2 + n cos 3) X + (Z sin fl t + sin 2 -f n sin 3) lr
sian coordinates
;
i
If
r (X 1? l j)
are
the coordinates of the point (a 1 from it to the straight line
length of the perpendicular
U
v
-r
where
{ (I
cos
fl
is
Z/!
1
2 -f
2
n cos
3)
+ (I sin ^ + m sin
X
the result of substituting Za 1 -f-m/3 1 -hny 1
=
Evidently
ITj
Now
cos 0j
(Z
+ m cos
1?
,
/3 a
,
the
yj),
is
2 -f
2
w sin
Tj for X,
F
3)
} *,
t r.
in
.
-f
m cos + w cos 2
-. 2
+ m sin 2 4- n sin flj) + 2 itn cos (0 a 3 ) 4. 2 w? cos (0 3 cos C 2
+ (I sin
3)
2 3)
X
2)
i
The length
of the required perpendicular is then (la !
+ mft + n ft)/ {*,*,*}
In areal coordinates the length of the perpendicular from is 2S(?a; 1 H-nii/ 1 -t-w^ 1 )-:-{aZ, 6m, en}. te-hwy-f^^ =
Cor.
(#i> Mi* *i) to
Note. If j>, g, r are the perpendiculars from the vertices of the triangle of reference to the straight line to + my + nz = 0, we have
p=
2 SI/ {al, bm.cn},
q =
2Sm/{al, 6m, en},
r=2Sn/{aJ, bm, en}, which verifies our previous result, I m n = p q r. If we substitute these values of /, m, n in the formula for the perpendicular from (xl y t 3t ) to the line Ix + my -f f is 0, we obtain pxl 4 gy t 4 rz t for its :
:
,
:
:
,
length.
Now the equation of the straight line is px + gy-f rz = 0, hence, applying the formula to this, we find that the length of the perpendicular from (*nyi.
i)
to
it is
2 S(pxl
Hence or
im
qyl
+ rwO
-5-
{op, bq, cr}
a*p*
which we
-f
-f
2>V
-f
c'r*
- 2bcqr cos .4
proved independently in
(a|), bq, cr}
=25,
2ca;y cos B 4.
O g
.
- 2abpg cos C = 452
,
TRILINEAR AND AEEAL COORDINATES
466 (iv)
I'a
To find
the angle bet^veen the straight
+ m'p + n'y =
Za + m/3 + wy
lines
= 0,
0.
Transform the equations by 2 (C) then in Cartesian rectangular coordinates the straight lines are parallel to ;
X
(I
cos
X
+ m cos
X (r cos 0{ + w'cos
2
+ n' cos
2
straight lines, Chapter (I
+ m cos
=- (wrc'
II,
2 -f n cos
sin 0j
(I
m'n) sin
-
(0 3
3 ) -f
the numerator
8,
sin
3 ) (I'
m sin
-f
Y(?sin0 1 -fmsin0 2 -fnsin0 3 ) Y (V sin O l -f m' sin 2 -f n' sin 3)
-f-
3)
2) 4-
2 -f
(nV
l -f-
w sin
-
tt'Z)
3)
sin
(0,
2
-
m'n) sin
(m/a'
A + (w
sin ^4
sin J?
I
m
n
r
m'
n'
sin
The denominator (I
cos O l
+ m cos -f-
= =
K'
r
ZJ
(I
2 -f
+ mm' + ^n
The tangent I
r
I
m) sin (0 2
0j)
C I.
';
n cos
x
3 ) (I
cos
l
+ m' cos
-f
3 ) (I'
(mn' -f m'w) cos u4
sinJB
sin
m m
2
sin
x
of the required angle
sinJ.
3)
Tm) sin C
+ (Zm'
m sin 2 -f w sin -f (mn' + m n) cos (0 b
+ mm' + *w'
n cos
is
sin 0! x
n'Z) sin J?
I'
2 4-
3)
+ (?m
=
=- 0.
is
+ n' sin 3 ) (V cos 0! + m' cos
m' sin
= 0,
between two
for the tangent of the angle
Using the formula cos 0j
+ n cos
2) -f
+ n' cos 3 m sin X +
)
x
n' sin
(w^ -f- n'l) cos (0 a -f (lm' + I'm) cos
(nV + w'Q cos is
2 -f
;J
3)
)
(0 2
-
X)
B
therefore
C
n n
In areal coordinates the angle between the lines px -f qy -f rz = 0, is the same as the angle between the lines whose p'x + q'y+r'# = trilinear viz.
= 0,
equations are
the angle whose tangent
/
op a
= 0,
+ bq'
is
p'
- 2 (qr' + g V) 6c cos A
}
.
TRILINEAR AND AREAL COORDINATES Conditions that two straight lines should be
parallel or
(i)
467 perpen-
(ii)
dicular.
Trilinears.
(a)
The
lines
straight
n'y
parallel if
sin
A
B
sin
sin
=
are
(7
=
I
m
n
V
w'
ri
0;
this is equivalent to
The
a
b
c
I
m
n
V
w'
n'
=
straight lines &re perpendicular
0.
if
2 IV - 2 (wn' -f m'ri) cos ^4 (b)
=
0.
Areals.
The
straight lines
px + qy + rz 1
= 0,
1
p'x + q'y + r'z
=
are parallel if
1
=0. r'
The
straight lines are perpendicular if
Sa2^/- 2 (#r' -f ^r)
&c cos
A = 0.
This result can also be written
-flrXf-*)}
== o,
2 ( 9 -r) (T'- /) cot 4
= 0.
or or
Note.
The reader can
obtain the results of
either of the transformations in
6.
Lr-f
The
(i)
my + nz
=
2,
(iii)
and
(iv)
by using
(A) or (B).
Let a straight line straight line at infinity. cut the sides of the triangle of reference, each
We have shown in 4 that 7), E, F. = = l/m. Now as the straight = CE/AE n/l AF/BF EDICT) m/n externally, at the points
;
;
line recedes in
any
direction each of the ratios
BD/CD, CE/AE,
tends to become equal to unity. Thus, when any straight line recedes in any direction, its equation tends to the limiting This is then the equation of the straight line form #-f ?/-f z 0.
AF/BF
=
at infinity
'.
The corresponding equation
= 0.
'
in trilinear coordinates is
TRILINEAR AND AREAL COORDINATES
468
Now
the
condition
=
the
that
should be
p is
px+qi/+rz
=
0,
parallel, is
111 this
lines
straight
= 0;
r
q
the same as the condition that the three straight liaes should be concurrent. 0, p'x -f q'y -f- r'z 0, px H- qy + rz
x -f y + z =
=
=
Thus, analytically, parallel straight lines
meet
at infinity.
=
Note i. The equation px -l- qy + rz -f X (x -f y + z) values of X a system of straight lines parallel to px
represents for different -f
qy + rz
= 0.
Note ii. The equation of a straight line through (x l9 ylf *,) parallel px + qy + rz=Q is (^x+ jy -f r^) (^ -f y, -f ^) = (# + y-f z)(pxl + qyi + rzi). Note iii. The point at infinity on the line px + gy-f rz = is
to
'
l
{g-r, r-p, ^~g}.
Note and in
The
iv.
trilinears
line coordinates of the line at infinity are in areals
(a:b:
(1:1:1)
c).
&
The
r
circular points at infinity, &, In Cartesian rectN=zQ is the coordinates + Lx+My general equation of angular 2 is the tangential equation of the a straight line and L2 -f (ii)
M
.
=
When the general trilinear points infinity ft, ii 0, is transformed to equation of a straight line, Za-fw/3-f My Cartesian rectangular coordinates we have seen that x
at
circular
.
=
The (I i.
Z
cos B l
Z
sin 0J -f
+ m cos
I
2
m cos + m -f ^ -f
2
2 -f 2
n cos
2 3)
{Z,
The corresponding equation
aV4-6
2
g
2
2
^c
-f (Z
2mw cos ^1
or
sin 6 l
(I
e.
separate equations of
cos O l -f
k
i$l
m cos 2 + 1''
+ me
-f
>w sin
2wZ cos J3
m, n}
=
2
ti ,
3.
2 -f
n sin
2 Zm cos
C
=
2 3)
=
0,
0,
0.
in areal coordinates
[ap, bq, cr}*
The
+ w cos w sin + 2 2
is
r 2 -26carcosJ[-2ca^cosjB~2abjp2cosC== 0,
or
i.
m sin
trilinear equation of the circular points at infinity is therefore
cos 6 l
e.
L= Jf =
3
cos
+ ne" =
ft, ft' 3)
3
= 0.
are
i (Z
sin 1
^ + m sin
2 -f
n sin
+ *-"* + we^"
0,
te""
0,
a#e-"*
3
3)
=
= 0,
or, in areal coordinates,
=
+ 1>qe-'*> + cre-"*
=
0.
0,
TRILINEAR AND AREAL COORDINATES The
trilinear coordinates of
&, Of are then 1
:":"),
(a**
fVi
ffl
/
or
469
(e-" 1
\
/
:-": e-" ~iA
\B
1
),
t ^
:e :-l), (e -1), (6 :e with corresponding results in areal coordinates. :
7.
The general equation
ordinates
of the second
degree in areal co-
is
= where
u,
t>,
w
replace
a,
ft,
c,
0,
since the latter have a special signifi-
cance in this chapter. The condition that this equation should represent a pair of straight lines can be found in a similar manner to that used for Cartesian coordinates, viz.
A=
u
h
h
v
9
f
g
,
f =
If the equation represents straight lines, tut?
0.
we have
+ vy 2 + tvz 2 + Zfyz + 2gzx + 2 hxy
=
The
(lx -f
my + nz)
(I'x -h
m'y -f
Comparing the form
coefficients in equation
(i),
we
(i)
get this condition in
Hence the equation ux2 -f vy2 -f wsP -h 2fyz -h 2gzx -h 2/tJcy and sents a pair of perpendicular straight lines if A = a2 w +
Note. In
2
b v
+ c2 w2fbccosA
2$rcacos JB
2/iabcos
C
= =
repre-
0.
trilinear coordinates, the equation U0(
2
-f
V& + wf -f 2//3y -f 2^ryO( -f 2&
represents a pair of perpendicular straight lines
if
j3
A
=
u 4- v + w - 2/cos A - 20 cos B - 2 A cos C = Illustrative
Ex.
n'e).
straight lines are perpendicular if
On Me Mree
and 0.
Examples.
of a triangle similar isosceles triangles are : tJte show that described triangle formed by tlmr vertices is copolar with the given triangle and when the base angles of the isosceles triangles are each
9,
Let
1.
sides
find the equation of the axis of perspective of the triangles.
ABC
t
coordinates.
the given triangle, be the triangle of reference and use trilinear
TRILILEAR AND AREAL COORDINATES
470
PN is perpendicular to BC,
If
PB =
since
{-PCsintf, PCsinC + 0,
The through
BQ
sin
equation
and
A
and
CR
PC, the coordinates of
or {-sintf
B+B
C+ 6 =
y sin
sin C-f B
:
P are
si
a
straight line the Symmetrically equations of
represents
AP.
P, i.e. the straight line
:
are
C+ 6'- 3 sin ^. + 6 = 0, a sin JT0 - sin iTf? = Hence ^LP, BQ, CR are concurrent at the point {cosec A + B cosec B + B cosec C + 0} and the triangles ABC, PQR are copolar. y sin
:
:
Again, since the coordinates of equation of PQ is
Q
0.
,
are {sinC-f#:
sin#: sin^ + tfj, the
y
ft
o,
and
meets the side AB, whose equation
this
is /
= 0,
where =
This point evidently 0(/ (sin
B
-f
6 sin
lies
on the straight
C + 6 -f sin 6 sin k 4- 6}
-f
line
^/ {sin
f y/{sin
0.
C -f B sin -4 -f 6 + sin
sin
B + 0}
A + Q ein JBT5 -f sin 6 sin CT}
== 0.
The symmetry of this result shows that the points of intersection of HP, and QR, BC also lie on this line it is therefore the axis of homology of
CA
;
the triangles.
Ex.
2,
the
//
the
diagonal points of a quadrangle are taken as the of reference, the coordinates of the vertices of
the triangle
vertices
of quadrangle are of the form (x l
:
yl
:
^),
(
x l :y x :^),
(x l
:
yl
:^),
(*i:yi:-*i). the quadrangle and A, B, C the diagonal points. the point (xl y * t ) then since PR passes through A, the point
Let
PQRS be
Let
P be
R is
:
(\xi y x s^ where X :
:
:
{
is
;
some constant.
TKILINEAR AND AREAL COORDINATES So the points Q, S are
Now ## v
=
1
and
{a?x
:
^y v ^J :
=
= = /
fi
1
orX =
inadmissible, for in that case P. Q, R,
The coordinates of the
Note.
Ex.
3.
y\
=x
\
:
MJ/i
1
i.e. X/i
;
similarly
R X
either
T/ws
of which are
:
1.
S Hence
{x l
passes through C, so that i>X
471
15
vertices
f/w simplest
/4
=
i>=
S would are
way of
1; the positive value
is
be identical, thus
then
(x l :y l :zl } t
(
x
l
:
ej,
:
t/j
denoting four points, no three
collinear.
If
the diagonals
of a quadrilateral are taken as the sides
of the triangle of reference, the equations of the sides of tlie quadrilateral are of the
Let
form px + qy + rz
PQKS be
px + gy + rz
= 0,
px
qy + re
= 0,
the quadrilateral and AB, EC, CA the diagonals. px -f qy + rz 0, then, since PQ and RQ intersect on
=
Let P<> be the line
whose equation
= Q,
is
z
0,
the equation of
Similarly the equations of
and px + ft^y -f r^
=
0.
RS
and
QR
is
of the form
PS* are of the
AB,
px + qy + vrz
form \px + gy-f
rz
0.
=
TRILINEAR AND AREAL COORDINATES
472
Now and
a?
PS,
= 0.
QR and BC are concurrent, Hence
px H- pqy + rz = 0, px + gy -f vrz =
ftv
= *= - 1. and vX = 1, so that X = ^ = = 1 or X /x makes the equations of the four sides the same and is inadmissible; therefore A = /A*=y*-l. The equations of the sides of the quadrilateral are therefore px'+qy + rz = 0, -px + qy + rz = 0, px qy + rz = 0, jw? -f qy rz = 0. Similarly
The
1
i>
i/
positive value
Note. three
Xft
=
i. e.
1.
j?%te
5
ffte
simplest
way of denoting four
no
li/ies
straight
of which are concurrent.
Examples XII
c.
1. Find the area of the triangle whose vertices are the feet of the perpendiculars from the vertices of the triangle of reference on the opposite sides. 2. Find the area of a triangle whose vertices are given by
a=0 wi/3-My 3.
Show
collinear, 4.
=0
=
I
OJ
tty-/a
=
i
OJ
y
=0
)
/- w/3 = Of*
that the orthocentre, centroid, and circum centre of a triangle are find the areal equation of the line.
and
Find the equation in
bisectors of the angles A,
trilinears of the line joining the points
B
meet the opposite
1 th
5.
Find the equation
the sides
AB AC t
to a straight line
where the
sides.
which cuts
off -
and the coordinates of the point where
i and -
tu
from
meets BC.
it
(Areals.) 6.
Find
lengths of the perpendiculars from A, B, C on the straight and circumcentre of the triangle of reference, and
th.e
line joining the in-centre
deduce the length of the perpendicular on it from the centroid. = 0, cy-f aOL 7. Find the area of the triangle whose sides are 6 + cy 8.
Show
the side 9.
that the straight line (x-y + z}coiB = (#4-y-s)cot of reference at right angles.
C
0,
bisects
BC of the triangle
Find the equations of the straight lines parallel to px + qy+rz = and it. Hence show that the length of the perpendicular from
distant d from
10.
Find the equation of the point of intersection of the lines (p l q l ,
,
11. Show that the line (Jf^-f wp a J^ + wgj, lt\ + mr^ passes through the intersection of the lines (p^ ql} r,), (/>,, g,, ra ) for all values ofl/m. 12. What are the conditions that the lines (p lt q lt i\\ (p^, ga ra ) should ,
,
be
perpendicular? Interpret the equations
(a) parallel, (b)
13.
:
(i) (iii)
g+
r0;
gtanB-f rtan(7= 0;
(ii)
(iv)
0-r=0; =
p-\ g-f r
0.
TKILINEAR AND AREAL COORDINATES
473
14. Prove that the equation <*pbq cr represents the in-centre and ex-centres of the triangle of reference. 15. Find the equation of the line joining the feet of the perpendiculars
B C to
from
bisecting
AB,
AC in the
point whose
{sin 16. 4?ind the
section of
x
Prove that this line intersects the line
the opposite sides.
y
C-B
trilinear coordinates are
sinC--4 *mA-B}. :
:
equation of the straight line joining the point of intery = z cos A to the vertex C.
= z cos B,
Find the equation of the straight
17.
parallel to the
parallel to
straight
line
px
-f
qy
4-
line
rz
through the centroid of ABC and of the line through A
= 0,
BC.
Use the formulae to find the distance between the circumcentre and
18.
orthocentre of the triangle of reference. 19. Show that a + /3 4- y = is perpendicular to the join of the in-centres and circumcentres. 20. Prove that
(a
+ (6 + d)p + cy
+ d)
~ 0,
are perpendicular for all values of d. 21. Find the equation of the line through (a lf ft,
toBC;
(i)
(iij
to
Ja
+ w,3 + My
=
yj perpendicular
0.
DPE
ABC is
is drawn parallel to a triangle; through any point P, to CA in and similarly FPG is drawn parallel cutting CA in in Q intersect are produced to and and parallel to BC.
22.
D
AB
E
BC
HPK
;
EH
DG
show that CPQ
:
a straight line. taken inside the triangle ABC\ AO,BO, CO meet the opposite sides in A\ J9', C'. If a line through A meets A'B*, C'A' in B", C" on B'C'. respectively, show that the intersection of the lines BB* y CC* lies 24. Find the equations of two straight lines through the mid-point of BC, 23.
A
point
is
is
equally inclined to BC, in aveal and trilinear coordinates. 25. is the orthocentre of a triangle ABC\ BO, CO meet AC and AB at and N\ meets BC at P. Prove that OP is perpendicular to the line
M
MN
joining A
to the mid-point of
BC.
Find the equation of the three lines joining the feet of the perpendiculars from the angular points ABC on the opposite sides and show that 26.
these lines
meet the corresponding
sides in three collinear points.
Find the equation of the lines through the centroid of ABC perpendicular to the lines joining the centroid to the vertices A, B, C, and show 27.
_
that they meet the corresponding sides of the triangle in three collinear points, 28. Prove that the lines
through
and through C
cos-4~l* + ycosjB =
parallel to
ft
B
parallel to j3cosC + ycosC-4
meet at the circumcentre.
29. Find the equation of the bisectors of the angles between the lines Oi
cos
A + cosB-y cos C = 0, sC= 0. ft
TEILINEAR AND AREAL COORDINATES
474
The mid-points
30. ft
=
y
0,
0,
of the diagonals of the quadrilateral formed
+ * fI + n in
+ my + nz
Ix {al,
=
0,
l'x
t
y
When Pis the
-
=
X
v
I*.
= p,
=
+ m'y -f n'z
Im, en} [al\ bm', en'} sin 6
=-
-
centroid,
= 2S
where p
is
.
*
0,
-
2 J(m'-n').
the distance between the
C=
2
2
prove that X fcccos^i-f ^rcacos J?-f afccos 34. Prove that the sine of the angle between the straight lines (p,
points
(a?,
y, z),
\
(P'*
is
oT,
26
y',
(a?',
l
r '}
0,
angle between the straight lines
32. If 6 is the
33. If
=
lie on 2a/(6/3 + cy-aa) = 0. of the lines isogonal with AP BP CP when Pis
31. Find the equations the point (<X 19 19 yv ) and show they are concurrent. find the coordinates of the point of intersection.
prove that
by a
I
|
j/
'),
p
q
r
!>'
?'
^
1
1
1
1.
g, r),
= find the coordinates 35. The sides of a quadrilateral are of the mid-points of its diagonals and prove that they are collinear. Find the equation of the line.
lxmynz
36. If (x, y, z),
(a?',
y',
z) are the points at xx cot A *= 0.
;
on two orthogonal
infinity
straight lines, prove that 2 37.
in the
The distance between the points form a&c2 (0( l - a a ) 2 cos A/4S*.
(o^
,
l
,
yj, (a 2
,
^2
>
Va)
can ^ e written
The general equation of the second degree. In the precise meanings preceding sections we have assigned to #, #, 8.
;
in the following
work the
results, except
where they involve metrical
quantities such as lengths and angles, hold equally well if we attach to #, y, z no more precise meaning than that x 0, 0, y the form & lines which three triangle of represent straight reference. The coordinates of a point are then connected with its
=
=
=
coordinates in a Cartesian system by three linear equations of the w3 Z Z forms + 2^+ W 2> * J 3 + 3 + 1 y 2 1
= X w r+i
= X w r+
= ^ W
;
but we need not attach any special values to the constants Z 1 tw 1 >^, &c., as we do in areals. The vertices of the triangle of reference are 0, given as the points of intersection of the straight lines x ,
,
=
y
=
0,
*
=
taken in pairs,
i.
e.
(1:0:
0),
(0:1:
0),
(0:0:1); the
actual values of the coordinates evidently cannot be found unless we know the identical relation which connects the three coordinates x,
y, 8) corresponding to the relations
aa-f
6/3 + cy =
An
2S
x+y + z~l
in areals and
in trilinears.
equation of the second degree in
a?,
y,
* transforms into an
TRILINEAR AND AEEAL COOEDINATES equation of the second degree in Cartesian coordinates represents a conic.
The general equation
it
;
475 therefore
of the second degree is
= ux* + vy* + w'2* + 2fy* + 2ggx + 2hxy = 0. f(x, y, g) We shall
use a notation similar to that employed for the general
Cartesian equation of the second degree, thus
N
uvw + 2fgh - uf* - vg* -
A=
H = fg-wh,
^gl-uf, G=fh-vg,
Y~
,
To find
(1)
(\> Hit
hx + vy+fz,
the equation of the tangent to f(x>
t/,
s)
=
at tlw point
*i)
Let the equation of the tangent be
*-*i I
where p
the distance between
is
Now
_ y^Ji\ m
x
substitute
= lp + x
f-fi n
_.
y, z)
(a?,
_
,
**
and (x^ y,
y = inp + y l9
lt
z
=
g^.
n + zl
the
in
>
i
equation of the conic, and we obtain an equation in ^> giving the distances of the points of intersection of the tangent and the conic
from
is
equation
Now
evidently both these values of /> are zero. The fi f( I, m, n) + 2 { ZXj + l\ + +/(^ ft ^) 0. r since is on the conic, so that one (x l3 ft, ^) ft, t)
y ly ^)
(x lf
;
,
The second value
is zero.
/>
lX l if (#,
Hence,
T/,
^)
is
^X at
-f
x
ft Tj
fa y l ,
,
-f
^)
(2)
To find
/>
,
being zero
,
=
we have
^ml\^nZ = 0. l
+ (if--fl
1)
YJ
^ Z =f(x iJ l
+ ^-^Z!
ft, ^4)
=
0,
-
0.
the equation of the
is
ay^+y 1^ + The method
of
]
any point on the tangent, we have
(s-.r J )X 1
tangent
M^
=
f(x l9
value of
Since
m
;
.
of Chap. VI,
4
(v)
^=
0.
(A)
can also be used.
the tangential equation of f(x, y, g)
= 0.
identifying the equation px + qy + rs=zQ with the equation (A) of the tangent, we can show in exactly the same way as in Chapter X, if 11, that the straight line (p, 4, r) touches the conic f(x, y, e)
By
=
JF(P, ft
This
is
= r)
CS*+
F^+ Wr + 2F(?r-h2Grp-h2r^ = 0. 2
(B)
therefore the tangential equation of the conic f(x, y, s) = 0.
TRILJNEAR AND AREAL COORDINATES
476
Cor. If the conic is a parabola it touches the line at infinity thus the coordinates of the line at infinity satisfy the tangential ;
These are in areals (1:1:1) and in
equation.
Hence, in areal coordinates, f(x,
we
K=
shall refer to this as
y>
z) =
is
trilinears
(ail:
a parabola
c).
if
0.
=
is a parabola if Also, in trilinear coordinates, /(a, /3, y) 2 2 2 c a 17+ 6 2bcF+ 2caG + 2al)H 0.
is
=
W+
7+
The point equation corresponding to a given tangential equation found in the same way by calculating the minors of its dis-
criminant.
For the tangential equation (B) we shall use the notation
R= To find
(3)
to
F(p
q, r)
9
that (x 1
:
is
=
be the equation of the point of contact, so
y l z^ are the point coordinates :
The equation This
of tJie point of contact of a tangent (p l9 q lf rj
the equation
= 0.
Let pXi+qyi + rZi
Gp + Fq + Wr.
of the point of contact.
in point coordinates of the tangent (p ly
q^ r^
is
therefore identical with
Now
Similarly,
From
HX^ VY
l
equation
(i)
then
Hence the equation
n
i>
+FZ = &y l
we
l9
&X + FY +WZ = l
l
l
get
of the
contact of the
point of
tangent
r\) is 0.
(4)
The equations of
y> *) =
We
>
awd of
the point (x l9
the polar
of
the pole
of the line
(C)
y l9 tj with
(p ly q l9
r,)
respect to
with respect
can show in the same way as in Chapter VI, 8, that Chapter X,
4,
to
and
TRILINEAR AND AREAL COORDINATES The equation
477
of the chord of contact of tangents from a point the conic f(x, y, z) is xX l + yYl + zZl 0. the of The of of intersection point equation (b) tangents at the line cuts conic where the the is (p l9 q l9 rj points F(p, q, r) (a)
=
t
(#i> y\ 9 *i)
=
=
pPi + qQi + rRi = 0, (cj
The equation
conic
y, z) =
(x,
and we deduce similarly that of the polar of (x l is
sXi + 03^ + (d)
The equation
the conic
F(p
The
Cor.
q
9
r)
9
,
y l9 e^ with respect
^=
to the
0.
(D)
of the pole of the line (p l9 q l9 rj with respect to
=
is
= 0.
pP + qQ + rEi l
centre of the conic
Y
is
the pole of the line at infinity
;
hence, (i)
If (#
,
coordinates,
x+y+z
y
,
its
=
the centre of the conic f(x y, z) in areal is the line at infinity polar xX + yY + eZ
#
)
is
9
= 0. We have
= X =Y =Z
therefore
Q
Q
.
=
The
areal coordinates of the centre of the conic /(#, y, z) Z. therefore given by the equations
are
X = F=
The
line coordinates (areal) of the line at infinity are (1:1:1); is therefore the equation of the centre of the conic JP(jp,
=
It follows therefore that the areal coordinates of the centre of the
conic JF(p,
(U+H+G)
and therefore of the conic /(#, :(H+ V+F):(G + F+ W).
g, r)
;= 0,
y, #)
= 0,
are
Corresponding results can be found in trilinear coordinates by = Ofor the equation of the line at infinity. using aa-f&/3-fcy
Note.
(6) Tfie equation
point (x l9
y l9 tj
of the pair of tangents which can le drawn from the
to the conic
f(x,
y
}
e)
= 0.
This equation caa be found as in Chapter VI, /(*i,
Jfi,
*i)
/(*, y, *)
p.
287, viz.
= (sXi + yYi + RZtf,
and in a similar way we get the equation of the points of intersection of the line (p l9 q l9 r^ and the conic F(p, q, r) = (v. p. 404), (E)
.
(6) (a) if (xlt
The Asymptotes. The Asymptotes are the y l9
#}) is
tangents to a conic from the centre, their equation is 9
y lt tj
=
its
centre
;
TRILINEAR AND AREAL COORDINATES
478
Since (x l y l9 zj ,
The equation if
is
we have
the centre,
X = Y = Z = X, l
the coordinates are areal, so that %i+yi +
w#j
Therefore
+ % t + jf^
g
k
v
f J w
X
/
\ ">
u
h
h fl
X
also
becomes
of the asymptotes therefore
Now
l
l
\
=
1-
= 0,
-0,
1111
or
&
i
/
i
f w
I
v
ft
g
1110
or
AJC=A.
or
The equation
of the asymptotes is therefore (P)
(b) Another form of this equation can be found as follows n y\> *i) is a point on an asymptote, the equation of its polar is :
The
point at infinity on this polar
and this
lies
on the curve
The equation then
;
hence
f{Y-Z, Z-X, X-Y},
Further, since written
X Y=
therefore
is
f\Y
of the locus of (x l}
l
y it
if
#j)
Z Z l
,
i.e.
l
X X l
,
l
Y = 0. l}
of the asymptotes,
is
or
(F
Z+Z
X), this equation
may
be
(w+ u-2g) (Y-Z)* + 2 (w-f-g + h) (Y-Z)(Z-X) from which form the separate equations of the asymptotes may he found,
TRILINEAR AND AREAL COORDINATES
479
~
If the
tangential equation of the conic, JP(p, #, r) 0, is since the touch the and curve then, given, asymptotes pass through the centre, their coordinates can be found by finding the common (c)
solutions of the equations of the centre and the curve, viz. !
^ PI*
ri
Pzi #2r r2 are th e equations of the asymptotes are
Cor.
i
;
The
i.
straight
lines f(x,
two
t/,
z)
= 0,
sets of values found, the areal
2
+X
(a?
-M/-f z)
=
are
perpendi-
cular if
s <7 i.
waa -f r62
e.
4-
w
A
2 pea cos
B-2
fcafc
cos
C=
0,
0,
for the coefficient of X
= = The
a 2 + Z>* -f c2
-2&ccoB^-2cacosJ9-2a&cos C
0.
condition, therefore, that the conic /(x,
should be a rectangular hyperbola is D == wa* 4- t?6f -f we2 cos ^4 - 2
2/6c
Note. -f t?-f
(in areal coordinates)
^a cos B - 2 fcafc cos C 0. = is a rectangular hyperbola
if
0.
We
have shown (Chapter XI, 2) that Conies are similar provided that their asymptotes contain equal
ii.
(a)
=
)9, -y)
^-2/cos-4~2pcos-B - 2fccosC=
Cor.
angles
In trilinear coordinates /(a,
y, *)
;
(b)
Conies are similar
and
similarly situated
provided that
their
asymptotes are parallel.
These conditions can be obtained from the equations of the asymptotes
;
in practice it is not necessary to find these equations, for by eliminating z from the equations f(x, y, z) Owe get the equation of a 0, #-fy-f-2
=
=
pair of straight lines through the vertex 9. 0*07
The
=
is
The equation of the circle, whose centre is obviously whose radius is
1>
is
/>,
equation homogeneous by means of the relation
this
- (a2 (i/# which
to the asymptotes.
Circle.
Vo9 #o) an(*
Making
C parallel
we have -i-
2 ey Q ) + 6 (#z
evidently
-f
^
+ (a2 yo^ oHhe form
)
+ c2 (xy Q + x Q y)}
(x -f y -f z)
TRILINEAR AND AREAL COORDINATES
480 Note.
In trilinear coordinates this equation a fry + byOi + cOCft -f (7<X w# + ny) (a
is -f
-I-
6/3
4-
Tbjprove, conversely, that the equation
a*yz + l*zx + c*xy + (lx + 'my + nz)(x + y+z)
always represents a
The equation
and
circle,
to interpret
I,
=
n.
If the conic is a
that of a conic.
is
m,
0.
cy)
parabol^ it touches the line at infinity, hence the conic d*yz + lP zx + &
=
>
;
6c =
impossible. The conic let this
therefore a central conic,
centre be (X Q
Now,
if
zQ )
(XQ,
yQ
/fa
V, *)
,
is
f(x, y, z)
,
y
,
#
=
is its centre,
i,
e.
it
has a
finite centre
the general equation of a conic and
is
we have
+ 2 {(x-x )X + (y-y Y + (*-*JZ<>} +f(xQ )
since
Z =T =Z
Apply
;
).
and
,
a?
this to the case
and we have
/fa
y, *)
=
2
(y-yo) (^-
/fay, ^)= where
Hence
,
Hence OP 2
(ii)
If
P
the point (XQ y Q9 * ) and any point if P lies on the conic, then f(x, y, z] (i)
is
P
is
constant,
does not
lie
i.
e.
on the
the conic
circle,
is
and p
(x, y, *).
= 0,
a circle
is
;
and we have its
the radius,
radius
is
we have
JT
= pa -OP 2 = -PT
/fa
y, *)
the
circle.
2 ,
where
PT
is
the tangent from
P
to
TEILINEAR AND AREAL COORDINATES
481
A
In particular, the square on the tangent from to the circle is I and 0. /(I, 0, 0) or similarly for the tangents from
S
;
are the lengths of the tangents If, then, to a circle, its equation is t lf t 2 , t a
2
(t 1
from A, B, C
=
x + t 2 2 y + t3 2 z)(x + y + z )-(a2 yz + b 2 zx + c'2 xy)
0,
and the left-hand side of the equation represents the square of the tangent from any point (x, y, 2) to the circle. To find f(x, y,
a
the condition that the general equation
z) = ux* +
vij*
+ wz* + 2fyz + 2gzx + 2 lixy
=
of
the second degree,
0,
should represent
circle.
The
general equation can be written
= 0; the necessary and sufficient conditions that this equation should represent a circle are therefore
These conditions can be written
In be a
trilinear coordinates the conditions that /(Of, circle are /(O,
c,
-?>)
=/(-c,
0, a)
To find the tangential equation of the and t(;/ia^e radius is
/3,
=/(6, -a,
circle
y)
=
should
0).
whose centre
is
(#
,
y
>
*o)
/>.
If
(^?,
tangent
{?,
is
r) is any tangent to the circle, the hence we have px + qy + rs =
areal equation of this
;
or
^
homogeneous by means of the
this
Making \apj bq, cr]
=
we
2S,
get
ISM^o + ff^o+^o) = 2
The general a2 /* 2 + 6 2 2 2 its
-h c
centre is
Note.
We
2
(I
:
m
2 ftc^r cos A
:
n\ and
(A
2 ca/^9 cos J?
its
have shown in
the circular points at infinity, 1207
2
P
tangential equation of a circle r2
identical relation
^ is
2
cr}
.
therefore
2 aZjp j cos
+ m + n). = bq> cr}
C
radius 2 S/(Z 6 that [ap,
Q and
H
U'. li
2
is
the equation of
The tangential equation of the
TRILINEAK AND AREAL COORDINATES
482
=
w where it 0, v are the equathe equation of the centre the circle, as we have previously proved, touches OG, 0Q', the line at infinity QG' being the chord of contact. circle
w=
and
The
;
circles of the triangle of reference.
Here
equation of the circumcircle this
= 0,
<x
2
a yz + b
is
2
t
= 0, so that 4he 3 = 0. X{/ (In trtfinears
= 0,
2
#x+c
2
= 0.)
becomes a/3y-f ftya-f-ca/^
The
,
is
The circumcircle.
I.
1
form uo
therefore, in the
is,
tions of O, O',
<
tangential equation of the circumcircle is therefore
which can be written a Vp
I
Vq
Vr =
c
0.
The nine-point
Cor.
circle circumscribes the triangle A'B'C', where the are A', B', mid-points of the sides. Iff/, q, r are the perpendiculars from A', B', C' on any tangent to the nine-point circle, and p, q, > the perpendicular from A, B, C to the same tangent, we have p+qs=2r', '
g-f
;= 2p', r+p = Now
CCS2C'.
Also a *= 2 a', 5 = 26', 2q' with due regard to sign. the nine-point circle circumscribes the triangle
since
A'B'C' we have a'^/p' + b'\/q' +
c'^/r'
= 0,
hence
= the tangential equation of the nine-point circle.
is
The inscribed and escribed
II.
For the inscribed circle ^
(a)
equation
z
The is
(op
5
^^
^
a,
3
=s
c; its
x + (s-ltYy^(scY2](x^y^^)-a^yz-b^zx-c 2 xy
tangential equation, since r
-f
=
is
a)
{(s
circles.
&g
2
-f cr)
=
= ar/2/S,
br/2S, #
2/ u
=
0.
= cr/2/S,
2 \
op, bq, cr]
.
This reduces to (
or
jr cot J
(b) t
l
0,
A -f rf? cot J B +pq cot | C =
0.
6)rp +
The reader should deduce the
Note.
=
=
a)jr + (5
c)jpj
(
trilinear equation.
BC
For the escribed circle touching t
5, 2
{S
The so that
%
=s
e,
tA
= 56
;
its
equation
externally
x + (s-c)*y + (s-b)*z}(x+y + z)-a2 yt~b 2 0x-c2 xy
areal coordinates of its centre are its
tangential equation (
this reduces to
gr cot i-4
{
a^/25, brJSS,
is
2 qp -f bq + cr) =r
rp tan
J
%
{ op, (
we have
is
^>
cr j
2 ;
Bpq tan 4 C =
0.
= c
0,
TRILINEAR AND AREAL COORDINATES III.
The nine-point
circle.
= =
A
483
Here
or
ic
& cos
.
Scot A,
.,
The equation of the nine-point is
or
circle
therefore
S [xcoiA + ycotU-McotCj (x+ y + z) a*yz Wzx c*xy 2 2 a 2 a; 4 (a2 t/^ -h 6 2 ## 4- c 2 a?w) (x + V 4- ^) 2 (6 -f c )
We have already shown that its tangential The reader can
equation
= =
0,
0.
is
deduce this from the general tangential
also
equation.
IV.
The polar
polar circle,
if
is
p
P
The orthocentre
circle.
and
the centre of the
is
A
the radius
we have
-^2 = AP.PL = BP. PJt= CP.PN. N/
Hence
= AP. AL = 2Ii cos A = 26 cot A.
f
C
6 sin
.
r
The equation
of the circle
is
then 2
2S(xcotA+ycotB + zcotC)(x + y which reduces
+y2 cot # -h^ 2 cot
tangential equation of this circle
p* tan
The
=
radical axis of
A+g
two
2
=
0.
=
0.
is
tan B + r* tan
C
If the equations of
circles.
two
are found in the forms (te
+ my + ne)(x+y + 8) /
/
a*y&
(ro;4-m y+n ^)(a?4-y-h^)--a it is
0,
to
a2 cot A
The
-d*yz--b 2X--c*xy
evident that points
common
2
^--&
to the /
= 0, = ^ 0,
b**x--c*xy 2
two
8r^---c
(
2
circles also lie
n )*J(* + y + *)
u h
2
=
0,
on
circles
TRILINEAR AND AEEAL COORDINATES
484 i.
e.
on one
of the lines
(J-Os+tw-wOy + fa-nO? = The
x+y + * =
o,
0.
the radical axis, and the second is the line at infinity, which is the join of the circular points at infinity. have previously shown that all circles pass through il, 1' we first line is
We
;
proceed to verify this.
t
The
Since the equation of any
circular points at infinity. circle can be expressed in the form (lx
4 my + nz]
(x
+ y + *)
+ Vl zx + c*ocy)
(u?yz
=
0,
meets the straight line. at infinity in two points given by the = 0, #+# + = 0, which are evidently equations tfyz 4- Wzx + e*xy it
the same points for all circles these points (Chapter V, called the circular points at infinity '.
1(>)
;
we
*
We
have from these equations 2 Z>
2
or i.
+ I2 -
= 0, & # + aV + 2 ab cos Cxy = 0, ic ic = 0. (bx + aye~ (lx + aye
*2 +
e.
ay + (a
2
2
c'
)
xy
2
)
)
Thus
ae
x\y:z
either
=
or
But
ae
c
=
a cos
c
ae~
C
b
:
l6
b
:
ae
b
:
:
,
b~-ae~
=
C
tC iC .
sin -4) hence the coordinates of the circular points at infinity are b
b
c
(ae
Note. line
The
:
-6
:
ia sin
ce~
u
c (cos -4
t(J
(ac"
),
:
-b
t
= ce"
fl ,
A :
ce'
).
tangential equation of Q, G' can be easily found thus : if the = passes through a circular point at infinity, this point
px 4- gy -f rz
must be the point at infinity on the line, i. e. (q-r: r-p :p-~q). This point therefore lies on the circle cPyz+Wzx + tfxy = 0, so that
We have
shown that
this equation is the
{ap, bq, cr}*
same as
-
0.
Illustrative Examples.
Example
i.
The equations
The
of these circles are
2
The
nine-point circle touches the inscribed circle.
- a)*x
2
{(fc'-f c
2
.
-a 9 ) -
i.e.
or
=
2
a (x -f y -f z) equation of their radical axis is.therefore (s
x/(b
- c) 4
2
c-a) } a? 2 (a-&)(c-a)#
= 0,
- a) -f */(a - ^)
= 0.
y/(c
(6-f
0,
TBILINEAR AND AREAL COORDINATES The coordinates of the
485
radical axis are therefore
{!/(&-) !/(-) !/(-&)}. The tangential equation of the inscribed circle is (s-a)c[r+(8-'b)rp + (s c)pq =* 0, :
or
(s
Now
2
(6
a)
c) (5
=
- a)/p -f
:
(s
- ft)/g -f (s - c)/r
inscribed circles touches the inscribed
nin-oint Tt
same
circle at the
=
0.
and
so that the radical axis of the nine-point
0,
circle
therefore touches the
it
;
point.
can be shown similarly that the nine-point
each of the
circle touches
escribed circles.
Example
The
ii.
ABC
of the triangle
CC' as diameters have l(m
common
the
The coordinates of A' are
m
I
.
(
\, "' 0-7
\
as diameter
m
:
2(w
I
f
-^ m n
f
- AA -n
centre of the circle on
m)zcot (7=0.
f
2(w~ n)
A
and of
1>
{1, 0, 0})
;
.'.
the
}
is
)
r
5-7
,
H)
BB',
radical axis
~^n
JO,
the sides
AA\
prove that the circles on
:
l)ycotB + n(l
n)xcotA + m(n
Q meets
Ix+my+nz
straight line
in A', B', C'
c
or
{m l
-n - n :
^ :
m}. J
)
Let the equation of the circle be cfyz + since
A
it
The
passes through centre is given by
i 2 2-f-c 3 y
(/uy-f vz)
=
=
(A# + py + vz) we have A = 0.
Wzx+tfxy (1, 0,
2
c .r-f
0)
a2 ^
(/iy-f-
vz)
(a?-f
- ^ (x -f y -f
=
y-f z);
z)
a 2 y 4- 6 2 a; - (
.e.
Substitute in these the 9 &3 m _ c n =
2 c'
(m
known
H)4-a
2 ^x (m -
Hence
The equation of the
x y
values of
:
m 2fi(wi -n) = m (a2 -f c - & 2
)
circle
i ,1.2 ,2 a 2 171/2? -f 6 2 2ra7 -f c*x\j J
=
:
z,
2 )
on AA' as diameter
--
mco ^^ oci/ 2 6^
Vm-n* (
and we get
= IP (m n) -a n = 2 ac cos Bin, 2
a
f/ 7
is
-m-n/
ncotC \, 2 )
,
-f (a? v
y J +
x
,
-);
is
radical axis of these circles is therefore
-
/
i.e.
n).
then
Similarly, the equation of the circle on CC' as diameter
The
2 v(m
/
cot
(m
A p.a?+
-
n)
m cot B ---
./-./'I
.i/-ncotC(
xcotA + m (n -l)ycoiB+ n
The symmetry of the result shows that circles on BB' and CO' as diameters.
it is
(I
- m) zcot C
=
0.
also the radical axis of the
TRILINEAR AND AREAL COORDINATES
486
Examples XII d. Find the equation of the
1.
circle
which passes through two angular
points and the in-centre of the triangle of reference. = is a circle. 2. Prove that 2 (m + n) a* + 2 2 1 fry cos
A
Prove that the circumcircle, nine-point
3.
circle,
and
polar-circle of a
triangle are coaxal ; show also that the circle circumscribing the triangle formed by the tangents to the circumcircle at the vertices of the triangle belongs to the same coaxal system.
Show that
4.
the equations
=
y( y cos C-acos^L-/3cos.B), j3y = a(acosu4~/3co8J?--ycosC) represent circles, and find their radical axis. 5. Find the lengths of the tangents from the ex-centres of a triangle to a/3
the circumcircle.
Find the equation of the common chord of two fundamental triangle as diameters.
6.
two
circles described
on
sides of the
7. Through the vertices A, B, C of a triangle straight lines are drawn parallel to the opposite sides and meeting the circumcircle in the points The straight line joining A' to the centroid of the B', C' respectively. in A", and IT , C" are similarly determined. Show that triangle meets
A
1
',
BC
AA" BV, CC" 9
8.
Show
are concurrent.
the equation of the
that
ex-centres of the triangle of reference
circle
which passes through the
is
a*yz + b*zx+ c*xy + (bcx + cay + abz) (x + y + z)
Find
0.
also its trilinear equation.
9. Prove that the equation of the nine-point circle can be expressed in the form a?/(bft+cy-aa) + l*/(cy + a(X-lp) + c*/(aO( + bft-cy) = 0. What is the corresponding equation in areals ?
10.
Show
touches
that the equation of the circle which passes through A and middle point is 4 (a*yz + b*zx + c*xy) = a 2 (y + z} (x + y + z).
BC at its
Find the equation of the AB, AC.
line joining the points other
than
A
where
this
circle cuts
Prove that the equation of the circle on the line joining the midpoints of AB, AC as diameter is 11.
12. If (a',
',
y)
is
sponding Sitnson line 13.
Show by using
a point on the circumcircle, verify that the correis
2
<*
C08
X
2<X sin A.
the tangential equations of the escribed circles that
their external centres of similitude lie on the line x/a + y/b + z/c 14. circle passes through B, C and cuts CA in the ratio I
A
ratio does it divide 15.
at
:
0. ;
what
in
BA ?
Find the equation of the
D so that ZBDC
= m
<X
circle
Hence
through l?and
C which
cuts
CA
again
find the locus of tie radical centre of three
which pass through B, C\ C, A; A, B respectively, and whose ments within the triangle contain equal angles.
circles
seg-
TRILINEAE AND AREAL COORDINATES Find the tangential equations of the
16.
487
whose centres are the
circles
vertices of the triangle of reference and which touch the sides opposite to their centres. Find the equation of their axis of similitude.
BC, CA, AB cutting CA, AB, BC respecshow that their radical centre is the same for all
Circles are described on
17.
tively in the ratio I/A
values of
;
A.
ABC
The
18.
AB
inscribed circle of a triangle touches the sides BC, CA, D, E, F; the escribed circles opposite to the angles A, B, and C touch these sides at l } P\ ; D^ 7?2 l<\ 3 8 F^, respectively. Prove at tlo points
D E ,
that
AD^ BE CF 2
19.
Z
,
,
D # ,
;
,
,
are concurrent; also the sets of lines
if the circles a*yz + lPzx+ c*xy = and 4 (t^x t*y 4 tfz) (x 4 y -f z) - a*t,z - &V# - cVy
AD^ BE^
t
CP\\
Prove that
=
cut orthogonally, then 2 8 8. tj sin 2 A 4 tf sin 2 B 4 f 3 sin 2 C [The tangent from the circumcentre to the second circle is of length /?.] 20. The sides BC, CA, of the triangle of reference are divided at
AB
A
E, JP*o that BD = 2DC, CE=2EA, AF*=2FB. Show that the 2 2 446 2 ). square of the tangent from A to the circle DEFis $, (c -2 21.
Show that the
radius of the circle fi3y-f
fe-jOK
4
cOf/3
=
A-a
(a34
?>^-f cy)
is
Conies circumscribing the triangle of reference. If the tcz* + 2fys + 2gzx + 2hry = passes through the
10.
ux* + *y* +
conic
vertices
then
(1, 0, 0),
u = 0,
of the triangle of reference,
(0, 0, 1)
(0, 1, 0),
v = 0,
w = 0.
The general equation of a conic circumscribing the triangle of reference is therefore
fyz+gzx+hxy
We ff
have in this case
?7=-/
=. 0.
F=
2 ,
(i) 2
-r/
,
TF=-fc 8
,
JP=flf/,
=//*, //
=^/ so that the tangential equation of the conic /V + 8 8 + V-2^ar--2ft/fy-2/flp7 = 0.
We may
write this
?
flf
fpgqhr =
i.e.
v^+ v^2
or
is (ii)
__ i
2\/ghqr,
v^Ar
=
0.
Note. The condition that the conic fyz 4 yzx 4 hxy = should be a parabola is that the line at infinity should touch it, i.e., in areal coordinates, that (1, 1, 1) should touch the conic (ii). Hence the conic is a parabola if + g 1 4 ft 2 - 2gli - 2hf-2fg 0.
The conic
ffty
+ g-)OL + kOLp
V4 V4 b
2
=
c h*
f
is
-2
a parabola if -2
%fc - 2 c
ft/
abfy
=
0.
TRILINEAR AND AREAL COORDINATES
488
To find #2> #2
#2)
9
equation of the chord joining the points (x l9
the
on
fyz+gzx + hxy^
th e con i
The equation
of the straight line joining the points (x l9
fy l e + gz xl + tej yx
But
v
l
/ya *2 + 0*2 a?2 + Aor2 y2
and hence f:g:h
= x^ (y^-ytf^ y^ :
L
(w*-e
no one of the coordinates of the equation of the chord can be written
The equation
^j,
y^
^),
= 0, = 0,
provided, then, that
Cor.
ylt
0.
of the tangent at (x lt
:
z
the given points is zero,
yt gj ,
therefore
is
o 2 2/i
equation of the tangent at
Since w a
= gx
l
= 0,
Now,
(/'#i
}
)
to the conic.
X =%
we have
1
1
-f^
since fyi^i
+ g^x 1
l
-{-hx l y l 1
,
V+
2
= 0,
we
v+
ii ^
+
^/i
which agrees with the last corollary must be zero when this form is used. The equation of
The equation
1
the point
^gq
=
2
r
l
,
is
+ /7*i) + !/ (^i + A) + ^ (^i +/>/i) =
i
the conic
ylJ z
so that the equation of the tangent
i-fy }J
a
^=0
v=0,
(sr l9
0.
can write this
=o
=,
^i ;
evidently no one
of contact of
of the coordinates
the tangent
(p lf q lJ r^
to
0.
of the conic written in full is
f*p* + g*tf + h*9*-2ghqr-2Wp-2fgpq
Thus Pl ~f(fpig
(**"i
=
0.
values for
-ffi -Mi)
Q 19
= 0.
B
lt
TRIL1NEAR AND AREAL COORDINATES since (p l9 q ly rj is a tangent to the conic,
Now,
/V +
P
a
a ffi
+*^i
hence
a
-Mjiri-2^
C#i-0i-**i)
fPl - Wl - fa-!
or
The equation
1 jp 1
489
we have
-2/flp 1?l
=
0,
a
=
of the point of contact can then be written, provided p ly q^ r l is zero,
no one of the coordinates
that
+ I Jwi + ~ Vft^ 0i
5 Sfa
PI
where Vfpu Vgq\,
0,
have the signs which make
Vh^
equation of the polar of (x l1 y v with respect to the conic.
These are x(hyl + gg
=
i
^)
of the pole of (p^ q lJ
af?fZ
+ y(fu +f0 ) + z(fix +fy =
and
t
l
)
l
l)
l
l
= Note.
The method we have always used
to
0.
show that the equation of
the polar of a point, which is not on a conic, is of the same form as the equation of the tangent at a point on the conic depends on the fact that the equation of the tangent is unaltered when the current coordinates and those of the point are interchanged.
It is
important to note then that
the equation of the tangent at (x l9 y lt z^ in the form
cannot be the equation of the polar of (a^, y t
Cor.
The its
To find
the centre
centre of the conic
equation
is
,
*
If
To find
j
+
=
-* zl
0,
of
the conic.
is
the pole of the line at infinity
(1,
1,
1)
;
=
0.
of a conic ivhose centre
is the
point
an( wAfcA circumscribes the triangle of reference.
^o)>
Therefore
the equation
l
Hence y + *o-*o
is
yl
areal coordinates of the centre are then
fyz + gzx + hxy
conic
*-
therefore
Example. QJ 2/o
+
zi).
f(f-g-h)p + g(ff-h-f)Z + h(h-f-
-
xl
/
-
(y
is
-f
^
=
is
=
the equation of the conic,
(^-^)
8
-/2 =
proportional to
- o? + )
-
(^
we have
() an ^ the equation
+ ^o - 2/o) +
TRILINEAK AND AKEAL COORDINATES
490
The asymptotes. Let the px+ qy+ rz = 0, p'x + q'y+ r'z = Comparing
;
= 9sf = 1* = + 2A = rp' +
/+2\ = gr' + g'r, (i)
or,
/
is
Q
= pq'+p'q.
h + 2\
one asymptote, the
is
.*
otner
/ = q^ + j'r- 2 A
2
proportional to p(q
It follows then that if
r)
.
=
px+qy+r&
is
an asymptote of a conic
circumscribing the triangle of reference, the equation of the conic q*jry
Example at
A,
JB,
A'B'C'
is
Q.
We have
(ii)
A,
r'j>,
px + qy+rz
p + y/q+z/r =
asymptotes be
we have
coefficients
if
the
we have
then
PP'
Hence
of
equations
A
i.
C form
=
is
0.
Mangle ABC, and the tangents Show that .the Mangles ABC, centre and axis ofhomology.
conic circumscribes a
A'B'C'.
the triangle
are coaxal, and find their
Let the conic be fyz + gzx + hxy
=
0;
its
equation can be written
fy* + x(g* + hy]
0.
This represents a conic circumscribing the quadrilateral whose sides are x = 0, gz + Jty = but since y =0, z = 0, gz+ hy = are y = 0, z = = + is the concurrent, gz hy tangent to the conic at the point of intersection of y = 0, z = 0, i. e. A. ;
The
;
sides of the triangle A'B'C' are therefore
= 0, a?//+ y/g = 0. The straight line (z/h -ha?//)- (#//+ y/g) = 0, e. z/h - y/g = A A'. Similarly, BB\ CO are x/f -z/h 0, y/g-x/f^O. y/g + z/h
=
z/h + x/f
0,
i.
##', C(7 intersect at the point (f:g:h). lines y/^r + z/h
The
*=*
and x
intersections of
=
;
is
therefore
Hence, ^4^',
Again, JJ'C' and JBC are the
=
0.
which
is
these intersect on the line x/f+ y/g -f /A
CA, C'A'
AB, A'B'
\
also lie
on
this line,
therefore the axis ofhomology.
Example ii.
The opposite
meet in three collinear points.
sides
of any hexagon inscribed in a conic
(Pascal's Theorem.)
AQCPBE be
the hexagon and take -4PCfor the triangle of reference, Let the coordinates of P, Q, E be (xl9 y,, ^)> fo, 2/a *t), (#s> ^s. *8 )The equation of AQ is y/y^ = z/z^ and the equation of is ar/a^ z/zl
Let
BP
these lines intersect at the point (a?,/^
:
y a /* 2
:
1).
;
TRILINEAR AND AREAL COORDINATES pair of
Similarly, the other r
\/V\
:
1
:
(1
^3/7/3),
:
opposite
intersect
sides
at
.
491 the
points
ya /;ra zs /#s). :
^^/
A
Rr
B These points are collinear
if
if
i.e.
i/y
But the equation of the conic on which the hence ffx + g/y + fc/s =
vertices lie
is
of the form
;
//*i
=
+ ^/yi + V^i
Eliminating/,
g,
//*
0,
g/y* +
-i-
V^ =
//* ^ ^/ys + V*s
=
-
h we obtain the required condition.
This evidently proves the converse proposition, viz. if the opposides of a hexagon meet on a straight line, the hexagon can he inscribed
Note. site
in
a conic.
Example
Find
iii.
points of the triangle should be concurrent.
The tangent These
i.
at
A
is
the condition that the to
of reference
normals at the angular
the conic
I/a
+ m//3 + n/y
=
(Trilinear Coordinates.)
n 8 + ny
=
;
suppose that the normal
is X/3
4- jiy
0.
lines are perpendicular, BO that
e.
X
Xn + pw (Xw-f /in) cos A = - n COB J) (n m cos ^L) -f /* (m
The equation of the normal
at
A
is
(n-l
(7
cos 5)
cos C)
(l-m The normals are concurrent satisfied
0,
are
= y-(/-n cos ^) a
a - (m-/
if
by the same values of a,
0.
therefore
(m-ncos^)3~(n-mcos^4)y Similarly, the normals at B,
0,
0. cos C)/3 these three equations are simultaneously
3, y.
Hence
(l~m cosC) (w-n cos^l) (n-l cos JB) (w-Jcos C) (n-wcos^) (Z-wcot J?) ? /. 2 / (n - w*) (cos A -f cos B cos C) 2 ^ (n 2 - m ? sin JB sin C ;
.-.
)
;
;
TRILINEAK AND AREAL COORDINATES
492
This condition can be written as
m
I
n
r 1 nr
=
ir l
l
Examples XII 1.
If X(X
+ >i/3 + vy
=
touches
f/3y
0.
e.
+ wyOC -f n|3 =
will touch >/9y + /i)a + ia0sO. 2. Find the locus of the centres of conies
0.
then
l(X
+ m|3 f /i 7 =
which pass through four given
points.
What is the equation if the points are A, B, C and the centroid of the triangle? 3. Prove that a conic which passes through the vertices and the orthocentre of a triangle is a rectangular hyperbola. 4. Show that the locus of the centres of rectangular hyperbolas which circumscribe the triangle of reference is the nine-point circle
a 8 / (y + z-x) + V/(z +
x - y) -f c*/(x -f y - z)
=
0.
Find the locus of the centre of a conic which circumscribes the triangle of reference and touches the straight line (p ly q 1\). 5.
l ,
Show
that the centre of any conic passing through the angular points of the triangle of reference and the centre of its inscribed circle lies on the 6.
conic 2 bcx*
- 2 a (b
0. -f c) yz that the equation of the tangent at any point of the circumscribing circle is of the form 7.
Show
/
z
(
i + a1
V c3
8.
a? \ 2
>)
-
.
tan
2x
A
.
f tan ^
a2
<
-f
( v
x --2 2 a
Prove that the tangents to 1/x+ m/y + n/z
BC are l(y + e) + (
y -f
=
triangle
\
3 a f) & /
=
A
0.
which are parallel
ABC, touches
to
the circum-
and passes through the centroid of the triangle. - a*/x = 0, or b/ft -f c/y - a/QL = 0. Find tf/y -f c*/z
circle at the right angle A,
Show
that
its equation is equation when the coordinates are transformed to rectangular Cartesian coordinates with AB, AC for axes, and hence find the eccentricity of the conic in terms of the angle B.
its
Find the coordinates of the centre of a conic which circumscribes
10.
the triangle of reference, and has the line (p lt g u t\) for an asymptote. 11. The condition that the straight line (p, q, r) should be an asymptote of a rectangular hyperbola circumscribing ABC is 2p (g r) a cot A = 0. 11.
conic
Conies inscribed in the triangle of reference.
up
2
+ vq* + wr* + 2fqr + 2grp +2hpr~Q
the triangle of reference, v
=
0,
w=
0.
(1, 0, 0),
(0, 1, 0),
Hence fqr+grp + hpq
=
is
If the
touches the sides of (0, 0, 1),
then u
=
0,
the general tangential
equation of a conic touching the sides of the triangle of reference.
TEILINEAR AND AEEAL COORDINATES The equation
of the conic in point coordinates is therefore
fW+f? + hW-2gh9*-2ltfgx-2fgxy = which can be written
The conic
i.
trilffl^ar coordinates,
0,
10)
(cf.
Note
493
is a parabola f/a -f g/b + h/c =
if /-f
0.
g -f h
=
we
or, if
0,
are using
0.
The equation of the conic can be written fqr + p (gr+ hq) = = are three collinear points, so that (Chap. X, 0, gr+hq 10, 1) the conic touches the join of q = 0, r = (i.e. BC) at the point hq + gr~Q. The areal coordinates of the points of contact of the conic Note'ii.
but q
=
0,
r
;
=
with the sides of the triangle of reference are therefore ,
To find
n
the
The equation
of
equation
Qi, r i), (Pt, 02,
r<2)
(I//:
the point
/:
:
the
= 0.
tangents
of the point of intersection of the lines (ft, q iy
= .Pijp* (it2-2*'i)
A
of intersection of
f/p + g/q + h/r
to the conic
fq^ + gr^ + hp^ = 0,
But
1/0:0).
:!//*), (I//:
:
and /J 2 r2 + yr
^+
t
te-'VPi)
^1^2
:
= 0,
hp 2 q 2
^1^2 (ft
r,),
hence
a-l>a2i)
;
provided, then, that no one of the coordinates of the given tangents is zero, the equation of the point of intersection can be written
.
is
The equation
of the point of contact of the tangent (ft a lf r^ ,
2 fp 'pf + gq/qf + hr/rt =
/
the conic
\ /a;-f \/gry-f
contact are {//ft 2
To find
v'W
If,
=
then,
ftoj+^y+r^
=
touches
the coordinates of the j>oint of
0,
,
the equation
Vfx+ */gy+ Vrz
0.
=
at the point (x lt
of the tangent
y lt z^
to
tlie
conic
0.
This equation can be found in the same way as that of the point of contact of the tangent (ft, q l
^
r
}
)
to the conic
= (see
10)
;
the equation
where ^/fx l9 Jgy\,
is
Vlie^
have the signs that make
TBILINBAR AND AREAL COORDINATES
494
(p^ q l} rt ) and of the polar of
The equation oftJiepole of with respect to the conic.
(x lt
y ly ^)
These are *(*8i + ^i)'+a(Ai + *ft) + r(flp 1 +/8 1 )
=
and f(fx l ^gy l -hz l )x+g(gyl -hz l ^fx l )y^h(hzl -fx l -gy l )e
The equation
Cor. infinity
(1, 1, 1),
of the centre,
0.
the pole of the line* at
e.
i.
=
is
B
The
areal coordinates of the centre are therefore
The
trilinear coordinates of the centre of
_
{(*+*) :(/+*):(*+/)}
j* (c
To find
Example. x ( ot
ZQ
1/Q)
=
+ f:^.*:? + a o c o of a conic
the equation
#o) an( l which toucJies
tiie
sides
of
ivhose centre is the
the triangle
point
of reference.
= 0,
then we have X Q = g -f h, y^ h +/, The equation of the conic is therefore
Let the conic be */fx 4- */gy + /hz =/+ g* Hence 2/ = y 4- ^ a? &c. ,
The asymptotes.
Vfoi+ Vgfi + */Uy
=
Let jM-hg# + r#
be an asymptote;
it
2
touches the conic at the point (f/p g/q* A/r2 ), but this must be the point at infinity on the asymptote, viz. (g r r p:p g). :
:
:
/=
Hence we have
Ap
2
(g
r),
g
==
It follows that if (p lt q iy r x ) is
2 fcg
(r
/i
jp),
=
fcr
2
g).
(j?
an asymptote of a conic inscribed
in the triangle of reference, the equation of the conic is
(
Pi
Now,
an asymptote, we have and //p2 + g/q* + h/r* f/P + 9/
+ W+/(0/2 + *A = h (/+ A) 2 + 2^ 2r + j (f+ g) r2 = (0/
or
22
:
2 Note.
;
q l9
0, 0.
r x ), (p 2
,
q z r2 ), ,
we have
=
r ?
It follows that, if is
)
two asymptotes are (p l9
then, the
=
a
a
so that
0.
r is
if p, q,
hence
If,
Sfa-pjy+rt *(pi-q* =
-**--
*px+qy+rz
=
is
ofie
asymptote, the other
a
The coordinates
equations fqr + grp + hpq
=
of the asymptotes can be found by solving the + h) p + (A +/) q + (/+ ) r 0.
0, (g
<,
TRILINEAR AND AKEAL COORDINATES Example
The join* of the
i.
Let
a,
e, c,
/
d, 6,
of a hexagon circum-
opposite vertices
scribed to a conic are concurrent.
495
(Brianchon's Theorem.)
be the sides of the hexagon and take
a, 6, c
for the
triangle of reference. Let the coordinates of d, e, f be (p l g lf rj. (p^ q>, rt ), (jp3 and of b, d is P/Pi^ r/ r\ \ these points lie on the straight line (Pi/r^ : g3 /r2 : 1) Similarly, the joins of the other pairs of opposite vertices are the straight lines ,
,
,
=
(Pi/
:
1
'
^s/^s)* (1
;
These
rs/Pa)'
:
lines are
concurrent
if
=
0,
i.
e. if
1/r,
l/r,
1/r,
Vft i/p
/PI
But the equation of the conic touching the hence
f/p,
i-
g/q v
Eliminating/,
Note.
+ h/9\
h
g,
=
//ft + ^/ga 4-
0,
we obtain the required
line** a, b, c is
Vr
2
=;
//> 3
of the form
4- /// ?3 4^
To find
ii.
=
equation of the
t/ie
which
is
parallel
to
tangent
0.
line parallel to
BC are
(1
-f
k
:
k
:
k).
l fk + gk(k+ -(g + h)/(f+g + h).
fc*=0, or
l)
This touches the conic
+ Wb(fc+
1)
=
the
conic
of a straight if
0,
The coordinates of the tangent required are therefore/: and its equation is fx = (g -f h) (y -f #).
Example
to
EG.
The tangential equation of the conic ia fqr + grp + hpq 0. The coordinates of BC are (1:0:0); hence the coordinates
reference,
=
This evidently proves the converse of the theorem.
Example
i.e.
/t/r,
condition.
(g
4-
h)
:
(<;*
),
The centre of a conic, inscribed in the triangle of on a fixed straight line parallel to BC; find the equation
iii.
lies
of the envelope of its asymptotes
in areal coordinates.
Let the given fixed straight lice be y of the conic be fqr+grp + hpq 0.
= (X+ 1) x, and let
+z
=
the equation
The coordinates of its centre are g + h h+f:f+g, and since this lies on the given line we have 2/ X (^-f h). If (P* & r *8 & n asymptote, we have f**kp*(q-r) g~ktf(r-p), h *~ kt*(p - q) hence 2p* (q - r) X {^ (r-p) -f r1 (p - q) } :
9
)
;
TEILINEAE AND AREAL COORDINATES
496
the factor q r = represents the point at infinity on the this, tangential equation of the envelope becomes
=
2p* + \(pr+pq-qr)
.
The point-coordinate equation of the envelope
BC
;
neglecting
0.
therefore
is
Examples XII f. 1.
The
areal equation of the inscribed circle
is
4
Show
that AA', BB', CC'
If a conic touches the sides of the triangle of reference at A', B' C', Find the equation of the axis ,
prove that AA', BB\ CO' are concurrent. of homology of the triangles ABC, A'B'C'.
is a parabola, find the coordinates 5. If the conic \/fx + */yy -f \/hz = of the point at infinity on its axis. 6. A conic touches the sides of the triangle of reference and the straight
line (PI, 7.
(ft,
rj).
Show
that the locus of
its
centre
is
2 (y + z
Find the condition that the conic
x)/p l
=
=
0.
should be
a rectangular hyperbola. 8. The locus of the centres of lectangular hyperbolas inscribed in a 2 2 fP bin 2 B -f > sin 2 C = 0. triangle is the polar circle a sin 2 A -f-
Normals are drawn
to a conic, inscribed in the triangle ABC, at its points of contact with the sides. Prove that they meet in a point if 9.
ft
2
sin
2
A
2 /3
sin 2
2 -y
sin 2
C
j
QicosA
/3cos#
1
1
-ycosC
=0,
1
where a, /3, y are the trilinear coordinates of the point of intersection of the lines joining the vertices of the triangle to the points of contact of the opposite sides. 10.
Prove that 6+b
if
the conies
a V(X tan 6 + b \/& tan
-f c
*/y tan
\ff
=
0,
^=
are parabolas, a circle can be described through their six points of contact with the sides of the triangle of reference.
a
\/tt cot
\/ft cot
-f
c
\/y cot
Prove 11. ABC is a triangle and P any point on a fixed straight line. that the envelope of the harmonic conjugate of PA with respect to PB, PC is a conic touching the sides of the triangle and the fixed straight line. 12.
A
conic touches the sides of a triangle at A', B', C', and passes Show that the straight line joining the points of
through a fixed point. intersection of B'C',
BC\ C'A\ CA\ ABC.
circumscribes the triangle
A'B',
AB
envelopes a conic which
TKILINEAK AND AREAL COORDINATES 12.
of
A
(1,
h
ux + hy+gz
= 0,
The
Conies referred to a self-conjugate triangle. 0, 0) with respect to the conic
= Q;
g = 0.
Similarly,
we must have
if
hence,
/= 0,
B
if
= 0,
g
The general equation
ft
this is the side
C
and
polar
=
w#2 + 1;#2 + wz* + 2fye+ 2gzx+ 2hxy is
497
BC
(#
= 0)
are the poles of
CA
we have and AB,
= 0.
of a conic with respect to
which the triangle
of reference is self-conjugate is therefore
=
ux* -h ty* + wz*
0.
Note i. One of the coefficients w, v, w must be negative if two are negative we can change the signs throughout. The conic can therefore be 2 This equation can also be written written J 2 a?2 + w* 2 // 2 = n ^ 2 ;
.
Px-
w
or
2
2 //
=
my) (nz + my),
(nz
= (nz
lx)
(nz+lx).
These forms show that the conic touches nz
my
=
0,
nz+
= 0, the nz~lx = 0,
wi/
BC being the chord of contact, and also that it touches = 0, the side CA being the chord of contact. The tangents from A and B to the conic are real. The tangents from C to the conic are imaginary, for the equation can be written n z = (Ix + imy) (Ix imy). side
nz + lx
2
Hence two vertices of a and one inside.
2
real self-conjugate triangle lie outside the conic
Note ii. If the equation is used in the form x*/P + y*/m* = z*Jn* we can use a parametric representation for the coordinates of a point on the conic, viz. [I cos 6 m sin 6 n}. The analogy to the case of the ellipse :
:
and the various equations are in a similar form example, the tangent at the point is x cos 0/1 f y sin 6/m = z/n. will be evident,
The
2
+ ww^ +
ti
vr 2
=
or
Tangent and point of contact.
JRPi/t*
Hence, is
is
it jpx
{#/?*, q/v,
+ qy + rz =
2 t>/ -f
wz*
The equation
at the point (x l t/j , ^) is uxxl -H vyy l point of contact of the tangent (p l9 q l9 ,
thus, for
= is p'*/u 4- q*/v + r?/w = 0.
tangential equation of the conic ux? -f
wcp
;
+ wzs = l
r^
+ 2Si/t> 4- rr^w
0,
of the tangent
and that of the
is
= 0.
touches the conic, the point of contact
r/w}.
Pole and polar. The equation of the polar of the point (x l9 uxx l + vyy l + wzz l = 0, and that of the pole of (p l9 q ly rj is rPi/u + qq^v + rrjw Thus the pole of the line Pi%+
I
i
= 0. =
is
(Pi/u, q\/v,
t/ 1?
^)
TRILINEAR AND AREAL COORDINATES
498
Cor. The equation of the centre is p/u+q/v+r/w ~0, and its coordinates are (1/u 1/v 1/w). Hence the equation of a conic with respect to which the triangle of reference is self-conjugate and whose centre is (xot y , # ) is :
:
The asymptotes.
The equation
(a)
of the asymptotes
is
(
8
(6))
2 (ux* + vy + w# ) = uvw(x + y+#)\ is an asymptote, to find the equation of 2
(uv + vw + tvu)
,
If px + qy+ rz =
the
conic.
px + qy+r0 infinity
=>
is
\p/u, q/v, r/w]
=
on px + qy + rz
TT
Hence
u
0,
kp ^
=
r
p :p
w=
,
q\.
Jcr
p-q
,
is
+ W*
=
:
this
Jcq
r-p
q-r
IfpiX+ qiy+w
r
= --
v
,
,
the point of contact of must be the point at
;
and
viz. {q
and the equation of the conic
(b)
=
ux2 + vy* + wz*
Let the conic be
w
I^L
4.
- 0.
an asymptote,
to
find the equation of the
other asymptote.
Let (jp x gj, fj), (^>2 ^2 r 2) be the asymptotes; then, since they touch the conic and pass through the centre, we have ,
,
,
p/u + q/v + r/w = 0, p*/u + q*/v 4- r*/w = 0, where ux* + vy 2 + wz* = is the equation of the conic.
Hence
*( t^Vt;
+ ^(i V VW
+ ^) + |(;/
h + air + t?/
VW
= +W I)' J
( Vt,
^
W VW +
;
h^a
W/
TT Hence
Now which is proportional to (^i J>i)2 (2i Hence 2 (M + w) is proportional to
j?j
Similarly for
p2 :q2
;
-r,
i>i
^i)-
2
2i (2i
:
r,
rx
jPi
"i),
so that
-^ - q.
so that the equation of the other asymptote
is
TRILINEAR AND AREAL COORDINATES Illustrative
499
Example.
Two
triangles are self-polar with respect to a conic ; show that their six angular points lie on a conic and their six sides touch a conic.
Let one of the triangles be the triangle of reference ABC, and A'B'C' be the other triangle. The equation of the conic is of the equation of the conic is of the I
form
Ix*
4-
=
my* -f nz*
form Ip* -f m
0.
Let the vertices of the triangle A'B'C' be (a?, #, *, ), (x* ?/, *,), ,
Then, since A', C' of B', we have tejtfj -f
my y l
,
,
a -f
lie
A'B'C'
,
on the polar
ngl z2
=
be
(,,
Then, since B'C
r, ),
(
triangle 9
,
a,
,
rJ,
7 ,
^.'J?'
pass through
we have
the pole of C'A', 0,
?p 2 ^ 3
and, similarly,
4-
m^ g 2
8
-f
nra rs
0,
and, similarly,
n^
Ix9 xl -f my3 y1 + 0. But these are the conditions that the points A\ J5', C' should lie on
the sides
the conic
touch the conic
But these are the conditions that
AB'>
B'C',
^9\
C'A'
should
^r^r^r^ ftft A v. r 3 This conic touches the sides of the
tyiP*P9
,
-f
, 1
|)
This conic passes through A, B C\ hence the six vertices lie on a conic. t
triangle
ABC\
hence the six sides
touch a conic.
Systems of oonios. Conies passing through four points. Let the four points be ( i, 2/i, *i), fa, -ft, *i), (i, 0i, -^i), so that the diagonal triangle of the quadrangle, whose vertices are the four points (i)
(
xu
(p.
y\> #1),
471), is the triangle of reference.
+2^0 =
2 be the conic; sub+ 2fy*+2gex the coordinates of the four points in this equation we find stituting
Let
/=0,
ux*+vy* + wz
= 0,
*
The equation
= 0.
of the conic
is
therefore ux* + vy2 +
=
0. ux^ + vy^ + wz
the condition that
whose
wz2 ='0, with
vertices
conjugate
quadrangle.
Let the straight lines (ii) Conios touching four straight lines. be #! # +i0 +*!*=: 0, 0, PiX^jf+r^ = 0, Pix + 2iy+ r^z
=
= 0, f*i*
i.e.
the lines
i
i
2
TEILINEAR AND AEEAL COORDINATES
500
so that the diagonal triangle of the quadrilateral they form triangle of reference.
With
same reasoning as
the
equation of the conic wpi
2
+ rq^ + wri 2
in
(i)
we
the
find that the tangential
up*-bvq* + wt& = 0,
is
is
with the condition that
= 0.
Evidently the diagonal triangle of a quadriself-conjugate with respect to any conic inscribed to the
lateral is
**
quadrilateral.
Examples XII g.
= is a parabola. **) Find the equation of the diameter of Ix 2 + Show
1. 2.
that 2 a 2 & 2 / (#
mf + nz* =
which passes
through (x, y', z\ 3. Find the equation of the tangents at the ends of the chord 2
of the conic
A
4.
fa?
-I
2 wy + n&
=
x-y
0.
conic, with respect to
conjugate, has
lx+my + nz =
which the triangle of reference is selfQ for an asymptote; find the coordinates of
its centre.
A
5.
point
which a given triangle is self-polar, passes through a fixed show that its centre lies on a fixed conic circumscribing the given
conic, to
;
triangle.
A
6.
ence
is
rectangular hyperbola, with respect to which the triangle of referfor an asymptote. Show that self-conjugate, has x + gy -f rz =*
2^aV(g-r)
A
8.
line
= 0.
Find the separate equations of the asymptotes of the conic
7.
conic has a fixed self-conjugate triangle and touches a fixed straight
find the locus of
;
its
centre.
Find the envelope of the polar of a given point with respect to a parabola which has a given self-conjugate triangle. 9.
10.
A
system of conies circumscribes a quadrangle how many conies of (a) pass through a given point, (b) touch a given line ? ;
the system
A
system of conies is inscribed in a quadrilateral; how many conies (a) touch a given line, (b) pass through a given point ? 12. The polars of a fixed point, with respect to a conic which passes 11.
off
the system
through four fixed points, are concurrent. 13. Find the envelope of the polars of a conies which touch four fixed straight lines. 14.
fixed point with respect to
A conic
(a)
The
passes through four given points find locus of the poles of a given straight line with respect to the
The
locus of the centre of* the conic.
;
conic. (b)
What
special points lie on
this locus? 15.
A
(a)
(b)
conic touches four fixed straight lines
;
find
The locus of the poles of a fixed straight line with respect to the conic, The locus of the centre of the conic,
TRILINEAR AND AREAL COORDINATES
501
Show that (b) passes through the mid-points of the diagonals of the quadrilateral formed by the given lines. 16. A system of conies is inscribed in a quadrilateral find the locus of ;
the points of contact of tangents in a given direction. 17. A conic is inscribed in a quadrilateral prove that the product of the perpendiculars from one pair of opposite vertices to any tangent to the ;
conic
is
in a constant ratio to the product of the perpendiculars
from the
jmir of opposite vertices. 18. A given triangle is self- con jugate with respect to a parabola; that the parabola touches four fixed straight lines. other
Two
19.
drawn
conies circumscribe the quadrangle ABCD, and tangents are from a point P, whose points of contact are Q, Q'
to the conies
P
If
R'.
JK,
20.
A
+
m\
:
;
on one of the one of two fixed
sides of the quadrangle,
lies
Q'R' intersect at
(I
show
QK
show that
points.
system of conies circumscribes a quadrangle whose vertices are find the condition that two given points should be conjugate u) ;
with respect to any conic of the system. 21. A system of conies is inscribed in a quadrilateral whose sidos are m: n)\ find the condition that two given straight lines should be (I :
conjugate with respect to any conic of the system. 22. A system of conies touches three given straight lines and passes through a fixed point; how many conies of the system will pass through a second fixed point?
Show
23.
that the conic
same points
at the
2(u+f-gh)x* =
as the conic represented
meets the
line at infinity
by the general equation of the
second degree.
13. I.
The general
To find
the equation
ux*
and
equation.
of the
director circle
+ vy* + W3 + 2fyz + 2gzx + Zhxy
of the directrix ivhen
the equation
the conic is
If (P> 2> r ) is a tangent to the conic
and
(#, y, #)
hence
Now
is
jp
of the conic
Z
=
0,
a parabola.
through the point
px + qy+rz
=
=
-y
:
q
(p, q, r)
:
r
(yz^
0, l
z]
:
px l + qy l ^rz l
= 0;
- ^ x)
-x
(zx l
:
(xy l
being a tangent to the conic,
2 2U(y* l -y l z) + 22F(*jc l -t l x)(xy l -xl y)
2 ( Ye? 4-
-c^),
=
;
6
(i)
(x y, #), i. e. the equation of the tangents from to the conic. Equation (i) can be written
locus of
(#i> V\ > ^i)
.
l j)).
=
hence the
yl
we have
2 Up*+ Fg + Wv* + 2Fg[r+2Grp+2Hpt
is
(x lt
any other point on the tangent, then
9
Wy? - 2 Fy^) & - 2 2 (Uy^ + *V - <*Wi - U*i *i) 0* = 0.
TRILINEAB AND AREAL COORDINATES
602
The tangents
are at right angles (using areal coordinates)
if
= If
0.
(ii)
we omit
the suffixes, this equation represents the locus of a point the tangents from which to the conic are orthogonal, i.e. the '
director circle.
The equation
can be expressed in the standard form for
(ii)
t
a
circle,
follows
;
(x+y+s) (Ix+my + n0)-(a *y2+b*ex + c2 xy)
viz.
collecting coefficients
we
=
as
obtain
2 ( Fc2 + TT6 2 + 2 Fbc cos A) x*
-22 (jFa2 + Gab cos C+ Hca cos B-Ubc cos 4) ye =
0.
Take the expression c
2
from the left-hand side of containing yz, zx, xy
left
;
this equation, and the coefficient of yg
we have
three terms
is
- (2JPa2 + 2 Gab cos C+ 2 Hca cos JB- 2 Ubc cos A)
= -a The equation
2
(#+ F+
of the director circle
is
W+2F+2G + 2H) = therefore
K=
Cor. If the conic is a parabola, we have the equation of the directrix of the parabola is 2 ( Fc2 + TFfc 2 + 2 Fbc cos A) x 0. The focus of a parabola can be found as the pole of the directrix. ;
=
II. (a)
Conjugate diameters. To find
px+qy+rz = If (Pi
&
the equation
of
the diameter bisecting clwrds parallel
to
0.
(#ii 2/D *i)
is
the mid-point of a chord parallel to the line
r)> its equation is
(
5,
Cor.
ii)
For the points of intersection of this chord with the conic the values of k are given therefore by the equation =, 0.
TRILINEAR AND AKEAL COORDINATES Since (x l9 yl9 z^
and
must be equal
the mid-point, the values of k
is
503
opposite, hence
Hence the locus
of (x lJ
bisecting chords parallel to
The
Coi. conic if
l
+m+n =
the equation of the diameter
i.e.
#j),
px+qy + r#
straight line
To find
(b)
y ly
= 0,
is
IX + mY+nZ
O
a diameter of the
is
0.
condition that the lines (ft, q lt r x ),
ttie
(jp 2 ,
gr a
,
r 2)
be parallel to conjugate diameters.
The diameter
bisecting chords parallel to (ft, q ly
(
we
If
write
the equation
is
for
WI,HI
7 lf
ll
(uli 4- /w*! 4-
This
- ft) Z = 2l -rjX + (r, -ft) r + (ft (ft
X + m Y+^Z = 0, (p 2
g2
,
vm l + /n x
gfZj
-f
m
2
,
w2
i.
e.
2 (M?! -f
hm l 4- flfn
ulJz 4- t;m 1
m
x)
2 4-
4-
4^
(h^
2
0.
1
for (ft
w
=
/Wj + ^M
1
(^2-^2)^
*a)>
this gives Z
respectively,
^
^2)
>
1 2,
ft)
gnj x + (W x + t;m x -f/w^ y -f (gr^ + /w x + wnj & ft^ 4-
J
(ft
ft),
0.
or
l
is parallel to
and writing
r x ), (rx
r^) is
ymj +/Wj) 4- n 2
(j?2
^
ffj)
4-/W?! 4-
respectively,
wnj
=
0,
wn
This is the condition that the diameter parallel to (p 2 # 2 r2 ) should bisect chords parallel to (ft, ft, rx ), and symmetrically it is the condition that the diameter parallel to (ft, ft, rj should ,
bisect chords parallel to
the diameters parallel to
The diameters
Cor.
where
Z
1 4-
m
x
4-
% = 0,
r z)
hence
(jp 2 ,
#2
(j>,,
ft, r x ), (^ 2
l
i
12
>
\
,
g2
it is ,
Z
l
=
?!
and similarly
Z
2
= (m
w x Z2 -f w2 ZlW2 4"Z 2
whence
also
27 1 I2
=
if
x
x 4-
wx
= Wj, m
2
4-
are conjugate
2wZ1 ?2 4-2/(rw 1 w 2 -f m 2 nx)
Now
the .condition that
r 2 ) should be conjugate.
X+m Y+ ^ Z = 0, X + m 2 4- n2
2
w x n2
,
=
0.
m 2 r + w2 Z = 0, if
AND AREAL COORDINATES
TRI LINEAR
504
Hence the condition
be written in
for conjugate diameters can
either of the forms
or
2(t; (c)
To find
the equation
l+m + n = 0).
(where
2/)(w n2 + m 2 n
+ iv
1
l)
=
0.
of the diameter conjugate
to
IX +
mY+ nZ =
Let the equation of this diameter be I'X + m'Y+n'Z we have V + m' + n' = and
Eliminating
we
w', n'
Z',
= 0*;
then
get
x
r
z
i
i
i
j
i
i.e.
SX
or, since wi-f
w
which can
+ 7 +/-^)n-.(v + t} t
=
= this becomes 2X {(cfZ+/m + wn)-(W + t;m+/n)} =
0,
Z,
0,
also be written
ul +
The
III.
[(n;
x
r
z
1
1
1
hm + gn
hl+vm+fn gl+fm+tvn
The axes
axes.
=0.
of the conic are conjugate diameters
at right angles.
Let the axes be the diameters
and suppose that they are parallel
to the straight lines
Then, since the diameter bisecting chords parallel to r + fa A) * + ( P* - flfa) ^ = 0, we have faa ^2)
(p.^ g 2
r 2)
,
is
x
_Ji__ (72- r 2
= w
-.
i
^a-JPa
-J!L_
?
an(]
^2-^2
Since the lines (p lJ q v
rj),
2
,
^2
e.
?! Z 2
cot
and further
^)Z!Z2
A + mimz cot JB + n
+ (t;+#
hf)m m l
_ ^2 ^i~(7i
r2 ) are perpendicular,
,
x
2
we have
= 0,
w2 cot
Again, since the diameters are conjugate,
(*+/
_
^2 'i-^i
2i-*"i (jp 2
2 (2i - rj (& - r,) cot ^ i.
_
(7
=
0.
(i)
we have
+ (w+hfg)n l n2 =
0,
(ii)
TRILINEAR AND AREAL COORDINATES Eliminating
Z
2
w
,
2
nt from
m
A
cot
0-*)Zi
l
theh, since
/
we
(iii)
(tv
1
w %
2,
2
m
x -f-
The equation
l
,
is
%
-f
= 0,
1
satisfy the
same equation.
any point on the axis ^ = 0, we have
i^
=
0,
of the axes is therefore
A
cot
=
w+Ufg
v+g-h-f cot
obtain
MJ cot G + h-f-g)^
1
if (x, y, z)
But,
and
B
+ g-h-jF)!
(v
1
and evidently
(i), (ii),
cot
505
B
0.
cotC
This equation can also be written
iV-Zf
(Z-Xf
=
>)-2f sin 2 -4 t/te
When axes of
Let tively.
the conic
fjt,
sin 2
0.
G
equation of the axis of a parabola.
symmetry A,
sin 2 J5
is a parabola, we know that the equation of the reduces to that of the axis and the line at infinity.
v denote
Now
u+fgh,
v+g
ft
w+
/,
hfg
respec-
the equation of the conic can be written
this can be verified
by immediate
simplification.
and Hence the common chords of the conies /(#, y, z) = Ao 2 + jmy a + vfi = and + # + * = 0; i.e. they are I,(fgh)x = meet the line at infinity at the same points. Thus if the conic /(#, y, si) is a parabola, so is the conic in this case therefore JJLV+ v\+ A/JL = 0. A#2 + /*i/2 + v#2 = The equation of the axes is ;
,
!
I
when
(z-x)(x-r) (x-r)(r-z) (r-z)(z-x) A cot
y
/ui
A
= 0,
and
cot^, cot -4
.e. i.e.
.
0,
cotC
cotJS
the conic is a parabola
=
this
= =
o, 0,
21/A(y-Z)cot4 =0.
becomes
TRILINEAR AND AREAL COORDINATES
506
Now
= Hence the
line
So? {A/mi;
J>A
JAI>/
=
/^X + yAr+A^Z
is
#+#+
=
0,
i.
e
.
line at infinity.
The equation
IV.
The
of the axis of the parabola is therefore
the
^
=
0.
^), (x%, ?/ 2 , # 2 ) are the foci of the conic, since the lines joining them to the circular points at infinity are tangents to the conic, we must have for some values of A and //,
A
foci.
If (x l9
2
(op, 60, cr}
We
+./x ( jp&i
+
yly
^+
=
2
C/p
r*i) -h
(
Fa
p^2 4- 2y2 +
2
-h
^
2)
Wr*+ 2Fqr+2Grp + 2Hpq.
(i)
can determine the value of A since
breaks up into two linear factors.
This gives (see V, below) 4S tfA -A0A+A 2 = 0; (ii) the two values of A give two pairs of factors, and two corresponding 2
2
pairs of foci
Or, in this identity put p,
g,
r each unity, then
provided that (xl9 yl9 ^), (#2 y2
are finite points. have also ,
Comparing
coefficients
we
,
#2)
U= 2G = 2 If
=
/x
(^y2 + 0^2^)
2 A aft cos
C
;
hence
Thus .
ana
x1 xz -= Jo.
Therefore xlt x2 are roots of the equation
Ex*-2(U+G+H)x+ U= Ac2
.
Similarly, the other coordinates of the foci are roots of the equation
Ks*-2 (W+F+ G)s+
W = Ac
2 .
TKILINEAE AND AREAL COOEDINATES
507
These equations can be written &c.,
*,
where XQ y Qj # are the coordinates of the ,
Since
tf-ff
we have
Sv
=
+#-#o+--*(>
/
+ //)
{(Z7+
2
centre.
(i)
0,
- UK+\Ka*} =
0.
Tl^s gives a quadratic for A, which must be the same as equation found,above taking either value of A, we have
(ii)
;
V{(U+G + H) 2 -UK+\Ka*},
K(x-xQ)=
&c.
;
the signs of the surds must be chosen so as to satisfy equation thus to each value of A there corresponds a pair of foci.
Note
i.
When
the conic
is
a parabola
K = 0,
(iii)
;
and the equations become
linear.
Note ii. The equations for finding the foci can also be obtained from the fact that the tangents from a focus to the conic pass through the circular points at infinity, and therefore that their equation satisfies the conditions for a circle.
Thus, if
f(x, t/, z).f(x,y',z)-(xX' + yY' satisfies the conditions for a circle, we have
+ zZJ
=
and two symmetrical equations. Incidentally, if we eliminate/ (#', y\ z) and A from these equations, we 2
2
v
(X-Y) (Z-X) (T-Z) + w-2f w+u-2g u + v-2h aa
|
hence the
foci lie
find
2
b*
0;
c*
on the axes of the conic.
Note iii. If the conic is inscribed in the triangle of reference, we have 17=0, F=0, JF = 0. 2 2 = 2 = 0; Hence 0, ^f/jt/j + Aft = 0, pz l Zi + Ac /uo^tfj-f Aa
(In trilinear coordinates this gives O^O^
If the conic is a parabola,
not conclude that fi
is
/i
= If =0,
i.
e.
K = 0,
c2
As
^^
then in equation
but fi(x + yi-fZi)(#2 + l
not zero, hence one focus is at infinity
where xl 4- y l -f z l = 0. Hence a 2/^a ^ ^/^ +
=
0,
;
we
still
or the finite
^2
have
+ **)
-
1
Of
=
=
(i)
we
could
^J evidently
^ = "^ C
,
focus lies on the circum
scribing circle.
Again,
if
=
is
the equation of a parabola inscribed
in the triangle of reference, the point of contact of the line at infinity is (f:g:h)i this point is its infinite focus, henco the finite /a a b 2 c\ f .
focus
is
V
(
-;
*
:
:
g
7-
1
k>
TKILINEAR AND ARfiAL COORDINATES
508
Illustrative
To find
the
Example.
equation of the tangent at the vertex of the parabola
inscribed in the triangle of reference, with its focus at the point If
fqr+grp + hpq
=
the equation of the conic,
is
x
:
y
:
z
=
:
'
&/g
we have
2
/
c /h.
Hence the equation can be written a*qr/x' -bWrp/y' point equation of the conic is therefore a^/x/x + b*/y/y'
The
directrix
-f
the polar of the point
is
c^/zjz
(#',
f/',
(#', #', ?').
-\-c*p
=
0.
The
= 0.
z) with respect to this conic
;
2
its
is
equation
2x
2 /
-f
(&
c
-a ) # =
2
2
0,
or 2
ax
=
cos A/x'
0.
The tangent at the vertex is parallel to the directrix, so that its coordinates are {a cos A/x' -f k b cos B/y' + k, c cos C/z' + k} and these satisfy the tangential equation of the conic. ,
t
2
^> a 2 ^x
1T
Hence i.
~,
2 a2
e.
But
cos I? (6---1
j I
.)
+k
y'
[ \
B + ky)
(b cos
a^y'z +b*z'x +c*x'y'
(ccosC \
(
(c
z
7
i.e.
/>' cos -B)
since
x'
/.
fc -f
/.
2
cos
P 2 sin A
cos J9 cos
2 (a cos
-
a cos
x
A+ ~
6cos
-f afec
+z =
cos jB cos
A sin J5 sin
The equation of the tangent
or
-f t/'
2a
^ 4- 2 jR sin
0,
C+kz) =
0.
becomes cos C = 0,
0, so that this equation
2 a 2 (cy cos C-f
A; -f
=
]
f
fc
,)
+k\
2 a cos
1,
C=
C= C=
at the vertex
-B
;
;
0. is
therefore
A /x -2R sin A sin B sin C)# = + -,ccos C = 2-Rsin^I z
y
,
sin
0,
B sin C(
Note. This is the equation of the Simson line of the point with respect to the triangle of reference.
V.
The lengths of the
Method
axes,
and the eccentricity
(#', y',
z)
of the conic.
Since the axes are conjugate and perpendicular, the tangents at the extremities of an axis are each perpendicular to it hence a circle, whose centre is at the centre of the conic and whose 1.
;
a semi-axis, has double contact with the conic, the pole of the chord of contact being a point at infinity.
radius
The
is
tangential equation of the conic is
Vf
-f-
Wr* + 2Fqr 4- 2 Grp -f 2//j^
;
TRILINEAR AND AEEAL COORDINATES
509
yOJ # ) be its centre and r the length of a semi-axis. Then the equation of the circle whose centre is (# y # ), and whose radius is r, is let (#
,
,
=
4S 2 (^o + ^o + ^o) 2 or,
writing r
= 2Sp,
the equation
(l>Zo+
% Hen/jo
where
X,
we
have, identically,
/x
are constants, and
x+y + z = 0. Now put j9,
g,
2 )
r 2 \ap, bq, or} 2,
is
=
(a;,
,
y,
2
p {op, 63, cr}
is
-e)
2 .
a point at infinity, so that
r each unity in this identity, thus
\K+l =
0,
or
We now have 2-f
Kp 2
2
[ap, bq, cr}
=
and since the right-hand side has two linear
The
H:abp
U+Ka*p*
!
factors, the left-hand
Hence
side also factorizes.
2
-
cosC '
coefficient of p Q is evidently zero
U j
;
acosC
H
0.
4 />
acosJS 6 cos
b
G
the coefficient of
=
ccosA
A
c
U+H+G
H G
The
=
coefficient of
SJTc
U
H
H
V -
G
F
A7f, where
cos
A
2
-acosJS
.
2c{r Aw
acosJS A(/
= 2a w-22/^cos A. 2
A2 XV+ A#fy2 + A 2 =
The term independent 4S2 Hence or
A
c
/>
=K
=
b cos
b c
of p
is
.
0,
bcosA A/]
TRILINEAR AND AREAL COORDINATES
510 Cor.
If
1.
rt are the semi-axes of the conic,
t\,
g
.
^10*rIf e
=
4A
Trot
flo
the conic
where
>
a
is
e is
the eccentricity.
and
circle,
8 in this case 16 ITS
2 ;
this condifr'on
can be easily verified by the conditions previously found. When thfl conic is a circle we have rt = ra hence the radius of the circle represented by ,
the general equation is given by r2 = 2 S&/K%. Cor. ii. If the conic is an ellipse, its area is 2nS&/K*.
Cor.
The conic is an and raa are both
iii.
values of rf i.
e.
K
according as
Method
;
positive or negative.
The equation
2.
f(x,
is
an hyperbola according as the one positive and one negative
or
ellipse
positive, or
of the conic
is
= ux* + vy + wz* + 2fyz + 2gzx+2hxy = 2
y, z)
let its centre
(supposed
be
finite)
(#
,
y QJ #
),
0;
and write #
=#
Then we have f(x, y,
=/(
*)
r]ro +C^o}+/(f, 1
?*
r?,
0.
=
Z =T =
* ) is the centre, 2 and +7)-f C 0. the coordinates to Cartesians referred to changed suppose the principal axes of the conic.
for since (#
,
y
,
Now
If f(x, y, a)
= A {X /V+ r A* - 1} 2
2
2
2
where A
is
an undetermined
constant, then /(*o, ^0, ^o)
But
/(,
where d.f(xQ ,y
Now conic
,
^
T/,
)
=
= AtO/r^ + O/r^-l} = -A. = /(a?, y, *)-/(a ^0, ^o) >
1.
the distance of any point
from the centre of the
is
Thus,
if
(
, ry,
$
and (X, Y) are the same -
and
(#, ^, ^)
point,
we have
TRILINEAR AND AREAL COORDINATES
511
=
The
2 right-hand side is a perfect square when ft r^ or r2 , hence also the left-hand side. Thus the squares of the semi-axes are the
values of
is
which
for
/*,
a perfect square.
&ow
if <J>(,
the expression in
This that
is
square
if
TJ
obtained
exactly the process
x+y+z =
condition
f,
by
is
a perfect square, so
is
+
in
substituting
we might
= z)
(x, y,
is
(
rj)
for
use to find the condition
should touch the conic $(x,
the same in both cases.
is
= 0,
+7j +
where
),
77,
y, z)
Hence $ (,
= 0, Q
TJ,
so that the is
a perfect
a parabola.
This gives us
+ 22 i.e.
i^~i&2 )(^dA-ic 2 )-/xdw(fcd/-ia
{(f
2
)}
=0
2{CVd2 + a2//xd-ia4 } + 22
where
2 2 {JP^ d
+ J/xd(a2 w-6 2 A-cV)+
=
0,
=
6
The squares
of the semi-axes are therefore given ,
Since /(#
\tf(?\
,
y
,
#
)
==
y
,
this result agrees
^/^>
^
by 2
)}
=
0.
with that of the
first
method. VI. Radius of Curvature.
Let
f
(x
,
y',
/) be a point on the conic
5 ux* + vy* + wt* + 2fyz+2gzx+2hxy = 0, = y-i/, C = ^ we have if we write f = x (/>
then,
since
The
x',
x
rj
^',
= 0.
circle of curvature at
at that point. line at 0, and
OP2 = 2pPN, in Cartesians.]
(x', #',
^0 touches the tangent
P
is any. point on a circle touching a straight the perpendicular to this straight line, then 2rx where p is the radius of the circle. [Cf. a?+y*
But
if
PN
is
=
TRILINEAR AND AREAL COORDINATES
512
The
or
equation of the circle of curvature
is
then
+ 4Sp(xX'+yY'+gZ')/{aX', IY', cZ'\ = a*rt+b*ti+c*fr + Sp(X'+iiY' + tZ')/{aX',bY', cZ'\ = 0.
0,
We may write this equation where 4SpA
=
{aZ'
f
Now thus 20=0 and the right-hand side t'actorizes since + n + = and these of must be a the straight lines, tangeat represents pair hence we have at the point of contact and the common chord
;;
;
identically
=
= Z'-X',
= X'-Y'; is allowable since the values chosen satisfy +?) + = 0, hence 2 2 2(r'-.z') +2S(/-a A)(z'-;r)(:r-r) = o, Jn
this identity
put
Y'-Z',
which equation gives the value of
Now we
and,
X=
have
if (x, y, e) is
on the
r?
this
A.
ux + hy+gz,
conic,
S M (r-Z)2 +22/(Z-X) (X-Y) 2 2(uY*-2hXY+ ;X ) + 22 (hZX+gXY-fX*-uYZ)
hence
= = 2 {z(t;X-*r)+ r(ur-wc)} +22 = 2 {Z(TFaj-flt*)+r(TTy
and, using the fourth equation above, this becomes
)-22y(GX+FY+ WZ)
Hence 2A
= - A/2oa (Z'-Z') (X - Y = - A/\aX', f
f
)
P
_ ~
/ ,
IT,
25A
3
cZ'}
fcY',
cZ'\*
;
TRILINEAR AND AREAL COORDINATES
513
VII. Invariants. The relations connecting areal coordinates with rectangular Cartesian coordinates are
= fTn(ptXcos03
* \
where
3
2
=
A,
TT
2
l
Now into p.
=
TT
Y8i 3
(7,
--0i
=
2
2
suppose that ux* + vy + w& + 2fy#+ 2g0x+2hxy transforms u'X* + v'Y* + w' + 2f'Y+2g'X + 2h'XY; then (Chapter XI,
433)
A =A X
X
^6
=AX =A
X
~ -AxX
^2
C(> S
^2
j? 3
cos
3
Sin ^2 sin 3
-- x (p sin J. -f-^ sin J?+j> sin C) 2 l 2 a 256S".R2
X
(ai
a A'
i.e.
= A/4S 2
.
Again, the squares of the semi-axes of the conic
ux2 + vy2 +wt? + 2fyz+2gzx+2hxy are given
by 3 r*
+ JfiCA^r2 +4S2 A2 =
=
0,
(i)
and the squares of the semi-axes of the conic are given
by
(' + /)A'Trr +A = 0, and (ii) must be identical, hence
so that equations
(i)
W' a"" _ (u'+v')A'W"
=A -'
or
Alao Also
1267
/2 ~ A
w =' A' f
/3
thus
/2
2
TF' 3 r* +
K k
AJ
(ii)
'
TRILINEAR AND AREAL COORDINATES
514 Note.
These two invariants can also be proved by substituting for and z and finding the values of u', v', and /' the simplification is somewhat tedious.
x, y,
;
Illustrative
Example when
it is
To find
i.
the latus
Examples.
rectum of the conic
a parabola.
If we transform to rectangular Cartesian coordinates referred to the axis of the parabola and the tangent at the vertex, the equation transforms into
Then
A'
hence
= -4 w
16 w'A'S
Thus
2
2
*=
-
*=
Example
ii.
v' = X = and X 0/4 S
X s and u' -f
;
2
A,
.
// the equation z
ux + vy2 + wz* + 2fy* + 2g2x + 2hxy
=
represents a circle, to find its radius. 2 2 suppose the equation transformed to X(#*-f # -p ) = 0; the 2 9 8 2 X and -X p and these give and A' are 2X, t/, ttV-/' the expression for the radius previously found.
We may
invariants w'-f
,
,
The following method gives other more simple expressions. wo^ + tyHtt* 1 * 2fyz + 2gzx + 2hxy =
The equation
can be written
(2/- v - w) yz +
Now sians,
(2 g
- w - u) zx
h - u - v) xy -f (<# -f v^ -f w^) (a: -f y -f z) = 0. when this equation is transformed into Carte-f-
(2
since x + y + z=* 1, (ux + vy + wz) (x + y + z) transforms into a linear expression.
Again, since x + y
+z
=
is
the line at infinity,
if (i)
is
a circle, then
is
also a circle,
(2f-v-tv)yz + (2g-u>-u)zx + (2h-u-v)xy and it is clearly the circumcircle.
We
=
(ii)
have seen that when these equations are transformed into Cartesians the terms of the second degree are the same. Hence if equation (i) transforms into X {(#-OK) 2 + (y-)* p*} = 0, - 7)s - -R8 } = 0. - x equation (ii) transforms into X {(x Oc )H (y
The discriminants of these equations are X s pa and .-. -XV'^A^.S8 and -X8 /? (2f-v-w)(2g-w-u)
= 4AJBV(2/-t?The reader can prove similarly 'that Hence
a
^
TRILINEAK AND AKEAL COOEDINATES
616
We conclude this chapter with some general illustrative examples. If two
Example.!.
triangles are reciprocal they are also in homology.
A
ABC
Let and B'C be the two triangles; take for the triangle of reference, and let the equation of the conic with respect to which they are reciprocal be
ABC
w*a
%
-I-
vy*
+ w# + 2fyz + 2 gzx + 2 hxy
0.
=
0; the point Since^ B'C' is the polar of A, its equation is uo? + hy+ gz of intersection of BC, B'C' therefore lies on the line 0. x/f+y/g + z/h
The symmetry of this result shows that the points of intersection of CA, CA' and AB, A'B' also lie on this line hence the triangles are coaxal and x/f+y/9 + z/h = is the axis of homology. Again, since A' is the pole of BC its equation is Up + Hq + Gr= 0, i. e. ;
-4'ls the point (U,
AA
Hence
BB\
Thus AA',
BB\
are then copolar,
Example is
Gy*=Hz\
= Fx, Fx =
Hz
CC' are
H, a).
the line
is
similarly the equations of the lines
Gy.
CC' intersect at the point {l/F, 1/G, l/H] and this point is the centre of homology.
To find
;
the triangles
the equation of the conic, one
of wlwse foci which the triangle of reference is selfDeduce the locus of the foci of conies which pass through ii.
at the point
conjugate.
(x ly
y ly ^),
to
four given points.
The tangential equation a*p
8
+ 6'g8 -f c'r2
of a conic whose foci are (x^ y lt ^), (x2
2 be cos Aqr
ya *a) 2ca cos Brp 2ab cos Cpq + k(pxl + qy l + /j) (pxs + gya 4- rzj = ,
,
is
0.
If the triangle of reference is self conjugate with respect to the conic, the coefficients of qr, >-p, pq are zero, so that
&
(#1*2
^ 2 be cos A,
+ y**i)
k (z^Xi 4-^0^)= 2 ca cos J&,
k (^iVa + ^2^1) y l ca cos
and
B
-f
8
Sjp is
2 ab cos C.
zl ab cos
hence, substituting these values of
which
=
Cx^bc cos A = z^b cos C + x^c cos A t^ca cos B xj>c cos A -f y^ca cos B z^Qb cos C
Hence
^ {a*y z l
l
X^ in
fcra , fcya ,
the 'equation,
+ (yl caco*B + ^aftcos C-a?l 6ccos-4)a?1} of the conic
2a?8/a?1 {aa y1 2f1 -f(y1 cacos
is
we have
= 0,
therefore
-B+^a&cos
C-
If this conic passes through the four points (/, the suffixes in the coordinates of the focus,
Vf*/x{a*yz -f (yea cos 3 4- zab cos is
,
the required equation.
The point-equation
which
ky l zl x2
C
a? 1
+
=0. we have, dropping i fr),
6ccos^.)a?1 }
<jr,
xbc co
JL) a?}
= 0,
the equation of the locus of the foci of conies which pass through
the points
(/,
g,
h).
K k 2
TBILINEAB AND AKEAL COORDINATES
516
Example the triangle is
self^conjugate
situated,
Two
iii
concentric conies are drawn, one circumscribing to the other the triangle bisecting tlw sides
of reference, and
and
prove that the two conies are similar and similarly
:
their areas are in the ratio
-8xye:(-x+y+0)(x-y+*)(x+y-z), where (x : y Let
:
z) is their
common
centre. r
ABC be
and
R the
mid-points of the sides. Project the figure orthogonally so that the circumscribing conic becomes a circle. Let A'B'C' and PQ'R' be the projections of the triangles and PQR. the
t dangle
P, Q,
ABC
The mid-point of a line projects into the mid-point of the projection line, so that F, Q' R' are the mid-points of A'B'C'.
of the
9
Now the
areal coordinates of a point are unaltered by orthogonal projecand the centre of a conic projects orthogonally into the centre of ^he conic. Thus (xiyiz) is the circumcentre of the triangle A'B'G\ it is therefore the orthocentre of the triangle PQ'R'. Hence the second conic projects into a conic to 'which the triangle P'Q'R' is self-conjugate and whose centre is the orthocentre of the triangle PQ'R', i. e. it projects into
tion,
the polar circle of the triangle P'Q'K. Since the two conies can be projected orthogonally into two concentric circles they must be similar and similarly situated, being evidently similar sections of .coaxal cylinders. Again, the ratio of two areas
unaltered by orthogonal projection, so
is
that the ratio of the areas, of the conies
= ratio of the areas of the circles = 7rl2' -7r#' COB A' cos B' cos C' = - 1 cos A' cos B cos C'. -x+y+z = y "* z * a
9
:
f
:
x
But
sin 2 A'
sin 2 B'
2x -x + y + z
hence
4 sin A' cos B' cos
sin 2 C"
:
cos A'
and two symmetrical results, therefore 8xyz:(-x + y-f z)(x-y + z)(x + y-z)
Example
iv.
Find
the area
:
cos
=
1
B :
f
C
'
cos C'
cos
A cos B' cos
C'.
of the conic
= 0.
We can apply the (x, y, z)
method used in the last example to this problem. Let be the centre of the conic, then 2l=sy + zx, 2m**z+xy,
2n =*x + y-z. Project the figure orthogonally so that the conic becomes a circle and let new triangle of reference. Then we have
A'B'C' be the
Area of conic area of circle inscribed in A'B'C' :
area of
AfC
:
area of A'B'C'.
Hence the area of the conic = nr^ S-rS' irflBT where a' is the semiperimeter of A'B'C'
W
r
f
TRILINEAR AND AREAL COORDINATES Now
517
the areal coordinates of the centre are unaltered by orthogonal
projection, thus
L= a'
.
x+ _ ~~ 1 __ ~~
b'
a'
c'
+z y_ + b'+c'
y + z-x + c -a'
_ ""
f
rr
Hence
is
a
8
8
therefore
c'
+
-b'
a'
m
I
the area of the conic
x + y-z z+x-y _ "~
__ ~~
b'
b
8
a'
+ b'-c
n S
nS\/ V +w+
'
xo 3
n)
(J
Examples XII
C
h.
1. A system of parabolas is drawn to touch the sides of a triangle. Pnjve that the locus of the point, in which the lines joining the vertices to
the points of contact of the opposite sides meet, is a conic passing through the vertices and having its centre at the centroid of the triangle. 2. If
to
LMNia
the pedal triangle of
ABC,
it is
self-conjugate with respect
any rectangular hyperbola through ABC. 3. The locus of points from which tangents to
OL*/l
+ fP/m + y*/n =
2 perpendicular is 2(w -f- n) a + 2 2 1 Py cos A = 0, which radical axis with the circumcircle is 2(m-f n) bcOC = 0.
4.
is
Find the tangential equations of the circumcentre, in-centre and
centres, orthocentre, centroid
are
a circle whose ex-
and symmedian point of the triangle of
reference. 5.
Investigate the character of the conies
(ii)
(iii)
aV sin (#- C)-f b*q* sin C-A + cV sin A - J9 = a*p
=
2
;
-AbcqrcoaA,
and
find their areal equations. 6. Show that it is possible for a conic to be described
7.
round a triangle that the tangent at each angle is parallel to the opposite side. Find the coordinates of the point isogonal with the orthocentre of ABC.
8.
Find the equation of the rectangular hyperbola through the four
ABC such
points (/, 9.
g,
h).
The mid-points of the diagonals
four lines lx + 10.
my
Show that
form a triangle
if
nz
=
lie
of the quadrilateral formed
on the line
J*#-f
w'y 4- n*z
= 0.
by the
the lines
self-polar
the triangle of reference
with regard to any conic with regard to which then
is self-polar,
1
JL
i
/t
m
t
n,
r *j
= OTg
IT '?2
l
l
l
-
TRILINEAR AND AREAL COORDINATES
518
11. Find the condition that the in-centre of the triangle of reference should be the focus of the parabola Ix* + my* + m* 0. 12. A conic is inscribed in the triangle of reference and has one focus at the orthocentre. Prove that its centre is the point
{cos
J9-C:
C-A
cos
cos
:
A-B],
and that the sum of the squares of its axes is J?*(l + 8 cos A cos B cos C). 13. Find the equation of the parabola which passes through A, B, tw/> of the angular points of the triangle of reference ABC, and also through the middle points of AC and BC. 14. If the line Ja + mj3-f ny * meets the sides BC, CA, AB of the triangle of reference in D, E, F respectively, prove that a conic may be drawn to touch DA at A, EB at B, FC at C, and find its equation. 15. The radical axis of two circles given by the general equations 2 (ax* + 2fyz) and 2 (a'x* -f 2fyz) = in areal coordinates is flag
+ fry +
cz
a'x+Vy+c'z + b' + c'f-g'
~~
a+b+cf-g~h
a'
h'
A rectangular hyperbola passes through
the corners of a triangle and Prove that the tangents to it at the in-centre and at the orthocentre meet in the point whose distances from the sides of the 16.
through
its
in-centre.
triangle are r(cosB-f cos C), r(cos radius of the inscribed circle. 17. If the
connector of 0,
(0( l
,,
,
C+ cos.4),
ya )
(<x a
,
j9 2
,
r (cos
ya )
A 4- cos B),
r being the
divided harmonically
is
by
then
All conies circumscribed to a given triangle which divide a given segment
harmonically pass through a fixed point 18. Show that the equation of a pair of conjugate diameters of the conic yz 4- zx+ xy = 0, may be written in the form
19.
Prove that the coordinates of the foci of the ellipse
J V jrr
a
c
ft
are respectively proportional to r 20.
Jl J&+ V V =o
+
r
, '
c
fc
, '
a
,
a-nd c
, '
,
a'
r b
The tangents drawn from the angular points of the wz*+ 2fyz+ 2gzx+ Zhay =
triangle of
reference to the conic wr'-f t# 9 -f
opposite sides in six points which
lie
VVTx*+ WUy*+ UVz*-% UFyz-2 VGzx-2 21.
Show
that
2 (x* + y*) -z* =
and the equation of its 22.
axis.
Find the focus of
directrix.
is
What
5 ^=
meet the
on the conic
a parabola.
is its
latus
WHxy
Find
its
0.
focus, directrix,
rectum?
and deduce the equation of the
TRILINEAR AND AREAL COORDINATES
519
The equation of that diameter of the circle 20t 2 sin 2 A = which is is 2 a(X cos A (cm - 6n) = 0. perpendicular to the line lot. -f m)3 -f ny = 2 24. Show that if A is a focus of the conic /a 2 -fw/3 2 + then n-y w = n and fc* + c1 a2 25. The squares of the semi-axes of the conic /x+
_____
.
>
26. its
A
i
arabola
focus are a',
is
#',
and the
inscribed in a triangle, /. Prove that its axis
is
trilinear coordinates of
the line C'*-p'*)
=
0.
m
Prove that (
a */& siiT(B - C) -f 6 y^
27.
is
c
28.
Show
/mp +
that the director circle of the conic v/to-f
\/ny
=
can be thrown into the form (a sin
A -f
ft
sin
B
-f
y sin C)
cot
A m
smB
sin
(1 01
sm^
cot
-f
B
-f
n y cot C)
A circle circumscribes a triangle circumscribed to a conic S, and has its centre on S; prove that it touches the director circle of S. 29. If p, q, r are the perpendiculars on a line from the vertices of the triangle ABC, show that the conies represented by the equations tan ^ qr+ tan
--
i
+ tan
-
pq
=
*/ap
0,
4-
Jbq -f
\/rr
~
are confocal. 30.
Prove that the foci of the conic
a(aco8^)* + 6OcoBB)* + e(ycosC)i
=
are the circumcentre and orthocentre of the triangle of reference, and that cos B cos C. the square of its eccentricity is 1 - 8 cos
A
Prove that the joins of the mid-points of'the sides of the triangle of reference to the mid-points of the corresponding perpendiculars of the 2 2 1>* c ), which is isogonal with triangle are concurrent at the point (a 31.
:
:
the centroid.
Show
that the vertices of a triangle, its circumcentre, orthocentre, and all lie on a conic whose equation is
symmedian point
2aV(cos2J3-coB2(7) = is
0.
32. When a conic degenerates into a pair of points its director circle the circle on the line joining the points as diameter. Hence find the
z 2 ] are the equation of the circle for which the points (a^, y,, ,), (.r 2 ;/ 2 extremities of a diameter. 33. Show that the tangential equation of a conic inscribed in the triangle ,
,
of reference and having a focus at (x, y, z) is Sqrxrf = 0, where i\, ra r3 are the distances of the focus from the vertices of the triangle of reference. ,
TEILINEAE AND AEEAL COORDINATES
520
Find the locus of the
84.
a conic inscribed in the
foci of
t dangle
of
reference and touching the line (j?, g, r). 85. The orthocentre of a triangle circumscribed to a parabola lies on the directrix. 86. The triangle of reference circumscribes a parabola whose axis is 0. the line (p> q, r) prove that ^pa 2/(q-r) (The point at infinity on the axis is the infinite focus.)
=
;
A
parabola has a fixed self-conjugate triangle show that the locus is the nine-point circle of the triangle and that its di/ectrix passes through the circum centre of the triangle. 38. The director circles of a system of conies which touch four fixed straight lines form a coaxal system. 37.
of
39.
;
focus
its
Prove that the equation of the asymptotes of a conic
substituting (yz^-y^z)^ (ZXO -ZQ X), (xy^-x^y) for p, equation, (# y z ) being the centre of the conic.
by
,
may its
41.
be found
tangential
,
Find the coordinates of the focus of the conic yz + zx+4xy
40.
is 2(fe
r in
g,
Show that the equation -c2 )(t/-*) 2 = 0.
of the axes of the conic
= 0.
*/x+ \/y + */z
=
2
42. A straight line passes through a fixed point. Prove that the line joining its poles with respect to two given conies always touches a fixed conic inscribed in the common self- conjugate triangle of the two conies. 43.
drawn
Prove that through three points A, B, C two parabolas can be so as to touch the circle ABC at A and that the axes of these
parabolas are at right angles. 44. A rectangular hyperbola circumscribes a triangle ABC. Show that the loci of the poles of the three sides of the triangle with respect to the hyperbola are straight lines.
45. If the internal bisectors of the angles A, B, C of a triangle meet the circumscribing circle in A'B'C', the trilinear equation of the conic inscribed in the two triangles ABC, A'B'C' is
/c(a+b)y 46.
The pencil subtended
at
-
0.
any point of the conic
by the quadrangle consisting of the vertices of the triangle of reference and (<X |3 y ) is harmonic. 47. Four conies circumscribe ABC and have the in-centre for a common Show that the centres of all conies which touch the four correfocus. ,
,
spending directrices
lie
on 2a cot
A= ^
0.
t
48. If B
= lOL + mB+ ny."
pencil formed (a, y), (8,
/3),
49. If the
0-f y+ 3<X
=
kl
then
mn
is
one of the anharmonic
ratios of the
by joining any point on OLd + kfiy = to the points where (a, ) means the intersection of OC = 0,
(<X, |3),
(8, 7),
triangle of reference is equilateral, prove that the a l 0. is a directrix of the conic /3
+/-8a
=
0.
lino
50. If
TEILINEAB AND AREAL COORDINATES 521 a = 2 be, prove thai two of the sides of the triangle of reference 1
are parallel to a pair of conjugate diameters of the conic whose trilinear
equation 51.
is <X
a
=
$y.
With a given
straight line as asymptote three conies are described is inscribed in 1, self-conjugate with respect to
such that a given triangle 2 t and circumscribed to 3.
and
centres of 1
The centre of 3 bisects the
distance between the
2.
52. If I, lt are asymptotes of a conic circumscribing ABC, l t 12 are asymptotes of a conic with, respect to which ABC is self-conjugate, ^, ?s are asymptotes of a conic touching the sides of ABC, then ?,, J8 are asymptotes of a conic circumscribing ABC. 53.
Show
bolas for
that the equation of the locus of the foci of rectangular hyper triangle of reference is self-conjugate is
which the
*
2 {a a 54.
Show
that
(
if
-a cos A +
cos
B+ y cos
C) + ap-y}-
the two conies *fix+ 4/my + */nz
1
=
0.
= 0,
ux(y + z-x) + vy(x+z-y) +wz(x+y-z) = are similar and similarly situated, they also touch one another* 55. A conic cuts the sides of the triangle at Z), D' \ E, E'
ABC F, F' and AD, AD' intersect the conic in dd\ BE, BE' in ee', inff. Show that the intersection of dd' with the polar of A, ee ;
respectively,
CF CF
f
9
with the polar of B,ff with the polar of C are collinear. 56. A'B'C' are mid-points of the sides, A"B"C" is a triangle formed by tangents at ABC to the circumcircle. Show that the six points of intersection of corresponding sides of A'B'C and A'B'C" lie on a conic. 57. A triangle is inscribed in a parabola with its orthocentre at the focus. Prove that its circumscribing circle touches the tangent at the vertex.
AB
58. The sides BC, CA, of a triangle are cut by the internal bisectors of the angles in and by the external bisectors in A' 9 B', C'. Show E, that A'B'C', A'EF, B'FD, C'DE are all straight lines, and prove that
D
9
F
the locus of the centres of conies circumscribing the quadrilateral A'B'ED is c (&?-$*) + y 0, a, /3, y being trilinear coordinates referred (aOC-fc/3) to the triangle ABC. 59. Tangents at the centres of the inscribed and escribed circles of a triangle to the rectangular hyperbola passing through these centres
meet in 60.
pairs
on the
sides of the triangle.
Through any point
vertices of
Show
a fixed triangle
P
PC
straight lines PA, PB, cutting the sides BC,
ABC
the perpendiculars through A', in a point, the locus of is the curve that, if
B
1 ,
are drawn to the
CA,AB
C' to
in A',B',
BC, CA,
f
C
.
AB meet
P
<X
a
sin
ABC 61.
a
A (3 cos B - y cos
C)
-f
a
sin
B(ycosC-Oi
cos A)
being the triangle of reference and the coordinates being trilinear. The normals at the points A, B of a parabola intersect at the point
on the curve.
Show
that tan9
C=
4 cot A cot B.
C
TRILINEAR AND AREAL. COORDINATES
522
ABC is cut by a transversal in DEF. On BC a point D D and D are harmonic conjugates with respect to B and C. The other tangents from D and D to an inscribed conic meet in P and AP meets BC in P'. If points Q' and R are similarly obtained, A
62.
triangle taken Bach that
in
l
l
l
f
show that
an<* -B' ar collinear. P', ') conic passes through the angular points of the triangle of reference and their centre of mean position. One of its axes is parallel to x = 0;
A
63.
show that
its
equation
a/x = ccos-B/y-f
*
is
wy + ** =
six lines
b cos C/z.
projection that the area of the conic
+ ty-f wz**=
t/ay'
The
65.
w's-f
is
Show by orthogonal
64.
(),
Vx
touch a conic. 0, r#-f m'y = Prove that z*-4xy** Q represents a parabola in areal coordinates
66.
and that the equation of the axis
is
+
67. If the equation Jy-f- m>a the equation of its directrix is
n0 =
represents a r arabola,
f (w + n*-2 mn cos ^4) fa-f- (n* + P -2 w7 cos 5) cap
4- (J
68. If the conic
wff*
+ ry* -f trz*
=
a
-f
show that
wa - 2 7m cos (7) ab y = 0.
touches at a finite point the conic
similar and similarly situated to it which passes through the angular points of the triangle of reference, show that t* + v + w ** 0, and that the conies
are hyperbolas. 69.
line
Through the angular point
AD is drawn,
in the points
the point Q. is
P
A
of the triangle of reference a straight
cutting the conic
and
P',
and
If a point C'
also cutting the line
is
taken on
harmonic, show that, as the line
AD
L == lx -f my -f nz =
in
AD such that the
range (PP', QQ') moves, the locus of Q' is the conic
70. Show that x, y, z being proportional to the areal coordinates of = at the a point, the equation of the circle osculating the conic x* 4- X yz
vertex
B
Show
of the triangle of reference
ABC
is
that if X varies the radical axis of the osculating circles at B,
C
passes through a fixed point. of 71. Prove that the equation of the conic inscribed in the triangle is the at circnmcentre, reference, and having one of its foci
linA^/oTcoiTA 72.
-f
sin
Find the condition that
by the conies w*-f v& + wy*
B/pco*B + sin C-/7 COR C >
+ /u|3 + i7- O should be 9 - 0. 0, u'tf 4- 1>'/3 + w'f s
0.
cut harmonically
If the triangle of reference is equilateral, prove that the conditions that the envelope of the line should be a circle are
TRILINEAR AND AREAL COORDINATES 73.
A
parabola has double contact , with
common tangent
touches the fourth
v^5+ v^h^-f- V^ Show conic
\flQL
528
+ \/w/34- \/ny
= =
of
\/J^cH+
its
chord of contact with the
^/m^*
\/*Vy
and and
0.
that the locus of the pole of
first
is
where
L = Wj na - w^
and similarly
,
for
M and
JT.
74. An ellipse circumscribes a triangle ^4BC and has its centre at the centre of gravity of the triangle. Prove that the radii of curvature at and that A, B, C are proportional to the cubes of the sides BC, CA,
AB
che product of the three radii of curvature is equal to the cube of the radius of the circle circumscribing the triangi, ABC. 75.
The
conic w^-t-t^-f
2hxy
=
meets the sides of
the triangle of reference in three pairs of points such that the lines joining them to the opposite vertices intei^ct by threes in two points: prove that 9
WfW-^fc-ll/ -!^*-^ 76.
The
8
= 0.
locus of the centres of rectangular hyperbolas with respect to
which a given triangle triangle. 77.
PQRS
self-conjugate
Show
th at the
si*
C
-X" is any its diagonal points. (XBCPS), (XBCQR), (XCAQS),
conics
common
a second
(XCARP), (XABRS), (XABPQ)*** meet QR, RP, PQ in. A, B, C v 78. The area of /L ue circle f
the circumscribing circle of the
is
a quadranf \i e and A, B,
is
other point.
is
=
>
A-ra
2
a
91
is
47rJ?
is
the discriminant of C. 79.
6 c
where
,
tf
is
(PS, QS,
RS
where
C =i- a*yz+b*z,<:+ c*xy+(fo+my+ *z) 3
point.
espectively.)
(x + y
+ z)
9
the radius of the circumscribed circle and
A
Obtain the expression f)c *
20
sin ^4
a (m + n' 2 mn cos A)* Imn
for the radius of curvature of the conic
~ <*
-f
77
+
p
y
=
at the vertex
A
of
the triangle of reference. 80. The triangle of reference being equilateral, prove that the envelope of the director circles of the conic whose equation is far 1
different values of
81. S110
^
fc,
is
the curve
that tlie radius of curvature of the conic
at the
'point
where
it
toucVs the
side
Oi
of the triangle of reference
is
TRILINEAB AND AREAL COORDINATES
524
82. Interpret
the equation a? -f y* = if in triangular coordinates, and chord joining two given points on this conic in its
find the equation of the
simplest form.
The chord PQ of the conic a?2 + y2 = z2 touches the conic or Show that P and Q are conjugate with respect to the conic
1
:*?'*
6~ ! y2
=
s2 .
A
83. system of conies is drawn touching the sides of the triangle o? reference and a straight line through the centroid. Show that the envelope of the centre locus, for different positions of this straight line, is the conic whose equation in areal coordinates is 0. >/y + 3-#-f ^8+x-y+ */x + y-z Find the equation in trilinear coordinates of the hyperbola -which touches BG and has AB AC as asymptotes.
84.
y
S
85. S,
&
9
a triangle and through that the pole of S& with the centre of either the inscribed or one of the escribed
an
S' are the real foci of
ellipse inscribed in
drawn another inscribed
is
respect to
it is
Show
conic.
circles.
Show
86.
that the equation of a 'onic circumscribing the triangle of
= -f + 0, where a, &, c are the sides of the triangle qp p(X ry and p, q, r are the focal chords parallel to these sides. 87. A pair of tangents containing a constan n e are drawn to the conic
reference
is
g\ inters* Action is .
utf + vfP + wy*
;
show that the locus <^eir
given by
S~
a constant and wtt 2 -f t>|3 2 -f tvy 88. Prove that any tangent to the conic
where
A:
is
(b
+ l)~
l
** +
(<*
+ 1)'
1
P
2 cut in a harmonic range by the conies Oc 2 + 13 2 = aO* -\-pfi* y 89. Find the tangential equation of a conic to which the triangle of reference is self-conjugate, and which has a focus &,t fa,
is
--
Show
that
if
one focus
is
^
.
BC the other is on :ko Jine B C to the opposite sides of
on
of the perpendiculars from
y
9
joining the feet the triangle of
reference. 90.
A
centre
;
parabola circumscribes a triangle and has
prove that 2 cos
A / -=
/
y^cos
.4
its
focus it the ortho-
= 0.
91. The major axis of an inscribed conic passes through tie point in which the external bisector of the angle A meets BC. Find the ^ocus of its
focus in trilinear coordinates.
The equation
92.
wo^-f t#
passes through the point
/-
15 \
+w -u v
K
5
-
w+u v
r
5
2
+ iosa
(1, 1, 1).
+ -w u
v\i. /
represents a parabola whi'e axis that its Vertex is the point
Show
TBILINEAR AND AREAL COORDINATES 93.
Find the equation of the centre of the conic up*
1 -f tjg
and the conditions that 94.
foci is
525
A
+ wi* + 2fgr 4- 2grp + 2 hpq
may represent a circle. conic touches four fixed straight lines
= 0,
it
;
show that the
locus of its
a cubic.
95. The triangle formed by the polars of the middle points of the sides of a triangle with respect to any inscribed conic is of constant area. = 0, where p, g, r are the 96. The four foci of the conic Lp* +
M
= 0, have as their equation B + tf sin* 0} - 22' {MN+ NL LM} + X*LMN = 0,
tangential coordinates of the line p(X
2 2 {L sin 8 A + where
M sin
2
-f
and 97.
+ g/3-f ry
2'=* +
Prove that the equation to the circle of curvature at the point = circumscribed to the triangle ABC is If$y + my0i + nOLfi
A
of
the conic a
(6
+ y2 +20ycos4)aZwn=
An
2
(tn
4-n2
-2wncos.4) (my-f nj3)(a
+ b,3 -fey).
described having the triangle of reference for a selfellipse and with its centre at the point whose areal coordinates conjugate triangle are (x, y z). Prove that the area of the ellipse = 2 nS*/xyz. 99. being the triangle of reference, three conies are described with 98.
is
%
ABC
B
iind G, C and axes &i,&s, 6 8 *
a2 &r 2
A
ami
y
Show
A and B respectively as foci and with semi-minor that the three will have a common tangent if
V V + V V V + # WW 2
*>*
*>{-*
- 2 ^i 2 cos ^4 - 2 ca cos B - 2 a& 68 8 cos C =
V
and that the equation of the tangent
will then be
100. Prove that the conic, which touches the sides of the two triangles formed by the pairs of tangents to any given conic from two vertices of the triangle of reference and the corresponding chords of contact, passes 2 2 =wW = 0. 1# through the third vertex if uvw 2fgh w/ 101. Prove that the polar reciprocal of each of the conies
with respect to the other, intersect in the four points :
102. Show that the points (-o^yi,**), (^i, -yi*i) (5i>yi -i) are 2 the angular points of a triangle inscribed in the conic a? -H;y*-f w = 0, = and for the conic \A^+ v^^y^ provided that
v^
self-polar
X#i + fiyi + y^i
= 0,
and w^-f ty^-f u^^O.
is circumscribed to a conic S and is selfan infinite number of such triangles can be drawn. drawn at the vertices of a triangle to a circumscribing conic are parallel to the opposite sides show that the osculating circles at the vertices intersect in a point lying on the circumcircle. 1
Hence show that
if
one triangle
polar for a conic S', 103. The tangents
;
TRILINEAR AND AREAL COORDINATES
526
The two points
(#1,3/1, Zi), (#2 >&>*) He within the triangle of Find the coordinates of the poles of the line joining them with respect to the four conies which pass through them and are inscribed in the triangle show also that an infinite number of conies, with respect to which the triangle is self-conjugate, pass through all four poles.
104.
reference.
:
105. Prove that all straight lines through a given point intersect the polars with respect to a conic Sa of their poles with respect to a conic Sl in points which lie on a fixed conic, and this conic circumscribes the common
self-conjugate triangle of the conies
S19 St
.
ANSWERS la. 1.
8* + 4t,~
3.
(4,
4. (5, 3.
12.
2.
(-K
4V3), tan-1
,,
(5^,
(-2, -3),
(6, 1),
^8"), Jir),
(4,
(4, ftr),
ir),
-1).
(2,
-A/2), (~Ja,
(a/^2,
(4,
Jw), (4,
V *).
(2,2), (-2,2),
(-2, -2), (2, -8). (8,2), (-3,2), (-8, -2), (3, -2). (2,0), (0,1), (-2,0), (0, -1).
3. 7.
Ib. 1. (1, 1).
5.
2.
Each side
xHy'-Ss-Gy =
7.
3. a.
-A
6.
Indnitely distant.
(1, 2).
10. 2>/2, 3>/2~, */2~6, 13. >/58, >/10.
Take AB, AC
i
9.
90.
4. 8x'-f
3y
2
156.
(~ 9 Jk 24f). 4:3; (-23,
11. (1,5).
15. a A 6 2
16. (a, 0), (ay's, 17.
is a.
0.
19).
12.
. a^j.
30), (2a, 60), (ay^, 90),
(a,
120).
for axes of reference.
1C. 1- (If, I), (2i, 3. JSfc 9j).
4. (2a, 0), (a,
-Jy,
2.
(0, 4).
(-1, ^3),
ayf), (-a, a^3"), (-2a,
5. (0, 0), (c, 0),
(c,
- C),
(0,
-C
)
0),
(0, 0),
;
(0,
-2),
-1),
(1,
(-a, -av/3),
(0, &),
(),
(a,
(
(2, 0).
-aV3;.
0)
(c,
;
0;, (0, 6),
.
6. (a, 0), (a, 60), (a, 120), (a, 180), &o. 9. (aa' + W/)-5- V(a* + b*)(a'* + 10. b'*). 14. (Jo, Ja). 15. 8, 15. 16.
^-8x-6y=0.
22. x
0.
23.
26
1(15^- 7) a*.
29.
4a
38.
{4(^-hx 2 -f^),
;
~
K^ +
33. xy yji
42. 4 (xj x,) (yf-f ya ) sin a>. 47. (84,51); 4: -3.
+ y,)}.
=
.
c2 .
41.
35. a + 6
.
55.
21. a:l^.
.
28 3 .
36.
c.
45. x - 4 y + 18
44. Equality.
:
C4x-a)H(4y-*6)c9
4ar.
:
1.
Bt. line.
x+y~l.
. .
2by
54.
13. 17. 5.
{(2a6)/(a+6) , 0}
2 y ; 9x (iii) 2 1. y ; 5x - 8y ; 3x (iv) 8x 50. ri r,tin(Ol -O.J+ r r 9 *iri(0t-$9 )+r9 rl &ia(es~O l ) 52. (aj-&)H(y-4)* = 25 ; circle. 53. 2ax+
49.
,
r*.
*
20. asec'0; a seca a ; y 24. x-y ccosecw.
80.
8- 150.
A =
234 2
7),
(x-a)H(y~6) 8
19. ** + =
7.
(-1,
{Sx-x^x^,
S
7y. 0.
0.
(v) 80, 16^, 15, 12*.
51. 2:1.
ANSWERS
528 56.
The orthogonal
50. x* + y*
63.
harmonic range
projection of a
60. 60.
I*.
is
62.
61. 3\/7.
5x-f6y9 +26ox+5a0; (-|a,
0)
a harmonic range.
(~5a,
3ax-3&y-f ab
0.
0).
64.
II a.
x-y -
2. (a)
(e) 3.
-
y
0,
0.
6. (a)
30.
8. (a)
80
12x-6y + 89 -
(b)
x . 0. x = 0.
-
y
(f)
5
120.
5. (a)
120. (b) 90. (b)
(c) x
0,
+y -
1.
x-2y -
(
0.
0.
80. 45.
(b)
90.
(c)
(d)
45. (d) 135. 2x + 3y - 12, 2x-f 8y -
(c)
7.
0,
lib. 1.
y-x
(i)
0.
(ii)
2. (8, 0), (0, 4),
4. i
f
9.
x
0,
80.
y
-
x/4-y/2
(ii)
1,
=
3.
7y-9x -
a.
(ii)
150.
;
y-f x
(iii)
5.
S&.
\/8y-f x
11.
=
0.
135% 45, 90. 126.
*
0,
x
=
12/5
=
2-a>/3
y
;
/6
;
17. (i) cA/8, c. (ii) c, -c. (iii) c, c. 18. The perpendicular from the origin on the line is 8.
30. (i) y y^8+x
0,
6.
= -3x/4 + 3. y = Jx-2. /52y/v 4/-V/5 = y\/8-x-3-/3 0, x 0. 8x/5 + 4y/5
1, x/ A
15.
-J.
0.
-4).
(0,
0),
x+y
0,
10. 150
13. (i) x/4 + y/8
14. 8,
(-8,
-2^, i,
|f f
6. (i)
y^/S-x -
4.
y\/3-f-x
(ii)
= -4
19.
y/3+x
(Hi)
**
i|. 10.
Ho. 1.
3. 5.
2. 8x-f4y-ffi = 0. 0. x-2-ffi(y--l) 4. 8x + t/ Q. x-f y-f 7y+fi /* 0; (b) (a)
5-f fix
0.
or x-f ^
0.
=
xcos(X-f ysinCX
6. y v^3
1.
x-f-^i
=
0,
Hd. 1.
2.
8(x42)-(y-8) = 8x-y+9 = x-f8y-7 (x-f 2)-f 8(y-8)
7x-4y-
3. (a)
(e)
4.
10.
(b)
4x-|-y+ll
(m-f n)y+2amn
ycot75x-5.
7. (a'?
0.
0.
x-2y4-l -0. 2x
0.
b'p)(cuc
=
5.
+ by+c) + (6p
xcos2a+ysin2a -
0.
(f) (x
0.
x-ya-&.
(c)
a)sin^
lOy+llx-lOO
(y
b)cos0
0.
0.
7
a^)(a x-f 6'y-f c')
*
0.
2p.
=
c'(ax+fty4c)-c(a'x-f d'y-f c') (ac'-a'c)x-f-(bc'-6'c)y -0 0. 12. xVS-f-y-4 - 0, x-y y/8+ 4 0, x\/8-f y-f 4 0. 16. x 2y - 0; x + y = 0; 2x+5y 13. adx + bcy+cd *(). 11.
0.
ANSWERS
529
2x-5i/ 2, 67x + 622/ = 569. 1, x + y 4(8x + 5|/-7)-3(4x + 6*/-5) = 2y-13 = 0. ~ 0. (ii) 6(8x + 5*/-7)-5(4x + 6i/-6)E-2.r-l7 ' l' 0. ro' 20. + m'; 17.
18. (i)
He. 1.
A,
3.
2. 10, 5^/J.
2.
ww: + nly-(rt-
i
lit I.
2.
-5), (i4, 80).
(-j|, J), (44,
II
6.
tan-'H-
afc-r-
x =
x
?/
2,
^3^
1 1(25^2), tan-
17. (0, aX),
21. (1,
(ta
-8).
g.
+ rofr + n^-r'^ + m*). 0. (4, 4). y _ 2.
- 3, * 4, x-2 + \/3 (y-1)
1,
=
7.
12. a|a a 4>& 1 &3 =
14.
3. (3,
3. (-8, 14), (-2, L4), .4. 6lx-20t/-l _ 0; 2^. 2 5. 56x-35t/+l67 * 0; 5x + 8?/-13 sin 8 + &cos a 0).
^. 4.
a
or (3,8.24).
2.
3.
11.
v^+1)
(8,
a;
^63 J.
10. 8f.
0. ;
a,a 2 -f
=
^^
0.
15. 100, 2^/6. 10. 63 + 18. a.
*= 0.
0.
16. (At 1&) 7y - 81.
(2^, 1A)
;
Jmn-M + w-fn).
23. 6u
av
25. u +
w=
=*
&i<-f
0,
0,
av
w-
M
r
=
2u
0,
a
=
0,
=
26. hy + kx
0.
20
6
=
24.
0.
wu
0.
t?
2xy.
II h. 1.
6. A-f/x II.
H.
2.
1J.
+
l/
3.
=* 0-
7.
Jor^
4. 77J.
5. (aft-f&c-fro,
ic^sinw.
3(8* + 4y-7) + 5(4 + 5.v-6) + 30(-y+l) - 0. 0, 39;c-18t/+9 = 0. lOx-y - 0, 29x-f 37?/~51 v-w = 0, M7-M 0, u-v ^ 0, 2u + v + u; = 0, 2r-M0 + t* = =
12.
0, v
13.
0,
i
0,
0.
IIL 1.
y + mx
4. (p-f
0.
3.
2.
(Z-m)(n-p)/(l-p)(n-m). -2r)j/-(2p-pr~Qrr)x * 0.
.r
s
+ y2
2a 2
.
5.
2.
3. 8x^(5^/3-3). 21x-77j/+53 = 0; obtuse. 4a-8y+70, 4* + 8y+l0.
4. 76 45'; 8.
12. ^(ISv^S). 15. (i) (m-f n)y
17.
60).
= 2amn.
(ii)
14.
8x + 6y
x + mny
2x-y .
6.
-f, 1(44-2^3). 11. 8,
2x4- 2y
6x + 3y
(w + n)a,
x/aoos}(^4-^)4-y/68inK^-f^) - cos 1(0-<J). = 6, 5x + 2y 28. (8|, 2|). 5, x + y 17x4-llvf 9 0, X4-4V4-5 . 0, x-f y4-l - 0. (J, -H). 1867 L 1
(iii)
16.
13. (2,
2
5.
10. 7:4.
8.
=
19.
J.
ANSWEBS
680 18. ab/(a sin
-f
22. xVfc' + y*/*8 25. x>*+yV&* 27. x2 + y2
& cos
{a (6-&')/(b 32. (0,
31. (1, 6). 30. (80, 86). 34. 33. r cos 20 = a cos 0.
35. r
21.
(-,
28.
hx+ ky
+ &')
2 M//(b ,+ &')}
;
29. x + a
-
0.
-mn/p).
1 | (X, tan- (1 + e cos
,
a*.
24. (fc-
V)-
i^s^^i-a*)-
28.
a*.
^ \/68.
20.
).
- */(* + *)' - I26.
)/ sin
(X.
f>cos(0-(X).
36. ^
= i(a+0)
37. (i)
or
w+$(a-f-0).
rcos(0-iaO = pcos^a; rsin
(iv)
88. 81x2 -34xy-31y2 + 44x+92y
{/(cos^'-hecos^^a)-^}
44. (-7,8). 48. (i) 4ax-|.c2
45.
57.
psin^a.
39. Jp'sina.
-r-
- 02x+y+5 = 0, x-2y-f 10 =
(i)
ca (a /2
c /2 (,a 2 -f
49.
.
1/
+ 6' a ) =
(8V%
{^(l-hSe cos a +
120-35y-fl2 = 0. 2 = c~a 2 (ii) s* +
sec 8
a (14- sin 2).
0.
58 .r
^ii)
:
~8.
= 83
67t/; 42 x
-
).
0.
- Jir. rsin^ = 2\/8. V3.rcos0 + 2rsin06V'8- 0. 60. x + 2 = 0, 7x + 24y-fl82 * 0. 58. *2 -f y*-2ay = a2 66. y2 - 4x. 64. 19x-2y-f-19 = 0. 63. (1, 7), -1. .
=
67. a2 71.
m
68. x2 /a 2 -f ya/&2
4xy. nu?
73. x -f y cos
0,
v
**
mv + nw
77.
^
69. y2
1.
72. w + 80
= 0.
p + q cos <w, x cos ai -f y
75. rsinCsin(^-f -4)
1/.
0.
2
ft
A/8rcos^72rsin^i2A/8 *
iff),
1
)}.
53. 147a;~121y-f 570
52. Escribed to third line.
56.
&)
(0
0.
(i) (ii)
51.
pq.
0iir+i03-a); *-w+iO&-a).
(iii)
43.
r(cos0sin0) -
(ii)
^
dsinJB;
=
-
61.
;7, 9).
4 ax.
0.
p cos w -f 7, x
pole
(x-a)(x-&)c = xe/(a~b) where
0^4
=
a,
*^. p, y d given line.
=
r sin
;
OJ3
=
2 2
-
6,
PQ =
c
and AB,
LM
are
axes of reference. 78.
2hx + 2Jcy
(i)
=
/*
2
-ffc
2
.
(xA
(ii)
-
fci/
- A2 -
fc
)
Ilia. 3. (xa
-y2 )
2
(x
-a2
2
)
(y*~a
)
0.
{(*-y)*-2a } {(x-fy-2a) -2a }(x-a)(y~a)-0. 8 = a, y = a. x 0. 7. x* y a, as a, y 8. xf -f-2/ixy y8 = 0; the given lines are perpendicular. 2 2 - 0. 9. (a) x2 +y2 +2xysec20 = 0. (b) x -fy -2xysec2d
4
2
2
2
-
5.
x-y = 0, x--v/8y = 0, x4-\^3y 45, 80, 150. xtan, y = z tan (ft + 60), y xtan( + 120); Of, OC + 60, Oi -f 120. (ii) y The lines u mv = 0, w-f mt? 0; these are harmonic conjugates of M = 0, t>0. 2 - 0. (a) x -8xy + 6y . 0. (b) x*-8*y+ 6y-6x4-9y-|.9
10. (i) 11.
12.
13.
;
a(y-a) -2ft(y-.a)(x-&H&(x-&) - 0. (-2,0); (0, -1, (0, -4); (6, -6), s
2
16. (4,0),
x-y -4; 8x+4y-f6 17.
0.
a + 6-0.
18. (a) 0/6 (c)
a-fc-
6/c
-
c/ d-
0, 6-frf
=
(
0.
b ) (^
14.
B.
(2,
-2).
15.
ua -f2
0.
ANSWERS 19.
531
(i)
(ii)
20.
aS-a^-oae^ + cd2 + 2y2 - 0.
d*
;
+ d*b-5a*d-a?b
22. a 2 Z2 + a2
21. x2 +xj/
w2 -
28. x4 -4afycot40-6zV + 4zj/8 cot40-f j/4
0.
1.
-
0.
Illb. 2. tan-l (- W) 9x + 12xt/ + 4j/ -6o;-4i/+l - 0. 5. 3^ 3. (avfytana)(x-ycota) = 0. 7. \/6. xs 8. 8as + 0. 11. f 0, x-3y 12. (i) (a&'-a'&) 2 = 4(M'-6'/z)(a*'-a'fc). - aa^ 2 + 4 (6V + a' *)(/*&' + aft') =0. (ii) (66' 2 14. xa -2xj/-j/2 -2x-6y--7 * 0. 13. x + 2aty + 2 = 0. 16. a tan 2 (X -2ft tan a + & = 0. 17. 60. 2
2
ft.
.
j/
i/
21. 2jp v/ft2 -ac/(a sin 2 (X
2 6 sin
cos
(X
a -f c cos2 a).
and fc(J2 -w2 ) - (a-fc)/m, or a/2 4-2Wm-f 6m 2 = and (a + 6) 2 + 4(a6-/i 2 ) = 0. 23. (a'h-ah')x* + (a'b-aV)xy + (bh'-b'h)y* - 0. x2 -2xycotO(-y2 -f AcosecOt 25. l/{(Z2 -m2 )8ina-f2Zmcos(X} 26. a&' + a'fc 2M' a&'-f a'622. a-f &
=
;
0.
;
31.
IIIo. 1.
8JT+4r=0.
2.
nid. oa .
2. 2^l'F4-a
a
-
3.
0.
2Jr y-ca -f aa -62
0.
4.
* 5.
X-.P -
2 7. 2T
6. 20,
0. 2
(i-->/8)4-y
3 t/
x 2 tan 2 0.
(i-\/3)~2-srro.
Hie. 1.
2.
4.
* (ii) a + 6 2 w + (te -fmy *+n)sina;4-\/(^ -2imco8a>). x a (ft-acosa>) xy(a-6)-j/2 bcosw) = 0. (i)
ul^'-f BB' = (^B'-fB^^cosw.
1
1
2 A cos w. 3.
(/i
Illf. 1.
ax+hy+g
3. If
4. Yes,
6.
fec-f
6j/+/
+ 2fttan0-fa */(- 6 )-
0.
;
have a
finite solution
;
one way
;
when (fy-
&tan2 (c) tan 20 ab *.
5. (a)
0; any number.
as-f fcy+gr
0. atan 2 0-2ft tan0-f 6 a () Always ( a ) and C b ) if ;* < a6
(b)
-
L
1
2
possible.
0.
ANSWERS
632
Illg. 4. (a) 9. (a)
- W-
The
A
and
-
10. xa -8i/a 21.
-H
(b)
=
C
.
a+6
(b)
*
or 3xa -j/ a
W=
0.
9 or 4J.
11.
12 or 48.
(e)
A
= 0. x2 -y2 - 0. and
harmonic conjugates are xa -f 2xy
pairs of a
2
y
xa -f5xj/-f 6y a
0,
0.
= 0.
Ax{x + XoHHb-f C}
25.
MJ.
(o)
0.
aa' +
28.
32. A'*- (J-J';(w
w')
or oo.
0,
Va. 1.
(i)
(1, 2), 2.
i/a + x2 4-2/2 -6*-8y * 2
(iii)
(iv)
(v) 4. (a)
62
3.
(-a, -6),
(iii)
c.
(iv)
(1,U
4.
V^7/U. 2 2 0. 0, (0, 0), (7, 7). (ii) x + y + 4x-6j/4-12 x a + y2 -4xcos^-4/sin^ - 0, {2 (sin + cos 0), 2(8010 + 0080;}, 4x + 4i/2 -12y + 5-0. x* + y* = a*, (a/v/2, a/^2;, (-a/^/5, -a//2). xa H-x{/4-ya + 5x4-7i/ + 9 = 0. (b) xa -xy + a -x + 5y + 8 0. a x4-\/2x2/4-2/ + (2-f3V'2)x4-(6-fV 2)y-f 64-3/2 = 0.
(v) (fa, fb), 2. (i;
(3, 0),
(ii;
.
(A, *)?
(vi)
(0,0;.
2/
<
(c)
5.
60;
6. (i) (iii)
8. (i)
(o;~4)
x2 + ya
(v) 4 (x
a
+
_
^13.
(8,2),
(x-3)' + (y + $) a
2
(i/61)
2 .
(ii)
+(y-|) ^(lV2; - 4. (ii) x2 + ya 9. l/
2
a 4-6
;
2
+ ya -
2.
-
(iv)
a
a
2
(x-2;
a
x 2 H-^
(iii)
196(x +y ) 2
(vi)
.
2
c2 .
2 a 9(x + V )
-
1.
07.
Vb. 1.
5.
6.
x2 + ya + 12x~10t/4-9 - 0; x2 4-^ 2 -6x + 8y-9
is.
9. x 2 -f y 2
0.
2 2 (x- ft) + (!/-*)* a (1, 2)(4, 3); /10.
,
(/i, Ar),
=
a 2.
(-1,
1).
2; (5, 6). c.
10. (4, 1;, \/6.
12.
a.
4.
0.
x + y -6x-8i/ + 20=0; /5; (2,2;(2,6); (2, x2 + 20x + c~ 0; x a + ?/ 2 -|-2/y-a2 - 0. 7. /*
8. x 2 + ya ~ax-&i/ 11.
-
a
2
x2 + v2
-
4 a2/
2
14. 8x + 3y -f 10ax + 8a 0, (~8a, 0; (-Ja, - 0. xa + ya +12x + 16y + 75~0, outside, (-3, -4)(~9, -12). / 18. 8^-28^10 5x 2 + 5i/ 2 -8x-My-5 - 0. 0; 23iv 209:16. 0. x 2 -hy2 -8x-f 1 = 0, x2 + ^2 -5x + 4y-f 9 21 ^ * 1 x2 + y 2 -(a + cV) *-(*> + cV & ;^^ c2 " 2
2
15. xa -fva ~ac-5y-f4
16. 17.
19.
20.
(iii)
xa -f ya -2rx-2ry+ra
25. x a --^-(a-f 0)(x + y) + 0^3 27. Circle through
29.
-
-
22. (-1, i), i /TO, (-2i, i), ^2. 23. (i) (x-a) 2 + 0/-6) 2 - (A~a) 2 -f-(fc-6 ; a .
2
&(J
+
2
* )
B and
am + W.
C.
(ii)
-
x 2 + ya -2ajc-2ry + a 2
0. tf.
.
26.
x+y-7
28. x a + ya -10x
=
0.
0.
.
0.
0)
ANSWERS
688
Vo. =
0.
4o;-3y== +
3.
j/-4 = 0; 8x + y-f 2 0; 8x + 4y * 4. 3x + 4y = 0; (f, -2).
2.
10.
10.
6. (2, 1). 7. 7/V2 8. ?.x + 5y 9x-2y-5 0. 29, 10. 2x-5y + ll = 0. 83-y~8 * 0, 8x-y+2 = 0. 11. x*f ya -2x-2y + 1 ^0;ix s 4-v2 -l?a:-12j/436 0. 12. (8co*0 + i, 3sin0-l-2). 14. lOx = 17 y - 2. li. 8a?-4y-50; 3x-4y+15 = 0. 16. Circle; straight line circle. 17. Circle; 3(xa +ya )~14 (x + y) + 26 18. x'+y* - 25. 0; (1,8). 20. xcosOt-f ysina 21. 2(ai-a 2 )x + 2 (bj-^y r. a^ + ^-ag3 -^-. 2 a a a 8 22. Concentric circle. c 23. (x + y -rtx-fcy)* fixed point. (x +y ; (A, 24. x2 + ya ~30x + 80i/-|.225 *0; xa + ya -6x4-6y-f-9 = 0. 26. 8n 27. (x 4 Va ) (p3 +/ a - c) 5. Inside
.--
;
9.
;
;
=
fc)
,
Vd. 1.
2.
litf.
(10, 5)
14.
xainCX-ycosa = x9 -|-f/ a -f ay-ox = 0.
19.
PA
5.
(^5, 10); 8.
a.
3. (3, 4),
(5, 15).
13.
3f
(4,
-8).
za -y*0.
Ve. is
a diameter.
Vf. 3.
x y xa
4.
(x -fy
1.
=
5
0,
+y9;
a
a
a
xt/H-3^0. x3 -f ya
-ax-6y) a
==
+ ya -ox-&y) a
7.
(x
9.
The concentric
2.
a)sinO(
(x
(y
b)
cosCV
=
r.
49. 2 6. 2*-2y (r^-hKy-lZfryHC-Rx-ry-JSi) 8. xa + ya = 2r (xcosa.
8.
ra (xa -fya ).
circle
a
8(x
+ya )
Vg. 6.
a
n(xHy ) -2y2a.
7.
|(x~r)
(x~^
10.
xco8(a4-^ + 7-f8)-fysin(a + /347-f 8) - 2 r cos (X cos cos 7 cos 8. 1 2 is the origin and circle xa -f ya 2rx. y 2rx, locus is a;
12.
13. If 14.
A straight
line.
Vh. +ya -a a )
+ a -aa). (ii) xa + y8 -2ay = 0. 2>/8ay-a (iii) 3. xa -ya + 2fcxy = a2 4. x2 -f ya -4ax + aa ^ 0. 1.
(i)
fc(x
a
y(A
xa +ya
a
fc
a;
xa + ya +
14ay-
a*.
8
.
8.
(i)
(Mi)
Vi. 1.
(8, 0),
(-2,
0).
3.
xa + ya ~a(X a -f X-4 )x~2a(\-A- 1 )y-3a a * 0. { -a, xa + ya -7x-21y-f 120 = 0. 4. xa + ya -tt*-.4y+ 8
6.
xa + ya -6x-8y-119*0.
2.
L
1
3
a^-*- )]. 1
=
0;
(4, 4),
(2, 0).
ANSWERS
584
Vj. 1.
(j*
4V- !*)(* + *-!)
7.
(*x + fcy-r
+ y + 20&+2,/^+2c-0*-./* 8x+4y = 7, i
4. 8?
2 2.
3.
)
5. 8cc2
0.
90.
+ 8y2 + 12x-18y+ 85
*. 0.
Miscellaneous, p. 174. 1.
0x-K/fc+c
5.
2x-f y z2 -f y2
8,
= =
= ;
0.
2.
x-2y =
2 j/
a2 + r2 where ,
7. (
-
3.
0.
A straight line.
13.
8x2 -8y2 -f 2at/\/3-3a 2 =
0,
where vertices
2
2
p(x -f )-Kp(n-ro)&-f (rwn-p 2 2 2 2p(* + )-(ro-n)px-(wm+|> )j/ - 0. mn/p)
of triangle are
2
t/
;
(15,20).
A circle.
+a, 0) are the given points.
10.
15. (0,
-*!/-x-f 4
0.
)
mnp
y
(a, 0),
a\/8).
(0,
0.
2/
3
16.
(ac1
18.
A
+y12 -a2 )(a;a2 4-|/22 -a2 )
2 2
(xj^-f j/,^-a ) circle touching a line parallel to given line at the given point. .
22. r8 -(a2 4-&2 H-c2 )r-2a6c
20. -1, -3.
23. zx' +
0.
t/y'
- a2
.
Vk. 1.
x-
3.
x2 + y2 + 9x + 72/-6
5.
x 4-y 81. e. x 1, x -f y 10x 2 -flOj/2 + 18x-30y-80.
7.
x2V2y
1,
2
2
=
= 2
2
2.
8.
4.
0.
x2 +y2 + 4x + 4y-l
*
=
0.
= i2\/2y.
24
21.
+ 2 -f2flrjC'f2/i/ + c)(aH&w-l) (a 2 + 62 + 2ga + 2fb + c) x 2 + 2 -f 8t/-f 9 13. A + /I = 0. ilO\/22/. 2 2 2 2 == 57 . -f + 67^7 ) 16(s 8(x ) xa -f t/~ 2(^(12 -f-b^Cai -f ag^ + b 2 = 0. 2(a ~a 2 )
22.
A
9. (x2 11.
18.
20.
|/
(to-f wj/-
j/
j/
.
2/
1
circle.
23. (x2 + y 2
~
24. (a2 -f b2 )(x2 + j/2 )--2c(ax-f by)
-
0.
VI. 1.
3.
4. 11.
x2 * j/ 2 -6a;-4y-U = 0. 2. 19x4-8^-12 2 2 x + + + a2 + 8a6 + &a j/ 2/) (a+6) -(2x {a+6, J(a-f6)};
(8, 2),
(-2, -1), (-gf-l-lX/a,
(0,
-8).
5.
v'^a' +
fc
2
- 0. - 0^
-
2 2 2 2 2 -/-fmX/a) where X y 4-tn )-f2X(flfH/m~an)-f ^ +/ -c
Vm. 1.
(x-l)(y~l) = 0; (x-f2/)(x+y-4) =
6. 7.
8.
0.
2.
56x+17j/-5 =
0.
(-2J 2 +(t/-S) 9 - 5. (2hm-la + lb)x + (2hl + ma-'mb}y + n~0. 2 2 a;2+^~86x-46y-|.824:0, 25x + 26y -80x-4942/-f 64 - 0. x2 +|/2_2(A;-f k')x + 2Kk'v/d + 2kk'-cP - 0. 0. ia 91x2 +91y2 -240x-866j/+504 xa^ya^ex-lOj/-.?
8. Circle.
p. 12. 4(na*
5.
0.
-
0.
ANSWERS 13.
4x+4y-ll*-llv-28-0.
14.
The angle in a semicircle is a right angle. S(0,g)(7,r)sin03-y) = 0. a 2(a + b*)(ab-c*)(bx + cy + a)(cx + ay + b) u + kw ~ 0.
16. 17.
22.
585
21. uv
0.
-
to2 .
Vn. ra
1.
rrj
cos (0
6.
9.
(i)
= 0. 2) a = 2#cos rcos(20 l ~0) 0,.
rr 2 cos (0
0j )
rwos(0-a) ra + 4arcos0
2.
R.
3.
=
12a 2
(8rcos0-2R)
(ii)
2.Rrcos0
L3.
A circle.
1.7.
The
-=
fc
2
12.
.
rcos^-f rf
of such circles lie
a==jRa sina 2a.
0.
circle circumscribing ult M 2 , r cos 0j cos 2 cos 3
The centres
_
0; 4a.
sin a a+9ra sin a 0cosa
a
9r2 -12flrcos0 + 4.Ra 8in a 7
LI.
cos (^
2) 4- r\ r*
=
;
us
{
-d
\/cP + 2ad, 0}.
is
a cos (0
by fours on
0j
circles of
2rcos01 cos0 2 cos0 3 cos04 = acos(0 There are 5 such circles, &c.
^3)-
2
which one
0i
2
is 4 ).
S
Miscellaneous, p. 217. 2. (a, 0),
4.
The
(aSnhtn,,
0).
= 0, te-y = 0. w. r. t. the circles lie on s + 2c between the equations of the pairs of tangents from these poles to
poles of x + ly
Eliminate
2
the
circles.
11.
{afcV(a
12.
Two
2 2 2 - c2 2 2 2 /(a 4-6 -c )}, fc c/(a + 6 ). coincident lines through the origin. 2
+ &2 -c2 ),
6/c 2
+ xt/-2cx\/5ct/ + c2 = + -.pae-gt/ = 0. - (a 2 + f 2 )/f, if x2 + y2 = a 2
14. xa + y2 2
22.
ax \/2
0.
i/
25. x
28
19. x 2 -f 2/2
0.
2
2 /ct/
=
r (A:-*),
35. y2 (49x 4
+ 49x2
0) is
(A;,
2 t/
-4y4 )
is the circle, (, 0) the fixed point. the mid-point of AB. 30. 6. - 0. 38. Circle. 39. x 2 + t/ 2 + 2gx + a*
=
0.
Via. 1. (i)
Parabola,
(ii)
Two
real straight lines,
(iii)
Ellipse,
2. 5.
6.
(iv)
(vi) Two imaginary straight (v) Hyperbola, 2 (-$, 1), (-6J, 3), 2x +16s+27 , 0. 2x + y = 0; 2* + y - 12. (1, -2), (8, 6), (0, 0), (4, 4) 2 /** + 6v+/= 0. or 8. 0. < C/ -^) >,
straight lines,
Two
parallel
lines.
;
'
,
9.
VI b. 1.
Parabola.
5.
Hyperbola.
9.
Two
2.
3. Circle.
Hyperbola.
6. Parabola.
parallel straight lines.
7.
10.
Hyperbola.
Hyperbola.
4.
Two straight 8. Ellipse. 11. Ellipse.
lines.
ANSWERS
586
VI o. No
2.
(-1&, -A),
centre,
-2), no centre,
(4,
(-1A, -&),
(<>>
1).
3. 1, 6.
Or-
+ 4a5y4V-2*-f 2y + 2-0. (v) (2fc + y-2)(sc + 2y-2) (*-y)(7* + y) - 0. (x) ( + y-l)(2*-y+l) - 0.
6. (ii) a (vii) 8.
2y^,
(vib) (4)
V%
(8) 0. (vib)
10. tan
tV^.
*/2/ll.
(8) asi+lly
20
-
-
1
(10)
/
no
A
15.
iV^tV^l;
Jy^,
(7)
(11)
iV^.
1^/7(24 + 8^
/
(11) (3-V 2)y' = S. 12. Centre and line at infinity.
2.
2ft/(a-&).
13.
2y^. ^2 + 2^10.
IV^,
(5)
0.
2)H(8-fv
*/l~^K. -i); (1, -1), (2i,i); *-y-2 - 0, Sff+8y~4-0; 0. 8a^+ 12a:y + 8va -6a;-18yH-2 0; z-f j/-l 0, *-f y-% 1,
value, positive, negative.
16. (2, 0), (0,
2;
(4,
17. af* +
iv'S.
1),
-1*), Vl; 6x-f 8y+l am* -2 Mm + W8 - a5-^9
18. (|,
19. (i)
x~2y~4 -
7
.
bg*~2fgh
0.
0.
.
(a6~^)(a?4-2Wm+5m2 ) - ^(^-m^-Ca-d)^] 2 8a;4-4y - 10. (H, If) (0, 0), (2f, 8J) 4x-8y -
(ii)
20.
.
;
;
;
22.
+ 4y-15 - 0. 2 = 4 2 2 a A(x ~y ) (a-&)a^; r (aa~/i )-(o-f6)r + l - 0. = 23. \/|. 2*-8y-2 0, 8a? + y + 2=0.
25.
(-tHt -1%); 20s-86|/7-l
a~7y+100. 8x+4y~50;
21.
7x + y
10
;
8
7 (v) 2s-f y C as axes; xy
0.
26.
+ 4y-l
-=
=
15, 2a-f-t/
5.
1. 4a-8y+ 0, (5/8, 18/8), 4*+ 4y-7 - 0. aV% iai/5; x-y - 0, s+y-0; (|a^5, |V^); IV^.
27. Referred to
28. 81.
The point
37.
3
1
of intersection.
Vila. 8
3.
4.
(acot ^, 2acot).
6. y
4a(to4-wy).
90.
10.
16. (i)
oi8.
4a?
(i)
-
r-4&y. -1J); x + 20;
1.
6. (ii) Z-f
l + am* am* = 0.
0. 7.
yV/4a.
12.
(-2,
(ii) (iii)
y-1;
(81,1); (4, 8);
(-2, -i),
(4,1); x
x-4 -
;
(4, 7J)
1-0;
(iv) (J, 1J); x-,,+ 19. (acot 9 ^, ~2acot<^>).
;
4y+9 -
0.
8.
2y-17 -
0.
(1,2); * + t/-l
-
0.
20. a-f ycosw-f acosec'w
2y-s-4 -
21. (4, 4), (i, -1), 25. (A, 4* + 8y+9
-
0,
2y + 4* + l
.
=
0.
0.
= 0. am8 (ii) aZ' + 2alm2 + t8 n 0, ac 0, y8 a(x-a). 16xa + 24y4-9ys -140x-f20y0; (1-6,1.2), 4. sum of perpendiculars f>om focus to Prove sum of radii
-A);
27. (i) In
28. 32.
,
normal.
Vllb. 7. 11.
2:1.
When
16. y'
9.
orthogonal the envelope reduces to the focus.
tangent and
ANSWERS
537
VII c. 3
+ to-crf-o*
=
2.
0.
1.
y
3.
(a;
5.
y-ao;-aft + 2a*-
-
-<*)(*+ cos (w) + (y-2<#)(l + fcosaO *
=
20. 6y2
(a
0,
xy + ak
+ &)' (* + a).
0,
0.
7.
y
-
2
16
{(-2)/15}
s .
^~6a; -JF.
23.
VII d. 1.
4a>/2;
3.
Ax-By + a + 4aA0.
(5 a,
-2a)
;
/2.
4a v 6.
2.
(3a,
z9 + y-10ax+ 9a 2
-
0.
+ 2a>s/3).
Vile. 2.
_ fy + at*)* -(! + <*)(?' -4 ax)
3.
(*
6,
&H y -20aa;-f 60a2
8.
a
4. A 2 + 40A-f 16
0.
=
0.
0.
sHt/ 3 -(8w a -f l)ax-f m(rn a -8)ai/ + 3a 2 m 2 = a (a~^) (a;-a).
0.
10. 2at/ 2
12. x 2 4-V 2
13.
-8ax0.
The point (-3a,
0).
VII f. 1.
8/125.
2. (i)
7.
-v/Sy'-f
=
0.
s
(ii)
t/
=
9.
(8a. 0), (8ia,0).
a;,
2
(iii)
t/
=
2x.
(iv)
-
VII g. 1.
A straight line
2.
A
3.
A
(2oc/a +
2y/u-l -
0).
0). straight line (as/a + y/ft-l straight line 2(b + acos
=
a6.
Miscellaneous, p. 290. 3.
17.
15. x-4a = fcy. 16. ai*-f 6iy + a?bJ - 0. x2 + y2 -6a;-4y.-.80, x 2 + y2 -22*+ 32y-f 13 = 0.
18. y*
+ y*(x-2a}(x-c)-a(x-c)* =
x+\+2y =
0.
0.
2 2 a a(x -fy*)-(y' -f 2a )fc + /(s -a)y+aa;'(2a--a;') - 0. Firfe } 7, 2 2 22 2 3 ax - 3 a 2 ) * 0. 22. -f3[* + y -2fc-fc(J<;-V)] = Oj if ft = a, (* -f a) (x -f 24. /c(y-^) 2 f2(ft-a)(^-ft)(y-/c)-A?(i;-ft) 2 *0. 45. Maximum or minimum according as the ordinate of the point is > or /
21.
^
?/
<4a. 47.
(i)
16a^-2a) -
y*.
(ii)
x+a
~
0,
48. (a + a')V + 4&2(a f a')x-4a-'.''>* 0. 2 2 60. y (y fc) (* ft)+* ( A^ + a (y fc)
To prove the second pnrt, find the 0. value of x for the foot of the perpendicular on the tangent at X, and then the condition that A should be real.
ANSWERS
588
Villa. 2.
2a-8y-l -
11.
17.
VI.
4,
0,
{\A/O + 1)
13.
V5/8
9.
0.
divides
;
AB in
ratio 8
:
5.
0},
(a
-
20. a(z 2 + y 2 ) 25.
7a;-12y+ 1 -
10s
+3
2
1/
A:l
28. If
>(a2 -p2 )(p 20* + 2li/ = 0. 21.
1.
=
2
187; (0, + -/1809/80) ; the ratio, 1-f 2/(X-l) =
is
2(l-e
2
2
2
sin2^),
)/(e
X greatest ,with
.-.
sin 2^.
VIII b. 8.
aa
:6
a
16.
.
2
16(a x
+&
2
2
y
=
2 )
2 2 2 (a -fc ) .
a + 52 sin2 a)*, (a2 sin 2 a + & cos 2 a)i. 20. p. 822. Pole of chord joining feet of normals lies on director circle. 28. * 2 /a2 + 2/ 2 /6' = 1/2. 82. a 2 a;V(a 2 H-& 2 ) 2 +2/2/4&2 - 1.
17. (a2 cos 2
37. tan**
43.
- (l-e4 )/(2e2 -!).
t(2mr-a), where n =
44. o 2 v/a 2 -262 /v/a 4 -6 4 47.
then (a, 48.
/3)
lies
on
y2a 2 -6 2 /\/a
62
;
The chord joining the
-
41. (* 2 + 2/2 ) 2
4 (a 2 * 2
+ b2
2 ?/
).
1, 2, 8. 4
~l> 4
.
two normals
feet of the other
is
0;
x/x'+y/y'+l
this.
The fourth normal from the point
;
aj/sin
J)+a2 b2 =
0.
VIII o. + v2 ) 2 -
3. (** 5.
11.
a2 x 2 -6 2 i/ 2
The vertex
xy\fa?-b* SMa =
12. (i)
15- (J/i
0,
2
26. sin 32.
2 2 -(a ~6 )2/ = 2
z^
xrf
2
2 2 )
(") S
-I2
2
20. z /a -j/ /6
cos i (0
-
2 )
a (a2 -*2)
-(a
2
2
2
2
(x /a -j/ /b
2
).
s
2
4-b )tan */6.
0.
*iW4 2
ifir).
*>
10. (a 2 -f & 2 ) sec 3 0/a,
=
21. axcosecfl-itysecfl
24. 62
4. (a2 +
.
(a, 0).
a 2 -f6 2
2
=
2
(a
3M
a f,
M ~62 ) 2
14 ^-l/2 =
-
2 2 2 -f6 ) /(a
-
B^/2.
.
23. (a2 tan 2 *-f 6 2 sec 2 0)f /a&.
.
a2 6 2.
- 0) -f sin
(0
+ 0) -
2
0,
(z /a
2
- y2/62 ) 2 =
x2 /a2 4- 1/2 /6 2
.
0.
VIII d. 8. (3
4c 2
+V
12. ax -by 27. a 2 62
0. 2
(x'
+y 2)
2
8. The hyperbola. The centroid of XJkftf is P. 2 2 4a2 z2 -46 2 (a 4-6 ) (b^xx^a^yy^)
(x
;
is equilateral
hyperbola 32. X 2 sin* cos*
-y2 ).
11.
/,
the centre of the circle
where
(B sin
+
1/
cos 0) a
=
is
|/
1.
8.
28.
8
(78v 5)-4y (2v 5) /
/
2
*V 4 +f/V&4 - V(a 2 -f ba ). 2 -y 2 -2a5ycot2(X - c2 .
12.
2.
6. Circle.
=
2
(a 4-6*)
the point P.
4/sin0cos0.
VIII e. a
* 2 /a 2 -i/V b2
= i1
-
14.
s2 sec a a 10.
2 .
When
the
ANSWERS
689
Miscellaneous, p. 371. 1.
A circle touching the
2.
ly*-mxy+nx
0.
=
3x* + 4j/ 2 -4ax
3. (i)
given circles at the point of contact. x = 0. (b) ly* + nx * 0. (c) y ** 0. 2 0. Sax. 5. The line BD. (ii) y
(a)
-w
w
+m
2
2 10. (1 + W! 2 ) (w a a ) (wia-wO : (1 8) (1 + s ) (n^-m,,). 11. If the fixed point is the origin, Z?(a + l>}-2gX-2fY+c(X* +
13.
:
(y--l)(4x+y--8)+X(a;-ft/-2)(5x + 42/+l); when
hyperbola. 14. 2x 2 -83/ 2 4-8x2/-5x + 20i/-12 16.
x a + t/2 + xt/-(a.f6)(x4-!/)-fa&
22.
2
34.
0; x-i/
=
0; 8x + 8y
tan-^Cl-e )-*. x2 + y2 - cos (2 a2 + a2 sin 2 B + 6* cos 2 0) x/a - sin (2 62 -f a 2 sin 2
39. 41.
\/f
;
foci lie
49. (xVa 2 +
2 /
45. 2
/6
-f 6
b2 3 2
4
F 2) =
(iii)
52.
ax-by =
is
an
+a 2 y 2 =
- 1) (cos O^y- cos
/3
a2 6 2
2
= 2a + 26.
0.
sides.
47. x 2 /a 2 + j/2 /& 2
.
1/2.
cos 5)
-f {x cos a/a -ft/ sin 0(/&~cos/8} {xcos7/a + j/sin7/6 Max. a 2 (x 2 /a 2 + !/V& 2 -l), m in. when OPQ a tangent. Min. b 2 (xVa 2 ^^ 2 /^ 2 - 1 )* max when OP9 a tangent.
50.
0.
the conic
cos2 0) y/b + (a2 -f &*)
on the bisector of the angle between the
43. 9(a 2 + 6 2 )/2.
=
0.
8fl5&sin^ + 8t/acos0 = 2a6sin0cos0. ~ 0. (Oi/x + p/y^^/at + yt/b^ + l
36.
X
cos8}
=
0.
-
65. ax 2 -f ty 2
56.
54. a 2 x^ 4- 6 2 yx l
0.
o/o'
=
(a
2 -I-
5 2 ) x,
^
.
+ V^-
A common chord of the ellipse and circle is a diameter of the ellipse.
65. x 2 -f 2ft:ct/
y
=
2
2/t/
(rectangular coordinates).
IX. 7.
Two
8.
^V2.
29. r 2
through each focus
(l-e
11. 2 )
Z
2
2
(rl
;
-
+ 2^rcos^-^2 =
0.
36.
ccXl-O^S^^+OC 1 -^)-^^^-*- 62 )^
44.
I
sec 2 J 8
=
r-f re sec 2 J S cos 0,
i.
e.
39
-
-
A circle.
a confocal conic.
F is the focus of the conic, which is a parabola, ellipse, or hyperbola the radius. according as the distance of the B line from F is =, <, or 0. 48. ra - 2 r U cos 6 4 Z 2 (e* - 1 ) 47. It touches at the point Oi.
46.
>
49.
c2
-tr' cos (0-0')>
ra
-2rccos0-fca (e2 -l).
Xa. 4. ^Za -f WiWa 6.
0.
5.
tan-U^iWa
7. (a)
8.
-
A point at infinity on kxhy
-cosa/p, -sin(X/p.
10.
0.
(b)
^m^l^m^ =
A line through the origin. 0.
ANSWERS
540
Xb. 8.
tan~ l (l 1 m f
**
6. *! (tea a -f w^-t-Sa^Xi + myi -f tUj) 8. as ty + l - 0, a'x-by+l * 0.
+my
=
+
10. (&!-<*!, 0).
:
27
:
{&-&', a'-a, afc'-a'fc}.
19).
a2
12.
IS.
11. (40
0.
0.
|-
14.
15.
Z;
17.
(2a-
'
16.
(&-&')
-
0.
18. 63
*S
19. 21.
20. (12, -1),
-
{/1
22.
m = 0. aw2 -Z-0. ^(^-f 2a a ) + w(b1 + 26 2 )-f 3 =
23.
-
24.
26.
-2&,)~l
0,
25.
0.
0.
Xo. 1.
- cos } (0 +^/r cos J
5.
Uo-f m26-f 3 -
8. (2, 1)
;
12. a 2 (P + 18. Centre
19. a*(J 2
m2 ) (c,
6. aa -f^
10. r* s 2 (p
2. 1,
+ m 2)
1
- g)* ^ + m*)
circle.
0), radius
-si
*-<*>);
0, c/3.
c,
13.
=
ra .
The points
common
7.
[(ps
1/2 a,
l/2a.
- ^r2 ) 4 s* -r2 ] 2 1
are (0,
.
the centre
c),
(ft,
point at the origin.
(see Question 18).
Xd. 1.
a2 d2
9. (1
1 (J
-f
m2
+ 4aJ)tan
4. Circle.
a2 H- 6 2
)
2
(X
6.
;
circle.
- 16a(am2 -J)
A parabola.
;
the poir.t (4a, 7.
A confocal
3. $ 2 -f
0).
2ap
parabola.
8.
10. 11.
12.
14.
-t 2/a(t-f2< 2 ),
3
<
/
A parabola whose focus is midway between the centres of the circles. A parabola touching the given lines. 13. am^ + = 0. a a lm(AtJ-Ar/i) + a -ftn -X-^w-0. l
Z
2
<4
(al+l)(bm + 1), a conic touching OJT, OF. + &-fc)?mH-2+m - 0. 6-f c)iwi l+m = 0, which fire parabolas touching OX, (iii) (c 2 17. 3a(? + w*)-H(aJ-l) - 0, (-a, 0).
16. (i) (ii)
c
Zw
(a
OF.
0.
0).
ANSWERS 18.
aP + b4 tn* - a*,
V^bVa,
{
541
0}.
19.
20. (*, 0);
..^-
(0,0);
a^^+^m,
23. 22. (4 a, 0) if the focus is the origin. 27. Coaxal conios. 28. v 2 -Mas+ 8a 2 0. 32. (4 a, 0).
29.
0.
A
point.
XI. 3.
(a
6.
(a-
8.
A conic.
18.
0. 18. x - 0, x+Sy + (ab'-a'b)y+2(tf -a'/) -
2(a*'-a'*)
(<^-a7)a*+2(/i/'-*y)fl*+(^-b7)
*
21.
a2 /"2 *!/ 3 / 62
24.
The hyperbola
26.
28.
30. 32. 37.
referred to
The vertex
= 20.
ax*-2bxy+by*
0.
-
o.
23. (0, 0), (4*, If).
ellipses.
*>
9
the
normal and tangent
of the parabola is (Jb,
at
point
+ y8 ~(a-f'Z>-fc)(a; + y) + (a&+&c+ca) = 0. - bV) 2 - 0. a' *) x* + (a' 6 a&') xy -f (*6' (afc' 2 a 2 2 + -f?/ 2art/-f2y -2) -0 where *2 -f 18fc+20 ^ 0. -2a;-2j/)+A:(x 4(a? 2 28 2 2 ~ 0. 8 34. (hx -f 6y) 2 - (ab - h*) 2 U + 86 K ^ /a + yy 6 The tangential equation is (oZ + ^n-f 1) (ai + bm-l) + X(! 2 -fm 2 = 0. a;
-ffl5t/
!/
t/
;
40. 2/tx-2ay-f/= 0. 2 c/(l-f cosw)}, 2c /(l
+ c6sw),
+ cosoj) where
a-f 6
^
4Z 2 + Jf2 -4.R 2 -
0.
48.
V2(d i +(X+/B),
^(^-fa 2 -^ 2 );
Note that a6
A
x
7i
2
Jc.
l)
.
45.
(a, ^),
(-a,
-
CS- A 0,^ =
-/5).
0.
(xV)
A parabola ABODE touching * a at B(o, So), y at s+v-0 at D(-a, a). -4B(-f + -l), Be(+-^), CD(4---),
56. 8a\/10/25. a),
0.
4c and use the tangential equation.
The conjugate diameter is a parallel to the polar of 53. See p. 428. the centre. through re
15-f5y~6a = 0. * 0. (ii) (*-V) - (a-a')(b-&')(i) (a-a )/*-2/ r(*-V) + (b-b =h ^ 2 (a52 -a 2 ) + (^2 -a'2 )v 2 - 0, where Xs -fy 2 - a a is the circle and -
Axis,
).
60.
2
)
- gF. /"Gf 42. 2ab; (c/(l 44. LM' + L'M
69.
is
2
39.
61.
the
Jb).
/
/
!
)fif'
fl
as
the
fixed line.
64. (^iCi-Bi 2 )* 2
-^^-* >!
2
C1 -2B1 JB2)*y +
are the minors of a ly b ltt . in f
I
I
111
U
a
C2 -Ba 2 )y
where ^,, Blt ..
(iv) Ellipse
and hyperbola.
No. "
2 2 abac a 2 (b-f c)y + a 2 bc = 0. 67. (oy v be ) 68. (i) Straight line, (ii) Circle. (iii) Circle,
p-tana 0, where
0,
20 is the angle between the lines, Jg distance of their point of intersection from the point midway between the poles.
ANSWERS
542
XII a. 1-
~i, *)
(J, i, I), (ti
4. (i) (1, 8 f -8). 7. (i) (i, i, i).
(ii)
2. (-4,6,0), (1,6, -8). 5. 2:8:4. (H, -A, -i).
3. (0, 1, 4).
6. 18
:
2.
(ii) (i, f,
8. 9. (i)
(1:1:1), (a:&:c).
A
cos B
:
(-1:1:1), (-a:6:c).
(ii)
cos C), (sin
2A:Bm2B: sin 2 C).
(iii)
(cos
(iv)
(secA :secB:secC), (tan ^4 tan B tan
:
:
10. (0
6 :c), (0
:
b
-c),
:
(0
1
:
:
:
(0
1),
1
:
C).
-1).
:
XII b. 26*-f 9y-5s
1.
3. a(X
5j8
4. vny +
m
(?7
5.
0.
7 -
6.
2.
0.
0, &/9
0,
7a
-
=
0,
cy-6** - 0, azcx c^-aa = 0. ys I'
^tn'y-fn's)
x
0,
r-o?
ay
=
0. 0,
(my-f ns).
O.
0,
0,
aai
0,
0,
XII o. 1.
3.
2. 86' 2 JWwn-r (6n-fcw) (an-f d) 25 cos A cos J5 cos C. (am cos .4 sin (B C) + t/cos5sin(0 ^-f^cosCsin (-4 0. E) a-f 3-7 = 0.
as
4. 5.
6.
V{^ =
d)
10.
p
q
Pi
Qi
(n~l)/(n-m),
{0,
*-(w-l)ji-.(n-.l)s0; BrSsinlJB-CcosecJ^L-r
a
0.
1
=
7.
-2J?r}. 1
12.
0.
-
0,
P* ' Mid-point of BC. (ii) Point at infinity on BC. Foot of perpendicular from A-to BC. (iv) Oentroid. '
13. (i) (iii)
cos
.40 cos 2?
7 cos C
xcosA
10.
15.
OK
17.
8(p-f3y+* )*(P + 9 + *')(-fy-f); y+0-0.
18.
J2V(l-3cosXcosBcosC).
21.
(i)
0.
ycosJ?.
>
a
7 t
1
3i
7i
cosC
cosJ5
0.
7 (ii)
3i
!
JmcosC ncoaB m
7i
ZcosC
ncos-4
n
ZcosB
mcos^i
24.
26. accot-lycotJ5^*cot(7-
Points 29.
0,
- 77U
36. (0
n3
:
0,
on xcot J^-f ycoti-B-f ^ cos J3 7 cos C 0. 77i
-m a
),
(n
!, 3
:
:
i
/3ft
-!), (w
a
:
&c. -= 0.
lie
31. 0ft :
0.
ycot J
27- a?(cot JJB+cot JC)
;
(a
-i2
:
:
6
0)
:
;
c).
0.
0.
ANSWERS
543
XII d. =
4. (Xcos.4 >cosC bcx(x+y+ai). 6 4-c &c. cotC = 0. 6. y cot 5. a)}, 8. a0i + byOL + c(X0 + (OC + + y)(aOt.+b& + cy) * 0.
*xy
aY(y + *-acH frVCsfs-vHcVCs+S/-*) -
9.
0.
3z
V{^A
10. x/a* - y/(4ca -a2 ) 15. 2ya(cosB-cosC)
+
/(46
a
~a 2).
14
16.
0.
0. (* 2
+ w ) c*-m& 2
aV, &c.
{op, bq, cr}
:
roi>
;
2 .
x/a+y/b+/c m
0.
-
0.
XII e. 2.
A conic;
0.
X
2
-f
2
+y
2
2y
-f **
r + 4 JT cosec 2 JB 2
10. ftCft 1 -^)-^!
2
(ri
-
2zx-2xy = 0. c-T+ &r 2 */2 cos
5. \/Pi * (V 4- s
-f
ij-f
;
-^ ):^ -^2 2
2
).
XII f. 2.
4.
cosi^V'-fsiniBv^ii-fsinlrV^-O. 2
fx+gy + hz *
0.
5.
(figih}.
3aV -l-256c^cos-4 11. If PA and the required harmonic conjugate meet BC in T 2", then the pencil A {BC, TT } is harmonic. The envelope is 2p 1 /p-2l /g-r /r - 0. 12. If (xlt t/u *i) is the fixed point, the envelope is V^aci + V^Wi-l' V^i^T- 0, or 7.
0.
t
f
1
V*+yi/+V-XII g.
8.
+ my) 2 + n (1 4- m) ** 0. 5. x */x + y'*/y+2f*/z - 0. {mn(m-n) fU(n-Q Jm(-w)}. = 0; +16y+21 8ac + 8y-7 0. A straight line, 2 sc-f m 2 y+na * -= whore (?, m, n) is the fixed line.
9.
An
2.
3. (Ix
0.
Xlx^y'-nz'}
/
4.
:
7.
:
Z
inscribed conic.
10. (a)
11. (a) 1.
(b) 2.
1.
(b) 2.
A conic inscribed in the diagonal triangle of the quadrilateral. A conic circumscribing the diagonal triangle, it passes through
13.
14.
mid-points of the sides of the quadrangle. 10. If the sides of the quadrilateral are
(p
point at infinity
xyz
18.
A
and (livnin)
r4 )
qi>
,
on the parallel tangent) the locus
then the tangents at
Q,
R
=
a 0, ?x
intersect on
=
+mj/2 +n
2
AB
ylmy
O p
(viz.
;
22.
4.
0.
Q(x ly
_
x^+y^+zf =
and Ixf -f my, 2 -f rwfj 2 2 n(J-w) (a^*,-!^) -
Hence QR passes through one of the points {0 Vw(n-J) 20. XtXt/P . ViV^m 2 . ! n'. 21. pa/i2 - ^ft/m :
:
+ \/^ Cz ~ m)} r^j/n
2.
XII h. 1.
xy+ yz+sx
4.
jpsin24-fg8in2J3-f rsin2C
0.
0,
ap-f 6g+cr
ptan-4-fgtan.B-frtanC- 0, p+g-fr 12s2 + 4t/8-8 2 - 0. (ii) Parabola. - 0. (Ui) x*bcoo8A + a*yz
5. (i)
the
0.
19. If the conies are x2 + ya + *2
which, since
is
is
pSx^mz-ny^ + qSy/^x-ty + rSz/Vu-mx) -
the six
0,
pa
0, a
-f
aptycr gb'-frc
9
0.
0.
ANSWERS
644 6. 8. 11.
U. 21.
7. (sin2A sin2 y*+3x+xy0. sin2C). 0. x\c*g*-b*h*) + y*(a*h*-c*f*)+*(b*f*-a*g*) a a * 0. 13. + aV(*-a) &V(s-&)+c /(s-c) :
:
=
0. 0y/J+7(X/ro+atf/n a a a a a a a a a a {2& -c :2a -c :2a* + 2& }, (2& -c )s+ (2a -c*)y-(a + 6 )*
0.
the
+ b*-ca )x-(aa + 8&a --ca )y + (aa -&a )* - 0. 4S2/(2aa -f 2&a -ca )!. - 0. Sa{& a (r-i>) + c 2 (i>--s)}/(3-r) a a ~s ^)(6^c )4-a2 (y a +y8 0}4-a v+^2Xf c2 xy 0. (x+y+)2{(y1 ya a a a 4S 40. + xy. (px+qy+ra) (pa?yz qb*tsx+rc*xy} (2a :2&* _ c ). xcot-4 + #cot.B + acotC = 0. The directrices (&4 - e.p) are (Xcosi-A + flcos JB7COstC 0, then ( 12) conies are Za a +ro/3 a + n7a = 0, where Zcos a J-4 + mcos8 J5+ncosa JC= 0.
50.
The
(8a 22.
32.
34.
44. 47.
a
1
1
:
line through
C parallel to
Sit &2 >
54. If the conies are
2x/(mn)
a diameter.
.4J3 is
2S1 + /S 2
^ nen
the corresponding line touches
;
^
^
4a6c2axa cosBcosO-|-(a 3 -f &a + ca ) ^a 63. The centre lies on the line bisecting BC 51
56.
factorizes,
and one
and therefore
factor
U
Sj.
0.
at right angles.
65. *p*/U'-Xpq(l/lm' + l/em) =0. 68. The other conic is (v+w)yz + (w+u)
2uo;(y+*) =
so that at a
0,
zx+(u + v)xQ 0, which can be written point 2ux' = 0, whence, comparing the
common
equations of tangents, the result follows. 70. (a :c : 6 2 ). 72. ^(tn^'-f v'^+^Cuw'+M'^-f ^(vu'+t/u) 73. Use Iqr+mrp + npq * k(pOL + q& + ry) for conic. a
2
74. a 3 /4S, 77. {/Vi
&/4S, 9
=
0.
s
c /4S. * f /*i}) where
0, + ft) are the vertices of the quadrangle. ( +/, the terms of second degree, when C is transformed to l a Cartesians, depend only on the terms a j/-f b*2X + c*xy. Thus if a'*yz+b*zx+c xy transforms into \(JP + r a B a ), then C transforms into
78. Since
Now
:
flf
/^
:
x+y+0 =
apply
A
1,
= A72Sa
to
both pairs of equations.
82.
84. 86. Focal chords are proportional to the squares of parallel diameters. 89. 1 1 ^aV-f-a&c5(^ 1 cotJB-f- 1 cotC-2a?1 cot4)XiV
2^y
00.
-
SA = 2^cos-4,
&c.
/.
=
2xcos^i
is
the directrix, and this
is
the polar
of the focus. 01.
a
/97.
The tangent ux+vy + wz = 0. 02.
at the vertex is parallel to Its coordinates are (w-f fc, v + k,
P*/u+?/v+i*/w = and the 08.
vertex,
i.
e.
the point of contact,
(V+ W-2F)/a?
04. See Ex.
ii,
2
(W+ l7~2Gf)/6
is
(1
(17+
the
of
polar
w+k), which
(1,
1,
1),
i.e.
satisfy
0,
+ fc/u
:
1
K-2H)/c
+k/v
:
1
+k/w}
2 .
p. 515.
A circumscribes the quadrilateral formed by the tangent at A, the line at infinity, and the pair of lines Aft, Art. Hence its = (a + 6)3 + 07) (my + n equation is of the form C 07.
The
circle of curvature at
and fcS-C has my+nff 103. See Exx. 6 and 97.
for a factor,
where
fc
is clearly
unity.
INDEX (The numbers refer
to the
pages.)
Abridged notationharmonic pencil 63;
defined 52; quadrilateral 59, 64; conies 198, 283, 412, 439.
circle
183;
Anglebetween two straight cartesians 50, 90
lines,
polars 66
;
between two tangents
;
tangentials 398
areals 466.
;
to,
circle 149, parabola 260, central conic 811.
between equi-conju^ate diameters 442. at which line cuts circle 187, line cuts parabola 298, circle cuts
Anharmonic
circle 184.
ratio
of range of points 61 conic 440.
of pencil of lines 62, 63, 93
;
;
of four points on
Area of triangle 19, 56, 94, 464 of hyperbola 330.
;
of polygon 20
of sector of ellipse 318,
;
Asymptotes defined 240 equation of 240, 477 separate equations of 242 coordinates of 406, 479 ; hyperbola referred to 340 properties of 242, 338, 343, 344. ;
;
;
;
Axisof similitude 193 of parabola 230, 505 length of axes 239, 435, 508.
radical 189, 483
230, 504
;
;
;
;
of conic
Bisector of angle between two lines 51, 91, 113. Boole on invariants 104. .
Brianohon's theorem 495. Central curves 229. Centre of inscribed circle 48 Ceva's theorem 462.
of wimilitude 176, 193, 400
;
;
of conic 228, 477.
Chordof contact 236, 477. defined 133 equation given mid-point 235 Circle or Circlesauxiliary 306, 316, 415; circum 129, 130, 199, 482; coaxal 190, 483; inscribed 482 director 238, 407, 501 orthogonal nine-point 153, 483 radical 193 of similitude 177; on given 184, 195; polar 173, 483; diameter 164, 186 equation of, polar coordinates 207, area! 479, tan;
;
;
;
;
;
;
gential 399, 481.
Coaxal and oopolar triangles 53, 490. Common chord of circle and parabola 280 two circles 182, 484 853 circle of curvature and conic 280, 325, 348. ;
;
;
circle
and
ellipse 325,
INDEX
646
Common tangents of two circles 178, 182
;
two
of
ellipses
312
of two conies 414.
;
Conditionson conic 402, 422
three points collinear 28, 458 four points 211 line touches circle 141, parabola 294, central conic 306, conic 402, 421 two lines parallel 30, 110, 467, perpendicular 35, 51, 467, 469 three lines concurrent 56; two pairs of lines harmonic conjunormals concurrent, parabola 273, ellipse 321, hyperbola gates 63, 92; two circles orthogonal 184 : 331, 332, 346, circumscribing conic 491 general equation represents lines 107, 108, 226, circle 127, 481, parabola 226, 476, ellipse 226, hyperbola 226, rectangular hyperbola 240, 479. , Confooal oonics 355 60, 416.
point
on
lies
circle 198,
;
;
;
;
;
;
Conjugatepoints 172, 308 lines 172, 308, 406 ; diameters 229, 308, 317, 343, 502 parallelogram 318, 340; hyperbolas 338. Contact of conies 285, 287, 417. ;
Corresponding points on conibcals Curvature
359.
centre of 281, 323, 326, 332, 347, 349 chord of 280, 325, 348 280, 325, 332, 349, 437; radius of 281, 326, 349, 511. ;
Diameter
;
;
circle of
228.
Director Circle 238, 311, 411, 501. Directrix of parabola 238, 248, 260, 293, 407, 411, 502 of central conic 304, 311 of conies 244, 247, 249 equation of conic, given directrices 444. ;
;
;
Distancebetween two points
463
13,
between
;
foci 246.
Double contact
Drawing
281, 285, 288, 416, 417. conies 231, 243, 343.
Eccentric angle 316. 330. Eccentricity 244, 247, 249.
Envelopes
66, 68, 253, 392, 402, 413.
Equi-conjugate diameters 309, 442. Evolute of -
parabola 274, 281
ellipse
;
323
hyperbola 333
;
;
rect.
hyperbola 349.
Focusof central conies 245, defined 244, of parabola 258, 261, 294,411, 507 relative to of conic given by tangential equation 407 303, 408, 506 circular points 250. ;
;
;
Harmonic conjugates 15, 92 range 15, 61 pencil 62, 63 property, of pole and of quadrilateral 64. polar 167, 237 Homogeneous coordinates 119, &c. in relation to circle 204, conic 241, general equation 421. ;
;
;
;
Infinitycircular points at 203, 251, 398, 400, 468, points at 121, 398, 397, 468 484; line at 121, 205, 391, 467," 477; position of infinite part of parabola 243. Intersections of two lines 49, 109, 393 line and circle 147 line and conic 237 circle and parabola 279 circle and ellipse 325 circle and hyperbola 332, 347. Invariants 102, 103, 432, 513 application of, 434, 435, 514. ;
;
;
;
;
;
;
Inverse points 172
;
of straight line 216
Ivory's theorem 860.
;
of circles 217
;
curve, defined 216.
INDEX
647
Iiatus rectum defined 257 of parabola 290, 435, 514. Length of perpendicular to a line 33, 45, 47, 434, 465 perpendicular to pair of lines 111 tangents to circle 139, parabola 268, ellipse 319 normal 257 segments of a chord 165, 436; chord of parabola Limiting points, 191, 209. ;
;
;
;
Locus, defined
Maximum
;
21, 150, 402.
triangle inscribed in an ellipse 317.
Menelaus' theorem Mid-points of
394, 462.
chord of parabola 267
chord of central conic 309.
;
Normal defined 133; general equation of 236; of parabola 272; of ellipse 320; of rectangular hyperbola 345 polar equation of, 381. of hyperbola 331 ;
;
Notation for general Cartesian equation 108 for areal equation 475.
Ordinate 257, Orthogonal
;
for general tangential equation 405
;
259.
points 397, 398
;
circles 184, 195, 212, 367
;
confocals cut orthogonally
356.
Osculating circle 280.
Parabola referred to pair Parameter, defined 145. Parametric coordinates
of tangents, 291-6.
circle 145, 156, 160 parabola 265, 292 ellipse 316 conic referred to tangents and chord of contact 439. Pascal's theorem 490. ;
;
hyperbola 330, 342
;
;
Pedal-^ line 154
Point -
;
equation of tangent to central conic 306.
equation of 393 of contact of tangent 405, 476 of intersection of line and conic 404, 423 of intersection of tangents at end of chord, 405. ;
;
;
Point circles
191.
Poledefined 170; properties of 171 405, 477.
;
of line at infinity 205, 477
;
equation of
Polardefined 170 of point 171, 236, 477 ; of centre of conic 205, 477 of conjugate diameters 309 reciprocal 214, 441, 515. ;
;
property
;
Quadrangle coordinates of vertices 471
;
properties 420.
Quadrilateral equations of sides 471 ; conic circumscribing 198, 499 64 properties 21, 59, 420.
;
harmonic property
;
Hadical Axis 189, 483
;
axes of three circles 193
Rectangular hyperbola
;
centre 193
240, 345, 349.
Self-conjugate triangle 172, 262, 362, 497, 499. Similar conios 436, 479.
Simson'B line
154.
;
circle 193.
INDEX
648
Subnormal and subtangent Syitemi of circles 162, lines 499,
257.
188,367,402; conies through four points or touching four
Tangent general equation
235, 475.
Tangential equation461 of circle 399, 481 defined 392 ; of point 393, forms of 412 ; of general conic, 421, 475. ;
;
of conies 409
;
special
Tangents pair of to circle 166, 168, to conic 237, 423, 477, to parabola 260, to central conic 311 ; polar equation of pair of 383 ; common to two circles 178, 182.
Transformation of coordinates Cartesian to polar 12; areal,
453
;
trilinear
and
Cartesian, change of axes 97, &c. areal to Cartesian, 456.
;
trilinear to
