be the eccentric angle of P; then [Art. 110] SP=a + ae cos 2 ) eos 2 = a 2 sin 2 + & 2 cos 2 0=C.D 2 .
[Art. 130.]
2. Let be the eccentric angle of P, and let A be ( - a, 0) and therefore
A' (a, 0).
Thus Y is the point of intersection of
x y
- cos 4> + r sin 0=1 andx= -a:
/. Oris -cosd>+^sind> + -=0.
a 00
mt _ ,. - .,„ . x-acosd> y-bsmA , ,
The equation of A'P is — -*- = * . r ; and CY and 4'P are
sec o, we have
c 2 seo 2 o=c 2 cosec 2 a(l-e 2 ), and therefore e=^/(l-tan a a).
VI.] CONIO SECTIONS. 79
4. Let the eccentric angles of P, Q, E be a, /3, respectively. Then the
coordinates of V are s (cos a + cos 0), 5 (sin a + sin 0) ; and the coordinates of
V are g(cos/3 + cos0), -(sin/3 + sin0).
Hence the equation of VG is
\x- - (cos a + cos9)lasin^-5 -!y--(sin o + sin 0)Licos — ,- =0.
Hence, at . G, # = - (coa a + cos 0) - ^ (sin a + sin 0) cot ^-=-
■UUMUU, U. U U ,
« 2 -& 3 ,
= — = — (cos a + cos 0).
Similarly at G'
x=— 5 — (cos/3+cos0).
Hence GG'
a 2 ~& 3 ,
-= —= — (cos - cos /3),
which is independent
of 0.
5. Since 50 2 =a 2 (l-c 2 ), SC=ae and XS=XC- SC=--ae, it follows
e
that BC 2 =XS . SO. . Hence the locus of B (or B') is a parabola whose vertex
is S and whose latus rectum is equal to XS.
6. Let the eccentric angles of P, P', Q be a, - and respectively. Then
the equations of PQ, P'Q will be
X a + V . a + a-0
- cos —- + £ sin — =-=003 -=- ,
a 2 6 2 2
, a; -a + 1/ . -a + 6 -a-$
and - cos — 3 h f sin — = — =cos — „ — .
a 2 & 2 2
Hence • CM=acoa^- ?r ^- / cos — - ,
= acos-g— / (
1-0 I a-0
r/ oos -2- ;
and CM'=a cos
.-. CM. Cll'=a 2 .
7. Let the conjugate diameters be y = mx and y =m'x; then mm'= - IP/a*.
The perpendicular lines through the foci are
y=--(x-ae), y = --, (x+ae);
V
hence, from the relation mm'— — ^ , we have
80 CONIC SECTIONS. [CHAP.
Thus the required locus is a concentric ellipse. The ellipse is similar to
the original ellipse, but the major axis of one ellipse lies along the minor axis
of the other.
8. Take the equi-conjugate diameters for axes ; then the equation of the
ellipse will be x 2 + y 2 = c 2 . ,
Let P be {x\ y'), then the equation of the tangent at P is xx' + yy'=c i ;
and therefore CT=-, and CT= -, .
x' y'
Now a TCP : a T'CP=y'. GT : x 1 . CT'=~, ;^p = CT* : CTK
9. Let Q be (x', y') and P be (x", y").
Then the equations of CQ, CP are -, - -, = and — , - -^,=0. Also the
x y x y
equations of the normals at Q, P are
x-x' y-y' ana s-s" y-v"
x' y' x" y'
a 2 b 2 a 2 ft 2
Since CQ is conjugate to the normal at P, we have [Art. 127]
_&_tf aV g's" y'y"_ n
a 2 x'' 6V" a* + 6* ~ '
Also CP will be conjugate to the normal at Q, provided
v" oV__6 2 „yx" , y'y"_ n
x"'b 2 x'~ a 2 ' or a 4 + ¥ '
which is known to be true.
10. Let P be the point (a cos 0, 6 sin 0) and D the point ( - a sin 0, .
&COS0).
Then P' is ( - a cos 0, 6 sin ) and D' is (a sin 0, 6 cos ).
tt .-. .. - t,t., . a; - a cos « - b sin
Hence the equation of PD' is .- — = t— %■ =— - — ,
a cos - a sin gr=6V (S + ^) ;
V \a 4+ *V
:. KG.PG=b 2 =SZ.S'Z'.
Hence KG : ZS ;. Z'S' : PG.
14. Let the conjugate diameters be PGP', DCD', and be the point.
Then OP 2 +OP' 2 =20C 2 +2CP=,
OD 2 + OD' 2 = 20C 2 + 2CZ> 2 .
Hence the sum of the square of the distance of P, P', D, D' from- is
40(^+2 (CP 2 +OD 2 ),
which is constant, since OC is constant and GP 1 + CD- is constant.
15. Let P be the point (a cos 6, 6 sin 9), and P' the point (a cos 0, -ftsintf);
then O is the point of intersection of the lines whose equations are
— 7. - -— a =a - * and 2 = ,". . .
cos 8 sax a cos - b sin
Hence at we have
x a 2 -& 2 „ , y a 2 -b 2 . .
-= „ , a cosg, and f = - , .. sing,
a a 2 + (i 2 6 a 2 + & 2
S. C. K. 6
82 CONIC SECTIONS. [CHAP.
Square and add ; then we obtain the equation of the locus of 0, namely
a 2 + W~~
/ a 2 -t 2 \ 2
V + 6V
16. The equation of the normal at P is — .™ =a 2 - 6 2 . Hence G
cos sin
( a s _ Tfi \
cos 8, I . Hence, if (x, y) be the middle point of PQ,
a 2 -ft 2
we have 2x=acos0 + cos0 and 2y = b sin0;
••V2a 2 -6 2 j + W '
the locus is therefore an ellipse whose semi-axes are — = and ^ .
2a 2
17. JJet 6 be the eccentric angle of P ; then the equations of AP, A'P
will be
-cosg + fsin^cos^,
, X 6 + TT y . + 1T 8 -IT
and -cos - 7r - + ^ sin— ^— = cos _ ■
a 2 J 2 2
Hence the equations of PJV, PM will be
(x - a cos 0) a sin 5 - (y - b sin 8) b cos ==0,
and (x-acos0)acos 75 + (v-&sin0) 6 sin s=0,
« 2
Hence MN= I a cos — sin cot - |
\ a 2j
~ I acos0 + — smtftan- l = — .
\ a 2J a
18. The directions of the two tangents from (x, y) are given by the
equation
2/=jnx+ v /(a 2 m 2 + 6 2 ). [Art. 113.]
Hence tan V tan 0„ are the two roots of (y - mx) 2 = aV + b".
Hence tan 0, + tan 0, = --*— -■ „ .
x 2 - a 2
« 2 -fc 2
and tan 0, tan 0~ = "-s — „ .
* • x'-a?
Hence, (i) if tan 2 + tan 2 = constant =k suppose, the locus of (x, y) is
given by
2xy = k{x 1 -a?).
VI.] COXIC SECTIONS. 83
(ii) If cot 8 1 + cot 0. 2 — constant = I suppose, the locus of (x, y) is given by
2*2, = I (j/ 2 -& 2 ).
(iii) If tan 8 1 tan 2 = constant = m suppose, the locus of (x, y) is given by
2, 2 -& 2 =m(:i; 2 -a 2 ).
19, Let PCP\ DGD' be any two chords of an ellipse, and let the
eccentric angles of P and D be 0, tj> respectively.
Then the eccentric angles of P', D' will be 8 + tt and tji + ir respectively.
Let pGp' and dCd' be the diameters conjugate respectively to PCP 1 and
BCD'; then the eccentric angles of p, p' are 0±-^ , and the eccentric angles
of d, d' are ± -~ .
Hence the sum of the eccentrio angles of p and d will b3 either 8 + h i ■
Y 003^(^-7) cos^(^- 7 )y
and so for the other points of intersection.
Hence the area of the triangle formed by the tangents .
cos g (£+7)
'■ 1
cos g(/3-7)
1-
cos -(7 + a)
1 1 '
cos -(7 -a)
1 — i '
cos-(a-/3)
ah
sin g (^ + 7)
cos g (£-7)
sin g (7 + a)
cos g (7-0)
1 '
cosj^a-/?)
1
2 cos g (/3-7) cos^-a) cos^(o-/3)
= "0 see 5 (/3- 7) sec 5 (7- 0) sec 5 (o-j&) . 2cos£ (£-7) sin \(p-y)
003^(^ + 7), sin 2 G8 + 7),ddsg(/3-7)
COSg(7 + a), sin g (7 + a), C0B^{y~a)
cos = (a + /3), sin^(a + /3), cos„(a-/3)
1,
=ab tan s C 3 - 7) tan 5 (7 - a) tan = (o - /3);
86 CONIC SECTIONS. [CHAP.
26. Let ABC be the triangle formed by the tangents ; then
S. 4A=BC . CA .AB.
I C0B = (a + 7 ) cos-(a+/3)|
Now BC*=la % a - >
I cos g (0-7) cos^(a-0)j
Bin 5 (a+ 7 ) sin„-(a + 0)
»— ? >— l — ,
C0B;j(a-7) cos^(o-/3)J
.■. ZJC 3 oos 2 |(o-7)eoB 2 5(a-/3) = ('
The eccentric angle of Q is
, NP ,&sin0
tan -1 J7S =tan- 1 - = (n).
CN a cos ti '
The equation of the tangent at Q is
x II
- cos + i- sin A = 1 (iii) .
a b '
Now (i) is perpendicular to (ii) since a tan 0= b tan 0, from (ii). Also,
since the tangent at Q is perpendicular to Cp, it cuts off from Cp a length
equal to the perpendicular from the centre on (iii) ; and this perpendicular
= a ± =CP
v /(6 2 cos 2 0+a 2 sin 2 ^) '
, . „ PN- 6 2 sin 2 , , _, CN* a 2 cos 2 8
for sm 2 =-(jpi= cps and cos 2 = -^ = cp2 .
and therefore
6 2 4 +2/'7* 4 )
_ s>*+y' 2 /& 2 -(s' 2 +y' 2 )/(a 2 + fc 2 ) . -
~ a-V+y"/* ' ' ° m W
= -a 2 6 2 /(a 2 + & 2 ).
36, Let (a:', y') and (a", y") he the extremities of any chord of an ellipse,
and let (£, i)) he the middle point of the chord; then 2£=x' + x", 2ri=y'+y",
and the equation of the chord is [Art. 114]
x {x'+x") , y{y'+y") _, *V' yV;
~a 2 + 6 2 _ + a 2 + 6 2 '
,.., 2af 2y,_ x>x" y'y"
that is ■^■ + -^ =1 + - a "5" + "b2-■
But, since the point ({, ?;) is on the line, we have
a 2 + 6 2 - x + a 2 + ■ ja •
Hence, by subtraction, we have for the equation of the chord whose
middle point is (f, 77),
(*-t)£+p=0 (i).
If the chord whose equation is (i) pass through the fixed point (/, g) we
have
so that the locus of (£, >;) is the ellipse -^ (a: -/) + K (y - j) =0 ;
or thus :
The coordinates of the point at a distance r from ({, 77) and which is on the
line through (£, r/) which makes an angle $ with the axis are given by
x=£ + rcos0, y=?; + rsin0 (i).
Hence the values of r which correspond to the two points in which
the line meets the ellipse are given by
(f+rcos9) 2 /a 2 + (ij+rsine) 2 /6 2 =l.
If (|, 77) be the middle point of the chord it is necessary and sufficient that
the coefficient of r in the above quadratic should be zero, and therefore
f cos 0/a 2 + 77 sin 0/4 2 = 0.
or cos 6 and sin 8 from (i
point is ({, 77), namely
£(3~!)/ a 2 + ,(y-,,)/& 2 =0.
Hence, substituting for cos and sin 8 from (i), we have the equation of
the chord whose middle point is ({, 77), namely
VI.] CONIC SECTIONS. .91
The equation of the required looua is then found as above to be
x{x-f)la?+y(y-g)lV=Q.
37. Let DGD 1 be the diameter parallel to PRO, let V be the middle point
ofPQ.
Draw PM parallel to GV, and let the tangent at P out D'GD in t.
Then we know that Clf . C«= CD*; henee as PR = Ct and PQ = 2PV-2CM,
we have PQ. PP. = 2CZ> 3 = ^(DZ)') 2 .
38. Let (|, tj) be the middle point of the chord and let 2c be its length,
and the angle it makes with the axis. Then the poin ts £ ± c cos 0, ?j ± c sin
are both on the ellipse, and therefore
(| + c cos 0) 2 /a 2 + (17 + c sin 0) 2 /6 2 = 1
and (f - c cos 0) 2 /a 2 + (17 - c sin 0) 2 /6 2 = 1.
Hence (f 2 +c 2 eos 2 0)/o 2 +(i) 2 + c 2 sin 2 0)/6 2 =l (i),
and £cos0/a 2 + ijsin0/6 2 =O (ii).
From (ii), cos0=a 2 ij/ N /(aV+* < l 2 )>
and sm$= -V%l«J{a i i? + >' i F)-
Whence, from (i)
{{V + T^/ft 2 - 1 } (a V + i 4 * 2 ) + c 2 (a V + ft 2 ! 2 ) = 0.
39. The equation of the chord whose middle point is (£, 7/) is [as in 36]
(x-Z)ya? + {y- v ) v IV=0.
If this chord is the polar of (x\ y') its equation is the same as
ix'/a 2 + 2/?//6 2 =l;
whence
£ V ?K +
But if the tangents from (x' t y') are at right angles x" 1 + y' !i =a a + b 2 , and
therefore
f + , 2 =( a 2 + 6 2 )(| 2 /a 2 + , 2 /6 2 ) 2 .
Thus the locus of the middle point of the chord is the curve of the fourth
degree (x 2 + ?/ 2 )/(a 2 + fc 2 ) = (x 2 /a 2 + !/ 2 /6 2 ) 2 .
40. Let a, /3, 7, S be the eccentric angles of the points A, B, C, D
respectively. Then, since AB, BG and CD are parallel to three fixed straight
lines, a+fi, (3 + 7 and 7 + $ are all constant ; and therefore a + S is constant,
from which it follows that AD is always parallel to a fixed straight line.
92 CONIC SECTIONS. [CHAP.
41. Let PGP', QGQ' be the two diameters, and let the eccentric angles
of P, Q be o, /3, respectively. Then the area of PQP'Q' is 4 aPCQ
= 2 | ocoso, fisina | = 2o6sin(a-/3) (i).
: I acos/3, &sin/3ll
The equations of the pairs of tangents are
x ii xw
-cos a + TBina= ±1, -cos/3+f sin/3= ±1.
a b a .
Now the area of a parallelogram is the product of the perpendicular
distances between its pairs of sides multiplied by the cosecant of the angle
between two intersecting sides.
The product of the perpendicular .distances between the pairs of parallel
/cos 2 a 6in a a\-4 /cos 2 /3 sin 2 p \-\
and the cosecant of the angle between intersecting sides is
//cos 2 a sin 2 o\ //cos 2 ^ sin 2 /3\ //cos a sin/3 sinacos/9\
V W ^j V W + ~TP~)I \a b 6 IT)'
Hence the area of the parallelogram is
4a6/sin (d ~ /3) ' (ii).
From (i) and (ii) it follows that the area of one of the parallelograms varies
inversely as the area of the other.
[The proposition can be very easily proved geometrically.]
42. Draw QN parallel to the tangent at P meeting GPq in N. Then,
since GN. Cq = CP 2 , and QN is parallel to Pp, we have
Cq : CP=CP: CN
. = Cp:CQ.
Hence Gq.CQ = Cp. CP, so that the triangles PCp and QCq are equal,
and therefore also the triangles TQp and TPq are equal.
43. If the eccentric angles of P, Q be 6 V 2 respectively, the area PCQ ==
a'cos0 1( Tisin^!
a cos ft,, isin ft.
= ^dbsm{6- l -8^.
Since the tangents at $ lt 2 meet in (h, ft), we have
h cos^ + ft,) k sin-(0 1 + 3 )
COS-^-ft,) COSgft-fl,)
VI.]
CONIC SECTIONS.
93
and therefore sin^-^2 ^/(J + g-l) / (* + **) ,
h« *««-V(5+5-0/(3 + 5)-
Again area OPCQ : A PCQ ::OG:VG, where F is the point of intersection
of CO and PQ.
The equation of PQ is xhja i +ykjb a =l, and CF : OF is equal to the ratio
of the perpendiculars from and on PQ ; .-. CV : 7= - 1 : ft 2 /a 2 + ftW - 1
and therefore CV:CO = l: A 2 /a 2 + ft 2 /6 2 .
Hence area OPGQ= aPCQx (fe 2 /a 2 + fc 2 /6 2 )
-s/(?*5-0-
44. Since CT bisects PQ, area CP^O. = 2 A CPZ'. Now P is
/ eos -(0 + 0') sin ^ (0 + 0') \
(a cos 0, 6 sin 0) and T is In " , b I .
cos -(0-0')
cos g (0-0')
Hence area required
acos0
cos g (0 + 0')
; 1
cos 3(0-0')
6sin0
sin § (0 + 0')
=a6sec-(0— 0')
cos g (0-0')
COS0
siu
0082(0 + 0') siu 2^ + ^
= a6tan-(0 — 0').
45. The tangents at P, Q meet on the axis at T the point I --, j ;
and the tangents at P', Q' meet on the axiB at T the point ( - — -5-, ) .
Also the tangents at P, Q', whose equations are respectively
- cos0 +r sin0=l and — cos0 --sin0=l,
a b a a
meet in the point t, where y= — =■ cosec .
a — b
94 CONIC SECTIONS. [CHAP.
„ n, ™ 2a lab 8a 2 i
Now parallelogram = 2 A TlT= . ; =r— ; — - = -. ., ■ „ , .
r coBip (a-b)sw (a-o)sin20
46. Take the centre of the circle for origin and axes parallel and
perpendicular to the given straight lines ; and let the equation of the circle
be x 3 +y 2 =a? and the equations of the fixed straight lines be x = ±c.
Let o, /3, ir + o, jr + /3 be the angular coordinates of the points on the
circle; and let the tangents at a, /3 meet on the line x=c; then
(i).
C0Sg(a + /5)
c=a
cos ^ (a -p)
The tangents at o and tt + j3 will meet in the point given by
cos;;(a+j9+ir) sinj;(a+/3 + 7r)
x=a y = a ;
cos 5 (a - /3 - it) cos 5 (a - fl - 7r)
.•. X sin g (o - /3) = -asin-(a + /3) (ii),
2/sin;j(a-/3)=acos-(a + /3) (iii) ;
.-. An*\(*-p) = a*\(x*+y*),
and therefore cos 2 - (o-|3)=(a; !i +j/ 1! -a i! )/(a; 2 + 2/ 2 ).
Hence from (i) and (iii)
c 2 (a; 2 + 2/ !l -a 2 )=^ !1 a 2 .
Hence the locus required is the ellipse
f «V 2
Since the minor-axis of the ellipse is equal to 2a, the original circle is the
minor auxiliary circle,
47. Take the fixed conjugate diameters for axes, and let (a, /3) be the
point 0. Then the equations of OP, OQ will be
j/-^=m 1 (a-o) and y- f3=m 2 (x-a),
where w -*.
The points P, Q are la- — , 0] and (0, (i-m^i). Hence, if (x, y) be
the middle point of PQ we have
VI.] CONIC SECTIONS. 95
j P-%y
and %=- -,
2» = o- — , 2y = p-m 3 a;
■'■ m l= - L -7T
and therefore the locus of the middle point of PQ is given by
P . P~ 2 V = _ b *
a - 2x ' a a 2 '
or a*li(£-y^ + V>a(j- s) = 0,
the locus is therefore a straight line which is conjugate to GO and bisects CO.
48. Let be the point (a, /3),
The lines CM, CN are bx - ay = 0, 6a; + oj/ = 0.
Hence OM is a(x-a) + h (y-/3) = 0, and 02/ is a (x- a) -6 (!/ -/3) = 0.
Hence M is the point a = a (aa + bp) / (a 2 +6 2 ), y = b(aa + bp) / (a 2 +6 2 );
also JV is the point a = a (aa - bfi) / (a 2 + 1 2 ), j/ = - 6 (aa - i/3) / (a 2 + b ").
Hence the coordinates of F, the middle point of MN, are
a 8 a/(a 2 +6 2 ) and 6 2 /3/(a 2 +6 2 ).
The equation of OV, which clearly goes through P, is therefore
— j,- = ?—£ ; and this is perpendicular to the line -» + ?? = !. which proves
oi 2 /3a 2 r a 2 o 2
the theorem,
49. Let a, 9, /3 be the eccentric angles of A, P, B respectively.
Then the lines through P parallel to the tangents at A, B are respec-
tively
„ % cos a , , . _. sin a
(a; -a cos 9) V(y-b sin 9) — =— =0 (i),
(z-acos0)^L£+(2/-&sin0) S ^ = O (ii).
Also the equations of CA, CB are respectively
._?_= L ( ui),
a cos a o sin a
and r£rp = b^p (iv)-
Then Q, the point of intersection of (i) and (iv), is
{a cos /3 cos (9 -a) I cos (p-a), b sin /3 cos (9-a)l cos (/3 - a)} ;
also ii, the point of intersection of (ii) and (iii) is
{acosacos(0-/3)/cos(a-/3),iBinaCQS(0-/3)/cos(a-/3)}.
96 CONIC SECTIONS. [CHAP.
Hence the equation of QR is
x - a cos ft cog (6 -a) / cos (ft - a)
a cos ft cos (0 - a) / cos (ft - o) - a cos a cos (0 - ft) / cos (a - /3)
_ y - b sin ft cos (9 - a) / cos (ft - a)
~ 6 sin p cos (9 - a) /cos (jS- a) - 6 sin a cos (0 -ft) /cos (a-ft) "
Hence Qi2 is parallel to
x ' 2/
a {cos ft cos (0 -a) -cos a cos (0-/3)} 6 {sin ft cos (9 -a) - sin a cos (0- ft)} '
and therefore to - cos + | sin 0=1: thus QR is parallel to the tangent
at P.
50. The equations of the normals can be taken to be
^L-*L=a*-b*=c* (i),
cos sm.0 Wl
and
= c J ,
that is * a *
K-)~-K) ;
cos sin =c : .•'("'•
From (i) and (ii) we have
sec cosec i
by~ax~ ax + by ~ aW-hWy* ' ■ ■ . -
whence lj(ax- by) 1 + 1 / (ax + by) 2 =c* / (a 2 x 2 + *y) 2 ,
or 2 (a 2 a; 2 + &y ) 3 = c 4 (A 2 - &y ) 2 .
51. H V 2 be the extremities of any chord parallel to one of the equi-
conjugates; then
I eos 2 ^ + e «) + 6 sin 5 (ff i + '«) = cos 5 Ci - »s).
must be parallel to one or other of the lines - ± ^=0 Hence
a b
"i + * 2 =!°r^.
The normals at B v 2 are
_£^_ Jy_ =c2and _^ »y__ c2 .
cos X sin X cos 2 sin a '
and hence the line through the centre and the point of intersection of the
normals is
ax
U>s 0i cos Bj by [em X " sInTj -0 $'
VI.] CONIC SECTIONS. 97
Now, if ^ + ^=5, (i) will become ax + hy—O; and if 9 1 + 9 i =~ , (i)
will become ax-by = 0.
Hence the normals at extremities of chords parallel to - + r=0 meet on
a b
the line ax + by = whioh is perpendicular to — - j- = ; and similarly for the
other equiconjugate,
52. Let PSP' be a focal chord of an ellipse, and let the normals at P,
J" cut the axis in G, G' respectively and intersect in 0. Draw OV parallel
to the axis to meet PP' in V ; then we have to shew that V is the middle
point of PP 1 .
From Art. 125, SG : SP=SG' : SP';
but SG : SP :: VO : VP,
and SG' : SP' :: VO : VP'. Hence VP=VP'.
53. Let OF be the perpendicular on the tangent at P ; then
CQ 2 =CP 2 + PQ 2 ±2PQ . CF (i),
the upper or lower sign being taken according as PQ is measured outwards
or inwards.
Since PQ=CD, CD . CF=ab, and CP^OD^^ + fc 2 ,
(i) becomes CQ s =a 1 + b !1 ±2ab.
Hence Q lies on one or other of two circles concentric with the ellipse.
Or thus :—
The equation of PQ is
x — a cos _ y — b sin 6 _ r abr
cost) = sing ~~ N /{cos 2 fl/tt 2 + Bin ;i fl/6 !i } ~ CD'
a b
since CD 2 =a 2 sin 2 + & 2 cos 2 0.
Henee, i!PQ = CD, the coordinates of Q are given by x—a cos 0±& cos 9,
and y = b sin 9 ± a sin 6 j whence the locusofQisone or other of the circles
jc 2 +jf 2 =(a±&) 2 .
54. The directions of the two tangents drawn from {x\ y') to the ellipse
are given by
2/'=mx' + N /(a 2 m 2 + 6 2 ), or
m 2 (z' 2 -a 2 )-2ma:y+y' 2 -& 2 =0 (i).
Hence, if be the angle between the two tangents, we have
(mj - mj) 2 _ (% + m 2 ) 2 - im 1 m 2
" (1 + m 1 m 2 ) 2 ~ (1 + Wj m 2 ) 2 '
= 4(&V 2 + ay 2 -a 2 6 2 ) / (aj' 2 +jf' 2 -a 2 -!/ 2 ) 2 .
S. C. K. 1
98 CONIC SECTIONS. [CHAP.
55. If 2' be the point (x\ y') the equation of PQ will he
xx?la?+yy'lb !! =l.
Hence the abscissae x lt x 2 of P, Q are the roots of
a: 2 / xx'y Z> 2
^ + V a 2 / 2/' a_ -'
.-. s 1 + z 2 =2a:'a 2 & 2 /(ay 2 +&V 2 ),
and x&^a* (6 2 — t/' b ) / (ays + ftV*).
Now SP . S Q = (a - 6%) (a - ez s )
= a 2 - 2o»e&V / (ay 2 + IV 2 )
+ eV (6 2 - j/' 2 ) / (ay 2 + 6V 2 ),
=a 2 6 2 <3/' 2 + (a;'-ae) 2 } /(«y 2 + 6V 2 ).
But ST 2 =/ 2 +(a;'-ae) 2 .
„ ST> x" «' 2
Hence sF7sq = ^ + P-
56. The directions of the tangents from (a;, y) are given by the equa-
tion y=mx+^J{phii i +b i ), or m i (x i -a?)-2mxy+y i -b i =z0.
, .. 4(6 2 x 2 +aV-a 2 6 2 )
Hence tan 5 B = , . , „ % ..,- / ;
con , g . (s 2 +^-a a -6 a ) a
cos - (a .2 +2/ 2_ a 2_ 6 y + 4 (6 2 x 2 +a y_ a 2 6 2j W'
Now SZ 12 . HT 2 ={ 2 / 2 +(a;-oe) 2 } {2/ 2 +(x + ae) 2 }
= (a 2 + j/ 2 - a 2 - 6 3 ) 2 + 4 (& 2 x 3 + ay - a 2 J 2 ) .
Hence from (i), ST 2 .flT 2 cos 2 9=CT 2 -a 2 -6 2 .
57. Let P be the point whose eccentric angle is 0.
Then the equation of Gil will be
ax by
cos0 sin0
The equation of SP will be
x-ae y
ae-a cos 9 — - 6 sin 8 '
,.(ii).
The locus of B will be found by eliminating 6 between (i) and (ii), that is
from
VI.] CONIC SECTIONS. 99
ax sin - by cos 8 = 0,
and b (x - ae) sin $- ay oos $ + aey = 0.
rr, , Bin COS 1
We have -=— = -j— = -5 ^ ■ ;
aoey a'ex a?x -¥(x- ae)
.: dWi/ 2 + aW = a 2 e 2 (aex + ft 2 ) 2 ;
.-. (x-aef+y'—a?.
[The proposition can easily be proved geometrically.]
58. Let S, S' be the foci of one ellipse, and H, H' the foci of the other,
C being the common centre. Then SHij'H' is clearly a parallelogram ; and,
since SH + HS' = HS' + S'H, the major axes of the two ellipses are equal to
one another.
Hence SC=CS'=ae, and HC=CH'=ae'.
Hence Si/ 2 =a 2 <; 2 + aV 2 -2aWcos0,
HS'' = a V + aV 2 + 2a W cos ,
and SH 2 + HS' 2 =2a 2 « 2 + 2aV 2 ;
.-. 4o 2 =(SH+HS') 2 =2a 2 « 2 + 2aV 2
+ 2ay (e 2 + e' 2 - Zee' cos 0) V(« 2 + e' 2 + 2ee' cos 9),
whence cos0= x /(e 2 + e' 2 -l)/ee'.
59. Let the conjugate diameters be PGP', DCD'. Then if P be 0,
P' willbe0 + 7r, D will be + £ and D' will be0- ~.
Hence, if A be 0, and A' be ir, and 23 be — and B' be -5- ,
the sum of the eccentric angles of A and P is equal to the sum of the eccentric
angles of B and D', and therefore AP is parallel to BD\
Similarly BD is parallel to PA', and the proposition is then obvious.
60. Let P be the point a, then the equations of AP, A'P are
X 1 u . 1 1
- cos - a + 7 sin r a= cos s o,
a 2 2 2
and 2 00S n( a + 1r )+ rsin-(a + 7r) = C0Sg (a-7r).
Hence cot a JP4'=^^^ysin 2 o.
r— 2
100 CONIC SECTIONS. [CHAP.
Similarly cot 2 ADA'=I -q-j—) cos 2 a.
fa?- 6 2 \ 2
Hence cot 2 6 + cot 2 $'■■
61. Since the normal bisects the angle between the focal distances
SP, S'P, tan e= P - , where p v p 2 are respectively the perpendiculars from S
on the tangent and normal at P.
Hence, if a be the eccentric angle of P,
„ (ecosa-1) 2 , , \cosa )
pr= i .1 and p.?= „ ' .
1 1 cos 2 a sin 2 a J a 2 6 2
a 2 + ~IF~ cos 2 a + sin 2 a
Hence tan 2 6 = — . = — „- sin 2 a.
pf be the eccentric angle of any other point Q on the curve, the
equations of QP, QP' are
|cos^(0 + tf>)+|sm-(0 + 0) = cos-(0-0),
-cos 5 (9+7r + ^))+^sin- (9 + 7r + 0) = cos - (0 + tt-0).
Hence cot 2 X = (-g^-J sin 2 (0 + 0).
Similarly cot 2 V = (^^-T sin 2 (e-? + are conjugate if ?» 1 m 2 = — ^ .
But y = vi,x and y=vi. 2 x are out by x = ft in points the product of whose
I 2
distances from (k, 0) is nijUioi 3 , that is — r> ft 2 , which is the same for all the
different pairs of conjugate diameters. This proves the proposition [Art. 61].
64. Let ABC be any triangle whose sides touch an ellipse and enclose
it, and let A', B', C be the points of contact of the sides BG, CA, AB
respectively,
A tangent line to the ellipse at any point P in the arc C'A'B' near to C"
will clearly be cut by AB, AC in points X, Y respectively so that XP< PY;
and a tangent at any point P' near B' will be cut by AB, AC in points X', Y'
respectively so that X'P>PY'. There must therefore be some tangent,
MDN suppose, whose point of contact is the middle point of the intercept
made by AB, AC; also, if BA' < A'C, B will be between A and M and there-
fore N between A and C.
Let N3I and BC meet in 0, and' draw Nii parallel to AB to meet BC
in n.
Then ND=Dm, and therefore OM 2 cos 2 9),
whence (a 2 - 6 2 ) 2 (ay - 6V) a = (x 2 + y*) (a 2 + 6 s ) (a 2 ;/ 2 + &V) 2 ,
the equation required.
Or thus :
Let 9, be the eccentrio angles of P, P' ; then W is the point of inter-
section of
» -3L = aS _ 6a (i) ,
cos 9 sin 9 v
- — -^- = a 2 -6 2 (n),
cos0 sm0 "
where 9, are subject to the condition
as 2 sin 9 sin + 6 2 cos 9 cos 0=0 (iii).
From (i) and (ii) we have
ax (cos - cos 9) _ by (sin - sin 9)
cos 9 cos sin 9 sin '
.-., from (iii), tan g (9 + 0) = ^.
so that ain5(9 + 0)=ay/ x /(aV+6 2 a; 2 ),
and cos^(9 + 0) = 5a;/V(aV + ^ !! ) ( iv )-
Again, from (i) and (ii) by addition
as (cos + cos ^) by (sin9 + sin0) _ o/ a „.
oos9cos0 sin 9 sin ~~
or, from (iii),
(w cos g (9 + 0) cos ^ (9- ^) + yi/ sin- (9 + 0) cos g (9-0)
= (a 2 -6 2 )cos9cos0
= (a 2 -6 2 ) |cos 2 |(9-0)-sin 2 i(9 + 0)i .
Hence, from (iv)
| J(«V + & 3 *=) cos | (9 - 0) = (a 2 - i 2 ) cos 2 \ (9 - 0)
-(a 2 -&Vy/(a 2 i/ 2 +& 2 x 2 ) (v).
104 CONIC SECTIONS. [CHAP.
Now (iii) may be written
(a 2 + i 2 ) C03 (8 - ) ;
.: , using (iv),
(a 2 + 6 2 ) cos 2 1 {$-) = a 2 6 2 (x 2 + y 2 )l(aY + IV) (vi).
We have now to eliminate cos- (8-tp) between (v) and (vi), and we
obtain as before the equation
(a 2 + 6 2 ) (a? + y 2 ) (ay + 6V) 2 = (a 2 - 6 2 ) 2 (i 2 x 2 - ahf)-.
67. Let 9, be the eccentric angles of the extremities of any focal
chord; then the equations of the tangent at 8 and the normal -at tj> are
respectively
-cos0 + 7sin0 = l,
a b
, ax by , „ ,„
and ^-=c 2 =a 2 -4 2 .
cos sin
The point of intersection is given by
x y -1
c 2 . " 6 a c 2 6 cos ti a Bin '
— sin $ + -. — - cos 6 -. — - - =-
sm cos tp a a sin b cos ya cos tf> ab sin cos
i 2 + c 2 sin 8 sin ~~ o 2 - c 2 cos 9 cos — o 2 sin sin = - {ft 2 cos cos + a 2 sin sin 0} .
Hence a;& sin = - a& sin cos ^ ;
.'. x= -a cos 0,
whence from the equation of the normal
2a 2 -5 2 .
V= 1 — sin0.
Eliminating 2
a;=acos0 + acos0 CO3 = cos 0. « = 2&sin0.
a a ' J
Hence the locus of Q is the ellipse
Va 2 + 6V + V 2 ^/
The eccentricity of the ellipse is given by
x, a 2 -* 2
whence ^^TF"
The tangents at P and Q are
^cos0 + ^sin0 = l,
cos + ~ sin = 1.
a 2 + 6 2 '21
106 CONIC SECTIONS. [CHAP. VI.
whence 2^ = ysing = ~*
11 all a
b 26 a 2 +6 2 a lab &(a 2 + & 2 )
cob 9 sin _ 1
ajo^+ft 2 ) ~"~2p~a 2 -6 2 '
x ^
Eliminating we have the required equation, namely
•& 2/
CHAPTER VII.
Pages 162—165.
1. Take OA, 00 for axes of x and y respectively, and let the coordinates
of A, B be ±o, 0, and those of C, D be 0, ±6. Then, if (*, y) be any point
on the locus, we have
J{(x-ay+y°-}.J{{ X +ay + y*)= l J{xi + { y -hy}J{x' + (y + bn;
whence x a - j/ 3 = - (a 2 - 6 2 ).
The locus is therefore a rectangular hyperbola.
2. Taking the asymptotes for axes, the equation of the hyperbola will be
xy = c 2 . Now any line y=mx + b will cut xy — c 2 =0 in points whose
ordinates are given by x (mx + b) - c 2 = ; and the abscissa of the middle point
of the chord is - ^— , which is independent of c.
Hence the middle point of the chord of the hyperbola is the same as the
middle point of the intercept by the asymptotes.
3. Take the fixed straight lines for axes ; and let the equation of the chord
in any one of its possible positions be .rjh. + ijlk = l. Then if the line always
pass through the fixed point (a, p) we have
alh + Plk = l (i).
And, if (x, y) be the middle point of the chord, 2x= h and 2y=k.
Hence, from (i), ■£- + £- = 1, or (2a; - 13) (2y -a) = a£.
&x &y
Thus the locus is an hyperbola whose asymptotes are the lines 2x ~/3 =
and 2?/ -o=0.
4. Take the fixed straight lines for axes; and let the equation of the line
be xjh + yjk = X.
Then, since the line cuts off a triangle of fixed area from the fixed straight
lines, we have hk= constant = c 2 suppose.
If (x, y) be the middle point of the moving line, we have 2x = h, 2y = k.
Hence ixy = c 2 is the equation of the required locus.
108 CONIC SECTIONS. [CHAP.
5. Take the bisectors of the angle AOB for axes; and suppose the
equations of OA, OB to be respectively
x cos a + y sin a = and a; cos o - y sin a = 0.
Let P be the point (f, 17) ; then
PM= I cos a + 1) sin a and PN= { cos a - 17 sin a.
The equations of PM, PN will be respectively (£ - x) sin a - (7 - y) cos o =
and (£ - a;) sin a + fa-y) cos a=0.
Hence OJ/ = £ sin - ij cos a, and ON = £ sin a + 1; cos a.
Now area FMONP
= l.OM.MP+loN.NP
= s (f- sin a - 7) cos o) (f cos o + rj sin a)
+ - (f sin a + 17 cos 0) (J cos a - 3; sin a)
= (£ 2 ~ 'f 8 ) s ' n a oos »•
Hence the equation of the locus of P is 3? -y i = constant, which represents
a rectangular hyperbola.
6. Let the equation of the hyperbola be x 2 - ?/ 2 = a 8 , and let (x', y') be any
point P ; then the distance of P from the centre is *J(x' 2 + y' 2 ) . The equation
of the polar of P is xx'-i/i/'-a 2 =0, and the distance of the polar from the
-a 2
centre is ,. .
Hence the distance of P from the centre varies inversely as the distance
of the polar of P from the centre.
7. Let x 2 /a 2 -j/ 2 /5 2 =l be the equation of the hyperbola, then the
asymptotes are given by x/a±i//6=0.
Let P be the point (x', j/0 S t^ eu ^ e normal at P is
x-x' y-y'
x' y'~
a* ~F
Hence G is the point ix'f 1 + ^J, ol; also Q is the point (x'.t — V
The equation of QG is therefore
, x'6
x-x 1 " a
fc a , ~ x'b '
and hence QG is perpendicular to - ± r = 0.
VII.] CONIC SECTIONS. 109
8. Let the equations of the hyperbolas be
x-la?-if-jb s =l &n&-x 2 la? + y*lV l =l.
Then 6 2 =a 2 (c a - 1) and a 2 =& 2 (e' 2 - 1); .-. l = (e 2 - 1) { e *- 1), or \ + i =1.
9. Take the asymptotes- as axes, and let (a:', ^') and (a;", y") be the points
of contact P, P' of the tangents.
Then the tangents meet the asymptotes in the points q, r and q', r' whose
coordinates are respectively (2x', 0), (0, 2y') and (2a;", 0), (0, 2y").
Hence the equations of the lines qr 1 , q'r are
_ + _|„ = land-- ;?7 + ^ = l )
and these lines are parallel since x'y'=x"y"-
Since qf is parallel to q'r, and P, P' are the middle points 2?-, q'r' respec-
tively, it follows that PP' is midway between the parallel lines qr 1 and q'r.
10. If the tangent at P (x', y') meet the axis in T, then T is the point
&"•»)
We have to shew that the perpendiculars from S, T, S' on the normal at
P are in harmonical progression.
The equation of the normal at P is
a?x h 2 y „ ,„
x y'
Hence we have to shew that
2
a:' 2 ■ x x
•2a;' 2 2a;' 2 (a 2 + & 2 )
(a 2 + & 2 )x /:! -a 4 (« ) + 1 ! ) ! j'»-«V'
which is obvious since a 2 + i 5 =a 2 e 2 .
11. Take the asymptotes for axes, and let the equation of the curve be
2a;«=c 2 . Then the equation of the tangent at {%', y') is - + -,=2 [Art. 153],
x y
or xy'+x , y = c'.
Hence, as in Art. 118, the equation of the polar of (a;', y') is xy'+x'y—c 2 .
The line y=y' which is parallel to one of the asymptotes meets the polar
of (a;', y') in the point Q whose abscissa is x= — -x'. Hence the abscissa
c 2 " c 3
of the middle point of OQ is ^-, , which is on the curve since 2y' . wr,-^-
110 CONIC SECTIONS. [CHAP.
12. Let (x', y 1 ) and (x", y") be the extremities of the chord. Then the
other angular points of the parallelogram are (x', y") and (x", y'). Henee the
equation of the diagonal is
^^^,^x(y"-y')+y(x"-x')+x' y '-x"y"=0,
which passes through the centre since x'y'=x"y".
13. Let P be the point (»', y 1 ) ; then A, A' being (- a, 0), [a, 0) respectively,
the equation of PA and PA' is
(xy'-x , yr-a?(y-y'y=a.
Hence the bisectors of the angle APA' are parallel to
*?-y* _ X V rArt oqi
1*-*+*- =W [Art - d9]
or a 2 - y 2 = 0, since x -2 - j/' 2 = a 2 .
Thus the bisectors are parallel to the asymptotes x 2 - j/ 2 =0.
14. Let (-a, 0), (a, 0) be the two points A, A' ; and let (acos 8, a sin 8),
(a cos 8, - a sin 8) be the coordinates of P, P'. Then the equations of AP, A'P'
are (x+o)sin 8-yao&8=y,
and (x-a)sin0 + j/cos0=j/.
Hence at the intersection of AP, A'P' we have
sing _ cos 8 _ 1
2y 2 _ 2ay ~2yx'
:. 2/ 2 -x 2 + a 2 =0,
so that the required locus is a rectangular hyperbola.
15. The equation of the tangents at (x', ?/') (x", y") to the hyperbola
2x^=c 2 are xy' + x'y = c' i and xj/" + x"i/=c 2 .
These meet in the point given by
x=
x'-x" y"-y'~y'x"-x'y"'
-e 2 (x'-x") _ -2o 2 (x'-x")(x'x") _ 2xV
y'x"-x'y" c 2 (x"a-x' 2 ) ~x' + x"
?-! i-
=,.-,, 2 11
and similarly - = - + -& .
y y y
16. Let the point (x', y') be on the hyperbola 2xy=cK Then the
equation of the chord of contact of the tangents from (x', y') to the hyperbola
VI1 -] COXIC SECTIONS. Ill
2xy=c" i will be [from 11]
xy'+yx'=c' i ,
Henoe the area of the triangle out off from the asymptotes by the polar
1 c' 4 c' 4
= ~ —rt sin w = -r sin w.
2 xy c 2
17. Let (x', y') (-x', -y') be the extremities A, A' of any diameter of
the hyperbola 2xy=c !! ; and let (£, 97) be any other point P on the curve.
Then the equations of PA, PA' are respectively
* (y' - v) - y (*' - 1) + m'n - y'Z = °.
and !E (_y_,)_y(_£ C '_|)_ a ;^ + ^'| = 0.
These will make equal angles with the axes provided
v+y' Z+x"
or iv-x'y',
which is true since (x', y') and (£, ij) are both on 2xy=cK
18. The equation of the normal at [x 1 , y') is
x-x' _y-y'
~x r '-~^t r -
Put each fraction equal to X; then, where the normal meets the curve,
we have
(x' + -hx')*-(y'-\y')2=a*,
or X 2 (x' 2 -^' i! ) + 2\(x' a +y 2 ) = 0;
hence at the extremity of the chord X=2— ^ — ~, If therefore (f, tj) be the
y — x
middle point of the normal chord
Ar'qi'2
2f=x'(2 + X)=^ ?2 ,
ix'hi'
and 2, =I ,'(2-X)=^| r2 .
Hence (ij« - | s ) s = 64a:'y 6 a~ 6 = 4a 2 fV-
19. We have a; (3% + 2y + 4) = 9 ; hence the asymptotes are
x = and Sx + 2y + i=0.
20. We have (x - 2) (y - 3) = 6 ; hence the asymptotes are
x-2 = and ^-3 = 0.
The conjugate hyperbola is
(x-2)(y~S)=-G,
or a^-3a;-2(/ + 12 = 0.
112 CONIC SECTIONS. [CHAP.
21. It is well known that tangents to a conic subtend equal angles, at a
focus | Art. 165 (i) and Art. 228, Cor. 2]. Let then S, S' be the foci of the
hyperbola, and P any point on the curve ; and let the tangent at the vertex
A, on the same branch as P, meet the tangent at P in T. Then
tan^S'SP :ta,nlsS'P'=tanAST : ts.nAS'T=S'A : SA.
22. Taie the asymptotes for axes, and let the equation of the hyperbola
be xy = X. Then equation of any circle is
x i +y ;> + 2xycosu + 2gx + 2fy + c = 0.
The abscissae of the points of intersection of the circle and hyperbola are
given by
X 2 X
x s + -= + 2Xcos u + 2gx + 2f . - + c = 0,
x* x
or x 4 + 2<7S s + (c + 2Acosw)x 2 +2/Xx + X 2 =0'.
Hence x 1 x 2 x 3 x i =\ 2 ,
and similarly ViVasVi-^-
Therefore the product of the distances of the four points from the
asymptotes, namely, x 1 x^c 3 x i sin 4 w and 2/i2/ 2 l/ 3 l/ 4 sin 4 w, are equal.
23. Let the equation of the hyperbola be xy = X, and the equation of the
circle
a; 2 + !/ 2 + 2(/x + 2/2/ + c = 0.
Then at the points of intersection of the two curves
x 4 + X 2 + 2gx 3 + 2/Xx + cx> = ;
■•• 2(«i + » 2 + a! 3+a'4)=-f (i)-
1 f
Similarly j (2/1 + 2/2 + 2/ 3 +!/i)= ~| («.).
Now the centre of the circle is (-g, -/) and the centre of mean position
of the four points of intersection is the point
hi x i.+ x 2 + x s + x i) < 5 (2/i+2/a + 2/s + 2/4)| •
Hence, from (i) and (ii), the centre of mean position of the points of
intersection is midway between the centres of the hyperbola and circle.
24. Let the four points be (x x , yj &c, the equation of the hyperbola
being xy=\. Then, if the join of (x v y^ and (as 2 , «/ 2 ) be perpendicular to the
join of (x 3 , 1/3) and (x 4 , y 4 ), we have
(x 1 - x 2 ) (x, - xj + (y x - 2/ 2 ) (jy, - yj = ;
VII.]
CONIC SECTIONS.
Hence
x 1 x 2 x s x i + \*=0.
Similarly
2/i2/2M/4 + * 2 =0-
Hence
x x ' x 3 ' JCj ' x t '
that is
tan a tan j3 tan y tan 5 = 1
113
25. The polar of (x', y') is
xxHaP-yy'ltf^l.
If this touch the circle
x*+y*=a? + V i ,
we have a;' 2 /a 4 +y 2 /& 4 =l/(a !! +6 i! ).
Hence the locus of the pole of ohords which satisfy the given condition is
the ellipse
= < e.
8. As in Art. 165 (3), the locus of the pole of PQ is a conic whose
semi-latus-rectum is 2sec45°=Z N /2; also the envelope of PQ is a conic whose
semi-latus-rectum is Z cos 45° = -£= .
9. Let P, Q be any two points on a conic, and let PQ cut the directrix
in K and let T be its pole.
Then, if the vectorial angles of P, Q be a + /3 and o -/3 respectively, the
equation of PQ will be
-=«cos0 + seo|3cos(0-a).
Hence PQ will meet the directrix
-=ecos0 where 8 = a+rr.
r 2
Again, the tangents at P, Q will be
-=ecos0 + cos(0-a-/3) and -=ecos 0+cos(0-a-/9).
VIII.] CONIC SECTIONS. 117
Hence the tangents meet where = a.
Thus the angle KST is a right angle.
Now suppose that the focus and directrix are given, and that T is a fixed
point. Then, if SK be drawn perpendicular to ST to meet the directrix in K,
the polar of T will pass through the fixed point K.
10, Let the equations of the two conies be
I I'
- = l + ecos0 and -, = l + e'eos(0-a).
r r
Transformed to Cartesian coordinates the equations become
(l-exf-x*-y*=Q (i),
and {l'-e' (bcos a + y sina)} 2 -s 2 -2/ 2 =0 (ii).
Now (l-ex) 2 - {Z'-e'(a;cosa + ysina)} 2 =0 (iii)
clearly represents some curve through the intersection of the conies (i) and (ii),
for (iii) must be true whenever (i) and (ii) are simultaneously true. But (iii)
is a pair of straight lines whose equations are
(I - ex) =f { V - e' (x cos a + y sin o) } = 0,
or in polars
--ecos T \-- e' cos (B -a) I =0 (iv).
Hence two of the chords of intersection of the conies are represented by
the equations (iv), and these lines clearly pass through the intersection of the
directrices whose equations are
I i>
— e cos = and -=e'cos (6 -a).
11, Let the equations of the conies be
I V
-=l + ecos0 and - = l + e'cos(0-a).
r r
Let = j3 be the straight line which cuts the parabolas; then the tangents
at the four points P, P', Q, Q' are
- = eco8 + cos(0-|8) (i),
- = ecos0 + cos(0-j3-7r) (ii),
I'
-=e'cos(0-o) + cos(0-/3) (iii),
;'
and - = e'cos(0-o) + cos(0-/3-7r) (iv).
118 CONIC SECTIONS. [CHAP.
Now the tangents at P and Q, and also the tangents at P', Q', intersect on
the line
l-V
= ccos 0-e'cos (0- a) (v)
r
for all values of §.
And the tangents at P and Q\ and also the tangents at 1" and Q, intersect
on the line
l+V
=e oos 8 + c' cos(0 -a) (vi)
for all values of /S.
The lines (v) and (vi) clearly pass through the intersection of the
directrices.
If e=e'. the equations (v) and (vi) may be written
l-V „ . a . ( . a\ .1 + 1' „ o f„ a\
• - = 2esm ^sin I 0-s I and = 2eco8g cos ( 8 -» 1 :
the lines are therefore at right angles.
12. Let the equation of the parabola be - = 1 + cos 8. Let the vectorial
T
angles of L, L', M, M' be a, ir + a, /S, ir+|3 respectively.
Then the equations of the tangents at L, L' M, M' respectively are
Z/r=cos0 + cos(0-a) (i),
Ijr = cos + cos (8 - a- tr) (ii) ,
l\r = cos 8 + cos (8 - /3) (iii),
and Ijr = cos 8 + cos (8-p-ir) (iv).
Hence 8=^ (a + /3) &tN, 8= g (a + p+v) at N',
1 1
= =(o+j3 + 7r) at K' a.n& 8=^(a + @ + 2t) at K.
Hence NK and A" A" pass through the focus and are at right angles.
13. Let the equation of the fixed conic be-=l + ecos0; and let the
r
equation of the moving conic in any one of its possible positions be
- = l + e'cos (8 -a).
Then, from 10, two of the common chords are given by the equations
l±V
= ecos0±e'cos(0-a),
VIII.] CONIC SECTIONS. 119
or = - cos + cos (0 -a) (i),
7 (*-*')
and = -, cos - cos (0 - a) (ii).
re' v ' y '
Now (i) is the tangent at a to the fixed conic
^— = 1 + -;COS0,
and (ii) is the tangent at t + a to the fixed conic
"? P_r) 1 « -
= 1 +-, cos 0.
r e
14, Let the two tangents be
-=ecos0 + cos(0-a) and -=e cos 9 + cob (5-/3).
The tangents will meet where
1 J 1
= -(a + /3) and -=« cos0 + cosg (a-/3) (i).
Writing the equations oi the tangents in the forms
-=cos (e + cos a) + sin sin a,
and - = cos0(e + cos/3) + sin0sin/3,
we see that the condition of perpendicularity is
(e + cosa) (« + coS|8) + sinasin/3=0,
or e 2 + 2ecoS5(a + /3)coS2(a-|8) + 2 cos 2 -(a-/3)-l = 0.
Hence, from (i),
e a - 1 + 2e cos 6 ( - - e cos J + 2 ( - - e cos j =0,
or j- 2 (l-e 2 ) + 2eZrcos0-2Z 2 = O.
15. We know that
PS + QS ~ I '
_L JL- 2
and PH + HB~ V
120 CONIC SECTIONS. [CHAP.
Multiply by PS, PH respectively and add ; then
= oonstant, since PS + PH is constant.
16. Let d be the distance of the focus from the directrix ; then the
equations of the conies may be supposed to be
— = l + ecos0 and — = l + e'cos(0-a).
r t
If the conies touch one another at some point whose vectorial angle is 8,
the equations
ed
— = e cos + cos (0-3),
r
e'd
and — = e' cos [6-a) + cos (9-/3)
will represent the same straight line.
Hence, by writing the equations in the forms
d „ /„ cosSN . „sin 8
- = oos0 ( 1 + ) + sin0 *-.
r \ e J e
d J cosS\ . ./ . sinSN
and - = cosfl(cosaH -f J + sin0l sinoH j- I ,
, cos 8 cos 8
we see that lH -=cos oh — -j-,
e e
sin 8 . sin 8
and — -=smaH ^- .
e e
Hence cos 8 I ; )=cosa-l and sin 8 ( , j = sina;
and by eliminating 8 we have the required result, namely
- . o 1 1
2 sin = = , .
2 e e
17 and 18. Let the equation of the conic be -=l + e cos 0. The equa-
tion of any circle of radius a which passes through the focus is r = la cos (0 - a) .
In the circle
(r- 2a cos cos a) 2 =4a 2 sin 2 a sin 2 0=4a 2 sin 2 a - 4a 2 sin 2 a cos 2 6.
Hence, substituting for cos from the equation of the conic, we have the
following equation, which gives the focal distances of the points of intersection ;
{er 2 -2acosa(f-r)} 2 -4a 2 e 2 r 3 sin 2 o+4a 2 (i-?-) 2 sin 2 o=0 (i).
VIII.] CONIC SECTIONS. 121
Hence, if r v r 2 , r 3 , r 4 be the four distances we have
■/•j . /, . 7- 3 . r 4 = coefficient of r° /coefficient of r , =4a 2 Z 2 /e 2 ,
which is constant if a be constant.
Also - + — H (- - = - coefficient of Wcoeffieient of ;■"
''1 »s r s r i
= -(-8a 2 Z)/4a 2 i 2
2
= 7'
19, Let the parabola be
a/r= 1 + COS0,
and the given circle j-=ccos0.
Let the equation of any one of the conies be
- = l + e cos (0-/3).
The directrix of this conic is
- = e cos (0-/3),
and this is a tangent to the parabola. Hence, comparing -=ecos (8 - ft)
with the general equation of the tangent to the parabola, namely
-=cos 0+cos(0-y),
r
1.
'(-»■
or -=2cos^cos
r 2
we see that 6=5, and that ae = 21 cos (3 (i).
2
Now the focal distances of the points in which the circle is cut by the
conic are given by eliminating $ between the two equations ; we then have
{c (I - r) - er* cos /3} 2 = c W sin 2 /3 - eV sin 2 /3.
Hence r x + r 2 + r 3 + r 4 = - 2ec cos /3/e 2
= -y.from(i).
20. Let the equations of the conies be
I I'
- = l + ecos0 and - = l + e'cos0.
r r
122 CONIC SECTIONS. [CHAP.
Let the vectorial angles of P and Q be a and a± 5 . Then the equations
of the tangents at P and Q will be
-= e cos 8 + cos (8 - a),
and -=e'cos + cos ( 0-a=p p).
Hence at their point of intersection
(--ecosfl) + (--e'cos e\ = 1,
or in Cartesian co-ordinates
(i - ea;) 8 + (V - e'xf=x*+y\
or ^ {1 -^){* + ^ 2 } S =^ + M.
(el+e'V \
- , 2 _ , 3 , J, and the
squares of whose axes are in the ratio 1 : 1 - e 2 - e' 2 ; hence, if E be the
eccentricity of the conic
1 - £ a = 1 - e 2 - e' 2 , or E* = e 2 + e' 2 .
21. Let one of the conies be - = l + ecos9.
t
The tangent at a to this conic is
- = e cos 8 f cos (8 - a) = cos (cos a + e) + sin 8 sin a.
Comparing this equation with the equation
in
- = cos (8 -A) — cos 8 eos A + sin 9 sin A,
we have, when p = Z,
cos4=cosa + c and sinJ = sina;
.'. e + 2 cos a = 0.
The required locus will be found by eliminating e between the equations
- = l + ecos0 and e + 2cos0=O:
r
and the result is I = - r cos 29.
22. Let the equation of the conic, referred to the focuB 5 as pole, be
- = l + ecos0.
VIII.] QONIC SECTIONS. 123
Let a - /3 and a + /3 be the vectorial angles of P and P' respectively, and
let be (r v 6 X ).
Then the equation of POP' is
I
- — e cos 8 + sec /3 cos (9 -a).
But, since is on the chord,
I
— = e cos X + sec j3 cos (8{ - a).
r
_. cos (0, -a) !
Hence „ = — e cos 0, = constant.
cos p r x '
tt i cos ( 9 i - o) - cos B
Hence also j-± '- £= constant,
cos {8 1 - o) + cos j3
and therefore tan - (6 X - a - /S) . tan - (0j - a + /3) is constant,
that is tan - PSO . tan g P'SO is constant.
23. Let any one of the conies be
-=l + ecos(0-a) (i),
where I is the given semi-latus-rectum.
Let the fixed confocal conic be
- = 1 + + l=0
128
CONIC SECTIONS.
[CHAP.
are at right angles, if 2 + 2ft + 2ft = 0, or ft = - = . Give ft this value ; then we
have
1\» 5
x+y ~2) i x ~y + i
1j2~J = V2 s/2 '
Hence the given equation represents a parabola of which the axis is
the tangent at the vertex
x+y- £=0,
x-y+l=o,
and whose latus-reetum is — ^ , the curve lying altogether on the positive
side of the line
x-y+g=0.
(3) The equations giving the centre are
5 5 3
2x+^y=0 and ^x + 2y+- = Q;
hence the centre is
( - ra)-
IX.] CONIC SECTIONS. 129
The equation when referred to parallel axes through the centre will be
2x"+Sxy + 2y"+H^\-2=0,
that is 2x* + 5xy + 2y 2 = 0.
Hence the given equation represents a pair of straight lines. These lines
intersect in the point l-r,-), and they cut the old axis of x where x= ± 1 :
they can therefore be at once drawn.
(4) From Art. 171, the axes are the roots of the equation
^_j^l _1 i_
r* 11 r 2 + 121 121 J
.-. r 3 =^ or -11.
o
The equation of the real axis is given by the equation
/ 1 3 \ 2
Vn - iiJ !B+ iT 3 ' =0, or x -y =0 -
(5) The equation may be written
(2x+Sy+\)*=(i\- 2) x + (6X- 2) y + W - 2.
The lines
2x+Sy+\=0 and (i\-2)x+(G\-2)y+>
S. C. K.
-2=0
130 CONIC SECTIONS.
are at right angles, if
8\-4 + 18X-6=0, or X=A.
The given equation is therefore equivalent to
[CHAP.
\ JVA J 13^/13 ( Jl'd )
hence the equation represents a parabola of which the axis is
the tangent at the vertex
2x + 3t/ + jjj = 0,
313
2
and whose latus-rectum = ■ ' , the parabola lying wholly on the positive
ld^/ld
side of the line
-3* + 2,-f = 0.
(6j The equations for finding the centre are
i-2y + 5=0 and -2x-2y + 2—0.
Hence the centre is ( - 1, 2), and the equation referred to parallel axes
through the centre is
IX -J CONIC SECTIONS.
x'-ixy -2y 2 + 5 (-l) + 2 (2) = 0,
°r x'-ixy-2f=l.
The semi-axes are given by the equation
-.+ i-2-4 = 0;
131
f=- s or r'= -- .
2 " ■ ~ ii'
Hence the equation represents an hyperbola whose real semi-axis = -«y2
and is along the line whose equation is
(l-2)x-2y=0 or x + 2y=0.
(7) The equations for the centre are
ilx + 12y-&5a=0 and 12a;+9y-30a=0.
Hence the centre is (a, 2a) ; and the equation referred to parallel axes
through the centre will be
41a; 2 + 2ixy + 9y s = 9a^
The semi-axes are the roots of
1 50 1 9.41-12 2
9a 2 i- 2 + "
81a"
= 0;
a" 9a"
,\ the squares of the semi-axes are -- and -=- . Hence the curve represents
5 o
9—2
132 CONIC SECTIONS.
an ellipse, whose major-axis is along the line whoso equation is
/41 5 \ 12 . „ , .
[chap.
It
4. Take the two straight lines for axes ; then if the equation be
ax 2 + 2hxy + by 1 + Igx + 2fy + c = 0,
the eonio will out the axis of x where
ax* + 2gx + c=0;
and, since the two points are equidistant from the origin, we have g = 0.
Similarly, since the two points in which the conic cuts the axis of y are
equidistant from the origin, we have/=0.
Hence the equation of the conic is of the form
ax* + 2hxy + by* + c = ;
the origin is therefore the centre of the curve.
5. The equation may be written in the form
( x-2y + l \*
' +
/ y-ay+i y
2
n/5
IX. J CONIC SECTIONS. 133
Since the lines
3
x-2y+l=0 and 2x + y-^ =
are at right angles, the equation represents an ellipse whose semi-axes are
^2 and 5N /2. Hence the product of the semi-axes=l.
6. The centre will be found to be (2, 2).
The equation referred to parallel axes through the centre will be
x*- scy + 2y* — l.
The product of the squares of the semi-axes is therefore [Art. 171 (iii)]
-'/(•-«)■
2
Hence the required product of the semi-axes = -= .
v '
Referred to the centre, the polar equation of the axes is [Art. 167 (iii)]
tan20 = j^-j=l;
2±
2 tan B x _ 1
' ' i - tan 2 _ y^~
Hence the equation of the axes when referred to the centre as origin is
x*-y*-2xy = 0.
Hence the equation when referred to the original axes is
(x-2)=-(i/-2) 2 -2( ; c-2)(v-2) = 0,
or x 2 -2.r2/-^ 2 + 82/-8=0.
=0,
.7. The condition is
whence \ = 1.
4,
\
-3,
X,
-2,
c,
- 3,
6,
-18,
8. The conic is
(2x + 3y-5)(5x+3y-8) = \,
where \ has such a value that (1, - 1) is on the curve.
Hence. X=(2-3-5) (5-3-8) = 36.
Hence the required equation is
(2a; + Zy - 5) (5a + Sy - 8) = 36.
9, The equation of the asymptotes is
3x" ■- 2xy - 5y* + 7x - 9y + c = 0,
134
where c is found from
7
3,
-1,
2
9
1,
-5,
~2
7
9
c
2'
2'
CONIC SECTIONS.
= 0; .'. c = 2.
Thus the equation of the asymptotes of the given conie is
3x 2 - 2xy - 5y<> + tx - 9y + 2 = 0.
The equation of any oonio which has the given asymptotes is
3a?-2xy-5y* + 7x-9y + 2 + \=0.
If the conic pass through (2, 2) we have
12-8-20 + 14-18 + 2 + \ = 0, or \ = 18.
Hence the required equation is
3x 2 - 2xy - 5y* + 7x - 9y + 20 = 0.
10. The equation of the asymptotes is
6x 2 - Ixy - 3«/ 2 - 2x - %y - 6 + e = 0,
[CHAP.
where
6,
7
i. - 1
2'
■1,
0, whence c = 2.
2
■„ -3, -4
-4, -6 + c
Hence the equation of the asymptotes is
6x 2 - Ixy - 3# 2 - 2x - 8y - 4 = 0.
Since the equations of two conjugate hyperbolas differ from the equation
of the asymptotes by constants which are equal and opposite to one another,
the equation of the conjugate hyperbola must be
6x i -7xy-Sy !l -2x-
11. From Art. 52 we have
■2=0.
Hence
a + b=a' ' + &',
a&-ft 2 =a'6'-ft' 2 .
(a + 6) 2 - 4 (ab - ft 2 ) = (a' + 6') 2 - 4 (a'b' - ft' 2 ),
(a - by + 4ft 2 = (a' - 1') 2 + 4ft".
IX.J CONIC SECTIONS. 135
12. If the axes be turned through an angle 8, we have [see Art. 167 (ii)]
g'=geos6 + fBme and /'= -gsin$ + fcoa0;
■■ <7' 2 +/' 2 =0 2 + / 2 .
13. Take the line joining the centres of the circles for axis of x and the
radical axis for axis of ;/, and let the equations of the circles be
x* + y* + 2ax+b=0,
x*+y*+2a'x + b=0.
Let kc+my = l be the given straight line, and let (x', y') be any point on
the line. Then the polars of (x 1 , y') with respect to the two circles are
xx' + yy' + a(x+x') + b=Q
and xx' + yy' + a' (x+x') + b=0.
Hence the polars intersect where
x+x' = 0.. (i),
and xx' + yy' + b=0 (ii).
But (x', y') is on the given line, and therefore
Vx' + my' = l (iii).
Eliminating x' and y' from the equations (i), (ii) and (iii) we have for the
equation of the required locus
Ixy - mafl + y + mb = 0.
The asymptotes of the locus are parallel to the lines
Ixy — ma; 2 =0.
Hence one asymptote is parallel to a;=0 and is therefore perpendicular to
the line joining the centres of the given circles ; also the other asymptote is
parallel to ly - rax = 0, and is therefore perpendicular to the given line,
14. Take the fixed point for origin, and the diameter of the circle
for the axis of x ; then the equation of the circle will be
r = d cos 8 (i).
Let the equation of the conic be
ax 1 + 2hxy + by* + 2gx + 2fy + e = 0,
or r 2 (a cos 2 $ + 2h sin cos $ + b sin 2 6) + 2r (g cos + /sin 6) + c = 0...(ii).
Eliminating 8 between (i) and (ii) we have
{ (a - b) r* + b(x, y)=0.
Hence the two conies represented by these equations are rectangular
hyperbolas ; but, from Art. 192 or Art. 193, these conies go through the four
foci of the conic (x, y) = 0. Thus two conies through the foci of tj> (as, y)=0
are rectangular hyperbolas ; and hence, from Art. 187, Ex. 1, all conies
through the foci are rectangular hyperbolas.
Or thus :
It is easy to shew that the most general equation of a eonic through the
four points { ± ^/(a 2 - 6 s ), 0} and {0, ± x /(6 2 -a 2 )} is
x*+2hxy -y 1 - a* + b*=0,
and this equation represents a rectangular hyperbola for all values of h.
2. The foci are given by
{ax + hyf - (hx + by) 1 _(ax + hy) (hx + by)
a — b h
=ax' i + 2hxy + by< 1 -l.
Now (ax+ hyf - (hx+by)'= (a - i) (ax*+2hxy + bif - 1)
is equivalent to
{h?-ab)(xi-y*) = a-b.
Also (ax + hy) (hx +by) = h (ax 1 + 2hxy + bxf - 1)
is equivalent to
(h*-ab)xy = h.
Hence we have
x*-y 2 _xy
■b ~ h ~K i -ab'
3. The foci are given by
( x-3y -l) 2 - (-3x + y -l) 2 _ ( x-3y-l) (-3x + y-l)
1-1 -3
= x--Gxy+y i - 2x-2y + 5.
Hence the equation of one conic on which the foci lie is
(x-3y-l)*-(-3x + y-l)*=0, or (x-y) (x + y + l) = 0.
Another conic on which the foci lie is given by
(x-3y-l)(-3x+y-l) + 3(x>-6xy + y»-2x-2y + 5) = (i).
Hence the foci are the points where x-y=0 and x+y+l=0 cut the
conic (i).
X.] CONIC SECTIONS. 139
It will be found that x-y=0 cuts (i) in the points (1, 1), (-2,-2); and
these points are the real foci. The two imaginary foci are the points of
intersection o! x + y + l=0 and (i) .
4. The foci are the points of intersection of the conies given by the
equations
(23-4^)2- (- ix- 4y -2) 2 _ (2a - iy) (- ix-iy- 2)
2+4 ~ -4
= 2a; 2 - 8xy - 4y 2 - iy + 1.
One conic through the four foei is given by
2 (x - 2?/) 2 - 2 (2x + 2i, + 1)= - 3 (z - 2i/) (2it + 22/ + 1 ) = 0,
or {2{x-2y) + 2x + 2y + l} {(x-2y)- 2 (2x + 2y + l)}=0.
Hence the equations of the two axes are
lx-2y + l=0 and 3a + 6y+2=0.
The foci are the points where the axes cut the conic whose equation is
(x - %) (2x+2y + 1) = 2s 2 - 8xy - iy* - iy + 1,
or 6xy + x + 2y-l = (i).
Now it will be found that 4x - 2y + 1 = cuts (i) in the real points ( 0, ^ )
and(-|,-|).
5. The foci of the conic are given by
{x + y-2)*-{x+y + i)* (x+y-2) (x+y + i)
1-1 1
=x 2 + 2xy + y" - ix + 8y - 6.
Hence x+y + l = (i).
Also {x+y) i +2(x + y)-8=(x+y) i -ix + 8y-6;
.-. 6x-6j/-2=0 (ii).
The focus is therefore the point of intersection of (i) and (ii), namely, the
point (-£,- 1).
6. The equation of any tangent to the ellipse is
xcos a+ysina- v /(o 2 cos 2 o + 6 2 sin 2 a)=0.
The product of the perpendiculars from {0, ±^/(6 2 -a 2 )} on the above
tangent is
V(& 2 -a a )sina- v /(a 2 cos 2 a + & 2 sin 2 a)} { - „y(& 2 - a 2 ) sin a- v /(a 2 oos 2 a+6 2 sin 2 a)(
= a 2 cos 2 a + 6 2 sin 2 a - (6 2 - a 2 ) sin 2 a=a 2 .
140 CONIC SECTIONS. [CHAP.
7. The equation of any tangent is
a; cos a + y sin a = J (a? cos 2 a + 6 s sin 2 a).
The equation of the line through an imaginary focus perpendicular to the
tangent is .
a;6ina-ycosa= ± ^(ft 2 - a 2 ) cos a.
Square and add : then we have
x i + y i =b i .
8. Let S = be the equation of the circle, and let % - a = be the equation
of the chord of contact of the ellipse and the circle ; then the equation of the
ellipse will be of the form
S-\(a;-a) 2 =0.
Hence for any point (x, y) on the ellipse we have
S=\(a;-a) 2 .
But S is equal to the square of the tangent from (x, y) to the circle, and
a; - a is equal to the perpendicular distance of (x, y) from the chord of contact.
Hence the tangent to the circle drawn from any point on the ellipse varies as
the perpendicular distance of that point from the chord of contact.
Pages 219—230.
1, Let P be (x', y') and Q be (jr.", y"), and let the equation of the conic
be ax 3 + by* + c=0.
Then the equation of the polar of P is
ax'x + by'y + c=0.
Hence the ratio of the perpendiculars from Q and C on the polar of P is
ax'x" + by'y"+c :c,
which is obviously equal to the ratio of the perpendiculars from P and C on
the polar of Q.
2. Let TP, TP' be any two tangents to a conic, and let the normals at
P, P' meet the axis in G, G' respectively ; then we have to prove that
TP:TP' = PG :P'G'.
From Articles 125 and 131 it follows that
PG:P'G'=GD:CD',
where CD, CD' are the semi-diameters parallel to the tangents at P, P'
respectively.
Also, from Art. 186, Cor. II.,
TP:TP'=CD:GD'.
Hence TP : TP'=PG : P'G'.
X.] CONIC SECTIONS. 141
3. Draw through the chords OQ, OQ' parallel to the axes of the conic;
then QQ' is one of the chords which subtend a right angle at 0, and it is
obvious that QQ' is a diameter of the conic. Hence the fixed point, through
which all chords which subtend a right angle at the point pass, is the
point where the normal at is met by the diameter CQ, where OQ is
parallel to an axis of the conic. The locus can now be easily found [see
solution of question 15, page 139].
Again, to find the fixed. point on tho tangent at through which any
chprd PQ passes which is such that OP, OQ are equally inclined to the
normal at O : take OP, OQ indefinitely nearly coincident with the normal at
; then PQ will ultimately be the tangent at the other extremities of the
normal chord, and therefore the locus required is the locus of the poles of
normal chords of the conic which is found in Art. 138 (4).
4. Let the equation of the ellipse be aa s + &#*=!, and let the chords
make an angle 6 with the axis.
Then, if be (o, j3), the equation of POQ will be
x-a_y-P_ r
cos 6 sin 6
Hence OP, OQ are the roots of
a (a + r cos 6f + b (/3 + r Bin 8f= 1 ;
2 -l
3 2 _ 4 (aa cos + fy3 sing) 2 „ aa? + l
•"■ »i +r 2 - (acos 2 9 + 6sin a "e) 2 "" a cos 2 6 -t
) + 6sin a fl
Hence if OP* + dQ?=k 2 , the equation of the locus of is
i (ax cos $ + by sin 0) 2 - 2 (ax* + by* - 1) (a cos 2 6 + b sin 2 6)
= ft 2 (acos 2 9 + 6Bin !! tf) 2 .
5. Take the fixed point for origin, and let the equation of the conic be
ax 2 + ihxy +by i + 2gx + 2fy + c=0.
Then, if OPP' make an angle 6 with the axis of x, OP and OP' will be
the roots of
r 2 (a cos 2 6 + 2 h sin 6 cos 6 + b sin 3 6) + 2r (g cos +/sin 6) + c = ;
' " UD 2 r x 2 r 2 a
4
= -5(0Cos0+/sin9) 2
2
— (a cos 2 e + 2h sin 6 cos + b sin 2 6).
c
Hence the equation of the locus of D is
4(gx+fy)*-2c (ax*+2hxy + by !! )=c 2 :
the locus is therefore a conic whose centre is 0.
142 CONIC SECTIONS. [CHAP.
6. Let the equations of the conies be
1 (x,y) = and .
2-e 2 a 2 + 6 2
9. Since the conic is a rectangular hyperbola, and PP' is perpendicular
to AA', it follows from Art. 187, Ex. 1 that PA is perpendicular to A'P', and
therefore PA and A'P' will meet on the fixed circle whose diameter is AA'.
The second part is question 14, page 164.
10. Let icosa+y sin a -p = be the equation of the fixed straight line.
Let (x lt yj) and (a; a , y%) be the extremities of any focal chord ; then we
know that
x i x 2 =a i and y 1 y i = -4a 2 (i).
_, PM P'M' _ x^ cos a + y 1 sin a -p x 2 coaa + y i sina-p
inen 7s + ~Ws~ x^, + 5^
ar,cosa + V] sina-p a 2 cos a -ay, sin a -ok, .
= — ^r — H =-" £ - i i from (l)
= (a cos a-p)/a=: constant.
11. Take the fixed point for origin, and the axis of x through the centre
of the given circle, and let the equation of the circle be
x* + y- + 2gx + c = 0.
Let the line y=mx meet the circle in the points [x v y,), (* 2 , y 2 ); then it
is easily seen that
2,9 _c
, 2mfl cm 1
J1 " 2 1+m 2 ' xlJt 1 + m 2
Hence the equation of the circle of which {x v yj, (x 2 , y 2 ) are extremities
of a diameter is
* + 2/ + l + m** + I + Sr s! ' + e -
The polar of the origin with reference to the 6ircle is therefore
gx + mgy + c (1 +»«*) = 0,
144 CONIC SECTIONS. [CHAP.
which may be written y= — ( x + -\
me
1'
from which it is obvious that the polar always touches the parabola
y
a \ a)
12. Take the fixed diameter and its conjugate for axes, and let the
equation of the conic be aa^ + 6y*=l.
Let the fixed point P be (x', y'), and let PQ, PR be the straight lines
through P; then, if (x", y") be the pole of QR, the equation of QR will be
ax"x + by"y -1 = 0,
and the equation of any conic touching the given conic at P and passing
through Q, R will be
ax* + by* -1 + X (ax'x+by'y -1) (ax"x+by"y -1)=0 (i).
The equation (i) will represent the straight lines PQ, PR if X be properly
chosen.
If the conic (i) cuts y = in two' points equidistant from the centre the
coefficient of x must be zero, and therefore x' + x"=0, for all values of X.
Thus the pole of QR is on the fixed straight line whose equation is x+x' = 0.
13. The equation of any chord which passes through the ./feed point (c, 0)
is y = m (x - c) ; and where the chord cuts the ellipse we have
1 fy+mcy ,«/',.
Also
, „ a 2 VW-o?)™?
•'■ yy = J~— T= w+aw (1) -
6 2 a 2 m 2
x 2 , M»(g-c)» _,
(x',y')- {ax'x + h (xy' + x'y)
+ by'y+gx+fy +{gx' +fy' +c)ylk}*=0.
Hence, if the lines OP, OQ are at right angles, we have
{ ft 2 (a + 6) + 2fk + c } 4> (x', y 1 ) - k* (ax' + hy' + gf - { k (foe' + by' +/)
+ (0a'-+/2/' + c)P=O,
the locus of (x', y') is therefore a conic.
150 CONIC SECTIONS. [CHAP.
27. Take the fixed diameter for the axis of x, and let (x', y') be the
point from which the tangents are drawn. Then the equation of the
tangents is
( x s+j,2_ a s) (x'2 + y'2- a ?)~{xx'+yy' -a*)*=0.
Then meet x=0 in points given by
(y 2 -a 2 ) (x' 2 +y' 2 -a 2 )-(yy , -a*)=0.
Hence the ordinate of the middle point of the intercept is
-oy/(x' 2 -o 2 ).
Again, the equations of the lines joining (x', y 1 ) to (a, 0), (-a, 0) are
respectively
x-x' _ y-y' jfi S-J _ y~y'
x'-a y' x' + a y'
Hence the middle point of the intercept made by these lines on *=0
2[ y x'-a * x'+a\~ x'*-a?
Thus the intercepts have the same middle point, whence the proposition
follows at once.
28. Take the tangent and normal at P for axes, and let the equation of
the conic be
ax 2 + 2hxy + by 2 + 2fy = 0.
Then the centre of the conic is easily found to be
( ¥ ~af \
\ab-h 2 ' ab-h 2 )'
and the other extremity of the diameter through P is therefore
c
2hf -2af\
ab-W ab-h 2 )'
Let (x', y') be one angular point of the triangle, then the equation of the
tangents through (x', y') is
(ax 3 + 2hxy + by 2 + 2fy) (ax 12 + 2Tix'y' + by' 2 + 2/2/') - {ax'x + h(xy'+ x'y) + byy'
+f(y +&?=<*■
And, if the. extremities of the base be equidistant from the centre of the
conic, the abscissa of the middle point of the base is the same as the abscissa
of the centre of the conic.
Hence ¥ - W+WM
ab-h 2 ~(ab-h 2 )y'+2af
whence x'=2hfl(ab-h 2 ).
Hence the locus of (x', ?/') is perpendicular to the tangent at P and passes
through the other end of the diameter through P.
X.J CONIC SECTIONS. 151
29. The two tangents to the parabola being at right angles the directrix
will pass through their point of intersection 0. Also, if S be the focus, the
two tangents from will bisect the angles between the directrix and OS.
[In the figure to Art. 98, the tangent BP bisects the angle SBM, and the
other tangent from 11 will bisect the angle SBO.]
Let SK be the perpendicular from S on the directrix ; then SK= 2a.
And, if XOS be 8, EOS will be 29, or tt - 2B. Therefore OS sin 20 = 2o.
Let P be any fixed point on the axis, and let SP=c ; then, if x, y are the
co-ordinates of P, we have
x= OS cos 8+e cos ( = - B | = - — -„ + c sinB,
\2 / sin 8
and y=OS sinB + csin (|- B ) = ^-HccosB.
The elimination of fl gives the required result, namely
x V = (« + c) s (a: 2 + y 2 - c 2 - 4oc) + 2ac 2 (a + c)
a^+j/ 2 — 4ac — c 2 "
The equations of the loci of the foci and of the vertices are obtained by
putting c=0 and c= - a respectively in the above : the equations are
x 2 y 2 =a 2 (a; 2 + y 2 ) and a V (a; 2 + y 2 + 3a 2 ) = a 6 .
30. If the tangents make angles 8 and 8+ a respectively with the minor
axis of the ellipse, we have
p 1 2 =o 2 cos a B + 6 2 sin 2 B,
and 2> 2 2 =a 2 cos 2 (0 + a) + & 2 sin 2 (0+a),
where p v p% are the perpendiculars from the centre on the tangents.
Now, if x, y be the co-ordinates of the centre of the conic referred to the
tangents as axes, we have x=p 1 cosec a and y=p s cosec a.
Hence
2a; 2 sin 2 a=a 2 + b'+ (a 2 - b") cos 2B,
and 2j, 2 sin ! a=a 2 + & 2 +(a 2 -& 2 )cos2(B + a).
.-. (ar s +y)sin 2 a-(a 2 +& 2 ) = (a 2 -& 2 )cos(2B + a)eosa,
and (a; 2 -J/ 2 ) sin 2 a = (a 2 - & 2 ) sin (20 + a) sin a.
Hence {(x 2 +y 2 ) sin 2 a-(a 2 + & 2 )} 2 + cos 2 asin 2 a(a: 2 - y 2 ) 2 =(a 2 - 6 2 ) 2 cos 2 a,
which is equivalent to
sin 2 a (a; 2 + y 2 -p 2 ) 2 - 4 cos 2 a (x*y 2 sin 2 a - q*) = 0,
w here p 2 =a 2 + b 2 and q 2 =ab.
152 CONIC SECTIONS. [CHAP.
31. If t v t 2 be the lengths of the tangents drawn from any point (a, j3),
and r v r 2 be the lengths of the parallel semi-diameters; then, from Art. 186,
Cor. in.,
tf.tflrf. r,'=(ay+ /3*/& 2 -l) 2 (i).
w 1 cos 2 sin 2 e
whence a 2 (6 2 -r 2 )tan 2 e + 6 2 (a 2 -r 2 )=0 (ii).
But, [Art. 113], the directions of the tangents from (a, (j) are given by
(a 2 -a 2 )tan 2 0-2a|3tan0+ i 8 2 -& 2 =O (iii).
Hence, eliminating 6 from (ii) and (iii), we have
{a 2 (ft 2 - r 2 ) (/3 s - ft 2 ) - 5 2 (a 2 - r 2 ) (a 2 - a 2 )} 2 + 4a 2 6 2 o 2 ^ 2 (ft 2 - r 2 ) (a 2 - r 2 ) = ;
a i 6 4 {(» a +^) 3 -2(a 2 -6 2 )(a 2 -/3 2 ) + (a 2 -6 2 ) 2 }
From (i), V«+*i'»= (« 2 /a 2 +/3 2 /6 2 ) V, ;
.-. * 1 t 2 +r 1 r 3 = N /{(« 2 +(3 2 ) 2 - 2 (" 3 -6 2 ) (^-^-(-(o'-S 2 ) 2 },
and it is easily proved that
SO 2 . H0 2 =(o 2 +/3 2 ) 2 -2(o 2 -6 2 )(o 2 -^ 2 ) + (a 2 -6 !! ) 2 .
Hence OP . OQ + CP' .CQ'=OS .OH.
[A simple geometrical proof is given in the Solutions of the Cambridge
Problems and Eiders for 1878.]
32. Take the given perpendicular lines A C and BD for axes, and let the
points P, Q be (a, 6) and (a 1 , V) respectively.
Let the equations of APB and GQD be respectively
lx+my=l and l'x+m'y = l.
Then, since the lines go through the points (a, b), (a', b') respectively, we
have
la+mb=l (i),
l'a'+m'b'=l (ii).
The equations of AD and BC are respectively
lx+m'y=l (iii),
l'x+my=l (iv).
Also, since APB and GQD are at right angles, we have
ll'+mm'=0 (v).
The locus required is obtained by the elimination of I, m, V and m! from
the equations (i), (ii), (iii), (iv) and (v).
From the first four equations it is easy to shew that
1= {b (a'y + b'x) - y (a'y + bx)}\ (bb'x 2 - aa'y\
m= {a (a'y + b'x) - x (ay + Vx)}\(aa'y"- - bVx%
X.] CONIC SECTIONS. 153
V = {&' (ay + bx)-y (ay + b'x)}l(bb'x* - aa'y"),
m' = {a' (ay + bx)-x(a'y+lx)}\(aa'y' 2 -bb'x 1 ).
Hence, from (v), the required equation is
J b (a'y + b'x) - y (a'y + bx)} {V (ay + bx)-y (ay + b'x) }
+ \a(a'y + b'x) -x(ay+b'x)} {a! (ay + bx)-x(a'y + bx)}=0.
If PQ subtends a right angle at the origin, aa' + bb'=0, and. the above
equation may be written
(a? + yt) [(ay + b'x) (a'y + bx) + aa' (a + a 1 ) x + bb 1 (b + b') y]=0.
The locus is therefore a point-circle, and the conic
(ay + b'x) (a'y + bx)+ aa' (a + a')x + bb'(b + V) y = 0,
which is a rectangular hyperbola since aa' + bb'=0.
33. Let the equation of the ellipse be ax 2 +by 2 =l.
Then the foot of the perpendicular from any point (x\ y') on its polar
with respect to the ellipse is given by
ax'x + by'y=l (i),
ax'(y-y')-by'(x-x')=0 (ii).
But if (x', y') be on the fixed diameter lx+my=Q, we have
lx'+my'=:Q (iii).
From (i) and (iii)
as' _ y' _ 1
m — I max — Iby '
also (ii) may be written
bx ay , „
---y + a-b = 0,
whence ( h -j- J (max-lby) + a-b=0,
which represents a rectangular hyperbola.
34. Let the equations of the conies be
ax>+by 3 =l and a'a?+b'y* = l.
The equations of the polars of P (x', y') with respect to the conies are
axx' + byy' = l (i),
and a'xx' + b'yy' = 1 (ii).
Also, if (x', y') move on the fixed straight line Ax+By = l, we have
Ax' + By'=l (iii).
Eliminating x' and y' from (i), (ii), (iii) we have the equation of Q,
namely
ax by 1 =0,
a'x b'y 1
A B 1
154 CONIC SECTIONS. [CHAP.
that is (ab'-a'b)xy+B(a'-a)x+A(b-b')y=(>,
which represents a rectangular. hyperbola whose asymptotes are parallel to
the axes of the given conies.
35. Let the equations of the given ionics be
ax> + Ihxy + by* + 2gx + 2fy + c = 0,
and a'x*+2h'xy + b'y*+2g'x+2fy + c=0.
Then the polars of (x', y 1 ) with respect to the given conies are
(ax' + hy' + g)x + [hx' + by'+f)y+gx'+fy'+c=0 (i),
and {a'x' + h'y'+g')x + (h'x' + b'y'+f)y+g'x'+fy' + c'=0 (ii).
If the lines (i) and (ii) are parallel, we have
ax' + hy' + g _ hxf+by'+f
a'x' +h'y' +g' ~ h'x' + b'y'+f"
and therefore the locus of («!, ■
= square of radius of director-cirole.
Hence SP . HP is constant. But SPHP' is a parallelogram, and there-
fore SP'=HP.
Hence S moves so that SP . SP 1 is constant.
The equation of the locus is therefore
{(z-c^+j, 2 } {(x + c^+y^^k*,
where ft 2 = SP . SP' and c = GP. The curve is called a lemniscate.
38. Take the two given tangents for axes, and let (a, 6) be the given
centre of the conies.
Then, if (x, y) be one of the foci, the other focus will be
(2a -x, 26 -y).
Since the product of the perpendiculars from the foci of a conic on any
tangent is constant, we have
x (2a - x) sin 3 a = y (26 - y) sin 2 w,
where w is the angle between the axes.
Hence the required locus is the rectangular hyperbola whose equation is
x 2 -y i -2ax + 2by=0.
39. The centre is given by
ax+hy = 0, hx + by = l;
a
hence y =
ab-W
The line through the centre parallel to the tangent at the origin cuts the
conic in points given by
2 2ah 6a 2 2a _ n
"* + ab=h* X + (ab-h>)*~ ab-h"-° ;
^y =
V ab
(a6-fc 2 ) 2 (a6-fc 2 ) 2 a6-fc 2 ab-W
Hence CD 2 = -r — ^ , and therefore the product of the focal distances
1
ab - ft* '
156 CONIC SECTIONS. [CHAP.
40. Let S be the given focus, and P the point of contact of the given
tangent ; also let O be the middle point of SP, and let OK be parallel to the
given tangent. Then, if S' be the other focus of the conic, and C be middle
point of SS', O will be the centre of the conic.
Then, if DGD' be the diameter parallel to the tangent at P,
CD"=SP.PS'=iSO . OC.
Since PS', PS are equally inclined to the given tangent, and OC is
parallel to PS', it follows that OC is a fixed straight line. Also CD is
parallel to the given tangent, or to OK.
Hence the equation of the locus of Z>, or D', when referred to OC and OK
as axes, is y' i =±ax, where a=SO. The locus is therefore a parabola of
which OK is a tangent and OC a diameter ; and, since the focal distance is
equal to SO, and 05 and OC make equal angles with OK, S must be the focus
of the parabola.
41. Take the two tangents for axes, and let the equations of the other
lines be x=a, y = b respectively.
Let the foci be (a, g) and (/, b); then, if (x, y) be the centre, we have
2x=a+f and 2y = b+g.
But, since the axes are tangents, the product of the distances of the foci
from one axis is equal to the product of their distances from the other, and
therefore af= bg.
Hence 2ax -a?=2by- b",
so that the locus of the centres is a straight line.
42. This follows at once from question 55, page Hi.
43. Let S, H be the foci of the fixed ellipse, and C be the common
centre. Let P be the point of contact of the ellipses, and let the tangent at
S to the inner ellipse cut the common tangent at P in T. Join CT, cutting
SP in V.
Then CT bisects SP, and is therefore parallel to HP. Hence CT and SP
make equal angles with the tangent PT; from which it follows that
VT=VP=VS,
and therefore the angle STP is a right angle.
Hence CT 2 = sum of squares of the semi-axes of the variable ellipse.
But CT=CV+ VT=CV+ yP=\ (HP +SP) = constant.
Hence the sum of the squares of the axes of the variable ellipse is constant.
Now let 0, O'be the two foci; then CS 2 + OS . O'S = sum of squares of
semi-axes; hence OS. O'S, and therefore also OS. OS is constant. Then
see 37.
X.J CONIC SECTIONS. 157
44. By question 23, page 164, if a rectangular hyperbola cut a circle in
four points the centre of mean position of the four points is midway between
the centres of the curves.
Let then A, B, C, D, E be five points on the circle whose centre is 0;
and let G be the centre of mean position of the five points, and a, b, c, d, e
the centres of mean position of four of the points excluding A, B, O, D, E
respectively. Then Aga is a straight line and AG=iGa; so also BG — iGb,
CG=iGc, DG=iGd and EG=iGe.
Hence a, 6, c, d, e are on a circle whose radius is one-fourth of the radius
of the given circle.
Let A v B lt C v Dj, E x be the centres of the five rectangular hyperbolas
which pass through four of the five points excluding A, B, C, D, E respec-
tively. Then OaA^ is a straight line, and 0A 1 = 20a; so also 0B 1 =20b, &a.
Hence the five points A,, B lt C v JD,, E 1 are on a circle whose radius iB
double the radius of the circle on which a, b, c, d, e lie and therefore half the
' radius of the original cirole.
45. Let the equation of the conic be ax* + by i =l; then the most general
equation of the rectangular hyperbola whose asymptotes are parallel to the
axes of the conic is
xy+gx+fy + c=0.
The abseissce of the points of intersection are given by the equation
■'(St)'- 1 -
or ax* + 2afx 3 + (ap + bg' - 1) a; 2 + 2 (bgc -/) x + 6c a -f = 0.
Hence, if (x v y^j, &c. be the four points of intersection, we have
x 1 +x i +x 3 +x i =-2f,
and similarly y 1 + y % + y 3 + 1/ 4 = - 2g.
Hence the centre of mean position of the four points of intersection is
( 2' a)'
The centre of the hyperbola is easily seen to be (-/, -g), and therefore
the centre of mean position of the four points of intersection is midway
between the centres of the two curves.
46. Let the equations of the sides of the triangle be
x = 0, y = 0, and Ix + my + n = ;
and let the three parallel lines be
x - a = 0, 2/-/3 = and Ix + my + n - y = 0.
Then the curve whose' equation is
xy {lx+my + n) -\(x- a) (y - p)(lx+my + n- y)=0
1.38 CONIC SECTIONS. [CHAP.
will clearly pass through the six points of intersection, whatever the value
of X may be, for the curve meets x=0 where (y-fi) {lx + my-tn-y)=0, and
so for the other sides. But when X= 1, the curve is a conic.
47. Smce 1 _ =J _ + _ ? , J
it follows that {PGOG 1 } is harmonic.
Therefore, as GOG' is a right angle, CO and OP are equally inclined to
the axes of the conic.
Hence, if £70 cut the curve in Q, Q', QP, Q'P will be at right angles to one
another.
But all chords which subtend a right angle at P cut the normal at P in
the same point, and therefore pass through 0, which is the point of inter-
section of the normal and one such chord.
48. Take the tangent at for axis of x and the diameter through for
the axis of y ; then the equation of the conic will be of the form
ax 2 +by* + 2ey=0.
Let the equation of PP' helx+my = X.
Then OP, OP' will he given by
as 2 + by 2 + 2ey (hs+my) = 0.
The extremity of the diameter through is ( 0, - ■=- ) , and the equation
2e
of the tangent at that point is y= - -=■■ ,
Hence at the points of intersection of OP, OP' and the tangent at 0', we
have
^t-J('-t)^
Hence O'Q . 0'Q'=x 1 x i ={ie% +
from which it follows that m is constant since O'Q . O'Q' is constant ; and
when m is constant the line lx+my = l cuts 00' in a fixed point.
49. Take the tangent and normal at P for axes, and let the equation of
the conic be
oa: 2 + ZJucy + by 1 + 2fy = 0.
Let the equation of LM be y = -q.
X.] CONIC SECTIONS. 159
Then the equation of PL, PM is
V
The equation of the bisectors of the angles LPM is
X 2 -;
as- J-
■y* _xy
M h '
V
or
h{x"-y^)=xy(a-b-^).
Hence where the bisectors meet y—ri the oo-ordinates satisfy the relation
h (x 2 - y*) = xy (a - b) - 2/x.
Thus the locus of B is an hyperbola whose asymptotes are parallel to the
lines h (x i -y i )=xy (a-b) ;
and these lines bisect the angles between the lines
ax i + 2hxy+by 2 =0,
which are parallel to the asymptotes of the original conic.
Hence, as the axes of a conic bisect the angles between its asymptotes,
the asymptotes of the locus of li are parallel to the axes of the original
conic
50. Let the equation of the given conic be ax 1 + by i = 1, and let (c, 0) be
the given point in its transverse axis. Then the equation of any chord
through (c, 0) is
y-m(x-c)=Q,
and the equation of any conic which touches the given conic at the ends of
this chord is
ax i +by !1 -l + \{y-m(x-c)} i = (i).
Now the centre of the conic (i) is given by
ax-\m {y - m (x -c)} =0 (ii),
and by + \{y-m(x-c)}=0 (iii).
Also, since (i) passes through (0, 0), we have
-1 + Xm 2 c 2 =0 (iv).
The required locus is found by eliminating X and m from the equations
(ii), (iii) and (iv).
From (ii) and (iii), m= - nxjby ;
.•. mc 2 ax-y+m{x — c) = Q ;
.'. c*a?x*+by 2 +ax (Bm 1 (^-a\ + 2gacoa(?-a}+2fb&m(^-a)+c=0,
a 2 cos 2 ( j + o)+J 2 sin 2 f j + a)+25ocos( j + a)+2/6sin ( j + o)+c=0,
and c=0.
Hence, by addition and subtraction of the first two relations, we have
a 2 + J 2 + 2J2ga cos a + 2^/2/6 cos a = 0,
and (a 2 - i 2 ) sin 2a + 2J2ga sin a - 2^/2/6 sin a =
or (a 2 -S 2 ) cos a + J2ga - ^2/5 =0.
Eliminating cob a, we have
a* -5*= 4 (flW -/"&»).
But the co-ordinates of the centre of the circle are - g, -/, and therefore
the required locus is the hyperbola whose equation is
4(a 2 x z -&y) = o 4 -J 4 .
52. Let the equation of the tangent at P be x cos a + y sin a - p = 0, and
let the equation of the chord through Q, Q', the other points of intersection,
be xcosa'+y sina'-j>'=0.
Then the equation
x* y-
~2 + j- 2 - 1 + X {x cos a+y sin a -p) (xcosa'+y sin a' - J>')=0,
will represent any conic touching the given conic at P and passing through
Q and Q'. The above equation will represent a circle provided
sin(a+o')=0 (i),
and
-5 + Xcosocosa'=iT+Xsinasina' (ii).
a 2 b' - l '
Also the circle will pass through (0, 0) provided
-l + \pp'=0 (iii).
And, since x cos a + y sin a -p=0 touches the ellipse,
j) s = a 2 cos 2 a + 6 s sin 2 a (iy).
x.] Conic sections. 161
From (i) and (ii) x = ±^_I);
hence, from (iii),
a*b* 1
. a _ ft , „ = ^ =i)' 2 p 2 =p' 2 (a 2 eos 2 a + 6 2 sin 2 a) =p"> (a 2 cos 2 a' + 6 2 sin 2 a") .
The coordinates of the foot of the perpendicular on QQ' are p' cos a' and
p' sin a' ; hence the equation of the required locus is
aW + bhf = a 4 6 4 /(a s - ft 2 ) 2 .
53. The conic
x (p + S:- 1 )-^- c )= «
■will go through the points of intersection of the two conies.
If X=c, the conic (i) will represent the two straight lines
?- 2 ^+f=° W-
If the curves touch the lines (ii) must be coincident, and therefore
c 2 =a 2 6 2 . Thus c= ±a6 is the required condition.
If c 3 =:a?b 2 , the lines (ii) will be - ± ^=0, so that the points of contact are
' a b
on one or other of the equi-conjugates.
The polars of (a, £) with respect to the two curves are
-2+^ = 1 and -E + ^-=±l,
a* b- ab ab
and these polars meet on the lines
a* + l*^\ab + ab)- V '
that is on -=ft=0.
a b
54. Let the conic which goes through the five points A, B, C, D, E cut
the circle A BE in G : then, AB and CD make equal angles with the axes of
the conic [Art. 186], and so also do AB and EG ; hence EG is parallel to CD,
and therefore G and F are coincident. Hence the conic through the five
points A, B, G, D, E will also pass through F.
The directions of the axes of the conic are known since AB and CD make
equal angles with the axes ; hence the axes can be drawn when the centre of
the conic is found.
Since CD and EF are parallel chords, the line joining the middle points
of CD and EF is a diameter. To find a second diameter, draw a circle
through D, C and E : if this cirole out the conio in a fourth point H, EH and
CD make equal angles with the axes of the conic, and therefore EH is parallel
S. C. K. 11
162 CONIC SECTIONS, [CHAP.
to AB, and therefore H is found by drawing through E a line parallel to A15
to out the circle DOE in H. Then the line through the middle points of the
two parallel chords AB and EH will he another diameter of the conic. The
centre of the conic is thus determined, and the axes can now be drawn.
55. The six points are always on a conic whose equation is
(ax V + by'y" - 1) (ax 2 + by 2 - 1) - (axx' + byy'-l) {axx" + byy" - 1) = 0. . . (i),
the equation of the given cpnic being ax 2 + by 2 -1 = and P, P' being (x', y')
and (x", y") respectively [Ex. 3, Art 187].
The conditions that (i) may be a circle are
x'y" + x"y' = , (ii),
and a {by'y" - 1) = b (ax'x'' - 1),
or x ' x "~ y ' 1J " = a~b <"')•
From (ii) and (iii), by squaring and adding,
(**+y'*)(x"*+y"*) = (j-iy,
that is CP 2 . CP' 2 =CS i , where C is the centre and S a focus of the conic.
It follows from (ii) that CP and CP' make equal angles with the axes ;
and, since ^-, (x" 2 + y" 2 ) =---, and - - T is positive if the major axis of the
conic is along the axis of x, it follows that y' and y" have different signs, and
hence the points P, P' are on different sides of the transverse axis.
Since PC : CS=CS : CP' and the angles PCS, SCP' are equal, it follows
that the angles PSC and SP'C are equal and that PS : SP'=^PC : P'C. Now
let the ellipse become a parabola; then C will be at infinity and from the
above relations we see that in the case of a parabola PSP' is a straight line
and PS = SP'.
56. Let T be (a/, y') and T be (x", y"), the equation of the parabola
being y 2 -£ax=0.
Then the equations of PQ , P'Q' will be respectively
yy'-2a(x+ x') = and yy" -2a(x + x") = 0.
Hence the equation of any conic through P, Q, P', Q' is given by
\(y*-4ax)- {yy'-2a(x + x')} {yy" -2a (x + x")} = 0.
If the conic pass through (x\ y') we have
X {y' 2 - iax') - (y 12 - iaxt) {y'f- 2a (x'+x")}=0;
.: \=yy"-2a(x' + x"),
and it is obvious that when X has this value the conic will also pass through
{x",y").
X.] CONIC SECTIONS. 163
Hence the six points P, Q, P', Q', T and 2" fill lie on a eonie whose
equation is , >
(y* - iax) {y'y" - 2a {xf + x") } = {yy'-2a \x + x') } (yy" - 2a (x + x") } .
In order that this conic jnay be a rectangular hyperbola it is necessary
and sufficient that
y'y"-2a{x' + x")=y'y" + 4<>?,
or - {x' + x")= -a,
which shews that the middle point of TV must be on the directrix.
57. Let T be the point of intersection of AA', BB' and CO'. Then
TA.TA'=TG. TC = TB . TB\
from which it follows that the point T is on the radical axis of any two of the
circles OAA', OBB', OCC. Also, since the point is common to all three
circles, it is on the radical axis of any two of the circles. Hence OT is the
radical axis of the three circles.
58. Let the point be (a, /S), the equation of the conic being
ax i + by' 1 =l.
Let (x' t y') be the middle point of one of the common chords, then the
equation of that chord is .
{x-x')ax' + (y-y')by'=0 (i).
Since the extremities of the chord are equidistant from 0, the line through
(x', y') perpendicular to (i) must pass through [a, /3) ; we therefore have
a-x' _ p-y'
ax' by'
Hence (a;'", y') is always on the rectangular hyperbola whose equation is
ax [y - 18) - by {x - a) = 0.
59. Take for origin, and let the equation of the conic be
ax 2 + 2hxy + by* + 2gx + 2fy + c = 0.
Then every conic of the system is included in the equaf ion
ax* + 2hxy + by 2 +2gx + 2fy + c + \(x i +y i -W) = a (i),
where X and k are arbitrary.
Now the centre of (i) satisfies the equations
ax + Jiy+g + 'Kx=0,
and hx + by+f+Xy = 0: ;
11—2
164 .CONIC SECTIONS. [CHAP.
Hence the centre of (i), for all values of X and k, is on the conic whose
equation is
y (ax + hy+g) -x {bx + by+f) = 0,
and this conic is clearly a rectangular hyperbola.
60., Let S, S' be the foci of the conic, and PSP' be the focal chord.
Then, if the tangents at P, P' meet in T and the normals in G, the points
T, P, G, P' are on a circle, and we have
PTG=PP'G=^-TP'S
=P'TS, since TS is perpendicular to PSP'.
But we know that PTS'=P'TS, and therefore TGS' is a straight line.
Gl. The equation of the normal at s,m
a?- 6 2
Hence CG= cos<4 (i).
a , '
If the normals at the points -fa, fc, i meet in any point (a, /3), these
must be the values of tp given by
, — r-C- =a 2 - 6 z =c s .
cos
Hence (aa - chf (1 - « 2 ) = 6 a j3»«? (ii),
Where z is put for cos (i).
The line through P parallel to CP' is
~* ~ T 7'
1*
or by£-axr) + (a-b)£iti=0 (ii).
Now the line (ii) will pass through the point given by
y _ « _ 1
bfi aa a + b
for all values of £ and ij which satisfy (i). This proves the proposition.
66. Let the eccentric angles of the four points P, Q, B, S the normals
at which meet in be 8 V 2 , 3 , $ i respectively. Let Ap, Aq, Ar, As be
chords through the vertex A perpendicular to the four normals, and let
U], o s> 03 , a 4 be the eccentric angles of p, q; r, s respectively.
Then Ap, Aq, Ar, As are parallel to the tangents at P, Q, R, S respectively,
and hence
a 1 = 2e l , a 2 =20 2 , a 3 =20 3 and a 4 =20 4 .
But we know that
e 1 + e 2 +e 3 +e 4 =(2ra+l)7r. [Art. 198.]
Hence a 1 +a 2 +a 3 +a 4 =2»Mr,
which is the necessary and sufficient condition that the four points p, q, r, 8
should be on a circle,
X.] CONIC SECTIONS. 107
67. Let the eccentric angles of the points of contact of the tangents
be 0j, 2 . Then it is easy to shew that the tangents at 8 V 2 intersect in the
point
ja cos - (6 1 + 2 ) I cos | (0 X - 2 ), b sin ^ (^ + 2 ) / cos i (8 1 - 6 A ,
and that the normals intersect in the point
|-COS0 1 COS0 2 COS-(0 1 + 2 )/cOS-(0 1 -0 2 ),
e"
- v sin e
o
i S 1 sin 2 sin ^ (0 X + 2 ) / cos - (0 1 - 2 )l ,
where c 2 =a 2 -Z> 2 .
Since the point of intersection of the tangents is on the ellipse
x 2 /a 2 +i/ 2 /& 2 =4,
we have 4 cos 2 s (8 1 - 2 ) = 1 (i).
Hence, if (a;, y) be the point of intersection of the normals, we have
-j = ± 2 cos 0j cos 2 cos q (0! + 2 ) = ± cos 5 (0j + 3 ) {cos (0j + 2 ) + cos (0! - 2 ) } ,
and
- -| = ± 2 sin 0j sin 2 sin - (0j + 2 ) = ± sin - (0j + 2 ) {cos (0 X + 2 )
-cos(0,-0 2 )};
.-. (aV+JV)/c 4
= cos s (0 X + 2 ) + cos 2 (0j - 2 ) + 2 cos (0j + 2 ) cos (0j - 2 ) cos (0 X + 0. 2 )
= j , since cos (0 X - 3 ) = ± - , from (i),
so that the normals intersect on the ellipse whose equation is
4a 2 x 2 + 4&y=c 4 .
68. The normals are the perpendiculars from the angular points of the
triangle ABC on the opposite sides, and therefore meet in a point. The
eccentric angles of two angular points differ by — ; and, by the preceding
o
2tt
-(y-y')y'=0.
Hence if the normal at (x, y) pass through (X, Y) we have
[X-x)x-\Y-y)y = 0, where xy=c.
168 CONIC SECTIONS* [chap.
Henee x lt ;r 2 , se 3 , 4 t are the four roots of
Xrf-at-Yxc + ct^O;
.: (c 1 + ac !! + a; 3 + x 4 =X, , •
and similarly Vl + Vi + y 3 + y t = Y.
70. The point of intersection of the normals at lt 8„ is given by
. ax=c" cos X cos 2 cos - [$ 1 + 2 ) / cos 5 (^ - 0.J)
- by = c s sin X sin 3 sin - (0 X + 0.J j cos - (0! - 2 ).
Now for a system of parallel chords
01 + 02= constant = 2o suppose.
Hence 2i c
i/ 3 g g _ a 2 -i 2 « 4 -Z> 4
a ;i & 2_ 2c a a-'fi 2 '
73. If the normals at the four points (x,, y,), &c. meet in the point
(£, ?;).; and if r t , r 2 , r 8 , r 4 be the lengths of the four normals ; then
&:,•=« (f-a; 1 )« + 2(7,-y 1 ) 2 =4(J 2 + ., 2 ).-afS* 1 -27 Z 22/ 1 + 2a; 1 2 + 2j/ 1 2 . ,
170 conic sections; [chap.
But, from the previous question,
2 ^ = S and 2j/l2= (S? { -° ! f 2 + 5 V + (a 2 -6 2 ) 2 }.
Hence :^=4(t 2 + „ 2 ) - ^ + ^
a 2 - 6«
a-* - o*
Hence the required locus is a conic.
74. The feet of the normals from (/, g) to the ellipse
a 2 /a 2 +j/ 2 /6 2 -l =
lie on the conic
xy {b- *- a 2 ) + a?fy - b'gx = 0.
Hence the equation of every conic through the four points of intersection
is included in
a; 2 /a 2 + 2/ 2 /i 2 -l-X {xy (W-a?) + a?fy -&V} = (i).
Hence, for some value of X, (i) will represent the same conic as
(ix'/a 2 + yy'jb 1 - 1) (ra"/a 2 - yj/"/6 2 - 1) = 0,
x'x" _ y'y" _ a' +5" _ y' + y"
'"' a 2 ~ b n - ~ Wd'g ~ -■hd 2 b i f~ ~
x'x'' 11'y" ,
Hence -^- = £-=-=-1 ifi)
and f(x' + x")+g[y' + y")=0 (Sit).
From (iii) it follows that the middle point of the line joining (a:', y') to
[x", y") is on fx+gy=0 ; this proves the theorem, since (j;', y') and (x", y")
may be the extremities of any one of the diagonals of the quadrilateral formed
by the tangents at the feet of the normals from (/, ( x s> Vt) and (^4. Vi) co-interseet. Then the four extremities
(a 2 & 2 \
— , — ) , &c. To
shew that these four points are on a straight line, it will be sufficient to shew
X]
CONIC SECTIONS".
171
that any three are on' a straight line, the conditions for which are of the
l'orm
6 2
1
1
= 0, or
y^'
*i
i 2
1
1
■Va
X%
J 2
i
1
yl'
*a
- s- '
=0.
.(i).
A, i
But, since the normals at the points (x v ■yj, &c. co-intersect, the following
equations are simultaneously true, namely
^_^ + 6 »_ a » = 0,&c.
*i Vi
By eliminating a 2 a;, - b"y and 6 2 - a 2 from any three of the last four
equations we see that the conditions (i) are satisfied.
76, The equation
X (x 2 /a 2 + j/ 2 /6 2 - 1) - (a s - ft 2 ) xy +a?fy - b*gx =
includes all conies through A, B, C, D, the normals at A, B, C, D meeting
at (/, g).
If the conic pass through the point (ae, 0), we have
X (e 2 - 1) - b"gae = 0, or X = - ga s e.
Also x=- is to cut the conic in co-incident points ; and therefore
-<*w(J+p-i)-(« , -a , !)fy+«*fr-i , fc;=o
has equal roots, the condition for which is that 8aV=a 4 (ae -/) 2 .
Hence the locus of (/, g) is the two straight lines ae - x= ± 2^/2^.
77. As in question 72, the abscissae of the four points the normals at
which meet in ({, ij) are the roots of the equation
{a 2 £ - (a 2 - S 2 ) a;} 2 (a 2 - x 2 ) = a»&V* s .
Now substitute for x, and we shall obtain an equation whose roots
are the values of the distances of the four points from the focus (ae, 0) : this
equation will be
{a 2 = cos l^- $ J •
.: ~{l + cos(9 1 + e 2 )} = l + cos(9 1 -0 2 ) (i).
Now ^A v /^|=<:os(9 1 + e i! ) + oos(0 1 -fl 2 );
.-,, from(i), ^ + -=cos(fl 1 + fl I ) (ii).
.-., from (i), . ? = sin(e l + « a ) (iii).
Hence the locus of (x, j/) is the circle whose equation is
\2
(--♦S)'
+r/ 2 =c 2 .;
X.] .CONIC SECTIONS. 173
80. If the normal at {x, y) to x>ja? + j/ 2 /o 5 - 1 = pass through the point
(I, ij), we have
£-*_V-V_ r .
*_ y /(*>, t\ '
a 2 6'- 1 V WW
where r is the length of the normal from (£ , tj) to the curve.
Hence J^VsV • Ife^VW
{a* - (a 2 - Z> 2 ) */} {a 4 - (a 2 - 6 2 ) a,'} {« 4 - (« 2 - « ! ) */},
where x lt x 2 , x 3 , x± are the roots of
{a 3 £-(a*-b-)x}*(a*-a?)-a*liW=0 (i).
From(i), a; 1 V^ 3 2 a; 4 2 =a 12 J 4 /(a 2 -6 2 ) 4 .., (ii).
Put £=£ -X in (i), then
{o 2 £+(a 2 -& 2 )X} 2 {a 2 -(|-X) 2 }-a 2 &V(f-X) 2 =0;
.-. (if - sjs (J - s,.) 2 ({ - s 3 ) 2 (I - sJ^VW
= & 4 £ 4 (a 2 & 2 -& 2 | 2 -aV) 2 /(a 2 -& 2 ) 4 (iii).
Now a 4 - (a 2 - 6 2 ) a; x 2 =a?(a- ex{) (a + exj) ;
and putting a - ex x =^ in (i), we have
o 2 (I - a« + /te) 2 {a"«" - (a - ji) 2 } - & V (a - /i) 2 = 0.
. Hence II (a - ea^) = ooeffieient of /t° 4- coefficient of /*.*
= -{(|-ac) 2 + ^ 2 }6 2 /« 2 .
Similarly n (a + ca; 1 )= - {(| + 2 /e 2 .
Hence
»i Vr,V = (« 2 6 2 - & 2 I 2 - a V) 2 { (I + ae) 2 + 1 2 } { (f - aef + ij»}/(a s - b*f.
Now, from question 81, the product of the squares of the tangents
= (a 2 6 2 -6 2 | 2 -aV) 2 {(| + ae) 2 +7? 2 } {(f-aeJ' + ijWF + ay) 11 .
Also the product of the squares of the perpendiculars from (£, tj) on the
asymptotes
= (6 2 i 2 +oV) 2 /(i 2 -6 2 ) 2 -
Hence the continued product of the four normals is equal to the continued
product of the two tangents and of the two perpendiculars on the asymptotes.
81. Since the lines #+Xi/ = and x+/iy=0 are in the direction of
conjugate diameters the equation of the curve will be
A(x+\y)"+B{x+^)"=l (i).
174 CONIC SECTIONS. [chak
. The condition that .either of the lines x + \y = ±p should touch (i) is easily
found to be Ap*=l; and the condition that either of the lines x + /ty=±q
should touch is 2?g 2 = 1.
Hence the required equation is
(s+X2/) 2 /p s +(a;+w) s /2 2 =l.
82. Let T be the point (x', y') ; then the equation of PQ is
xx'lcP+yy'ltf-l^G.
All conies through P and Q are included in
x 2 ja i +y i lb'-l-{xx'la' + yy'jb 1 -l)(lx+my+n)=0 (i).-
Hence (i) is the circle TPQ provided
l = lx' + my' + n (ii),
1 lx' 1 my' .....
a 2 a' ft 2 u 2
. hi' mx' .. .
°=i + ls- ; ( iy »-
From (ii), (iii), (iv), we have
la' mb* n i .
x'{tf-a2)~y'(a?-b2)~x'*+y' 2 "' (V);
Hence, if x' 2 +y' i =c*,
we have
JW + mV& 4 =n 2 (a 2 -& 2 ) 2 . (vi).
Now lx+my+n=0
is the equation of PQ', and (vi) shews that
lx + my+n=0
touches the conic
ic 2 i/ 2 c 2
a t+ ¥~(a*-by'
83. Let the equation of the chord be
y=m(x-ae) ;
then the points A, B are
{a, ma(l-e)} and {-a, -ma(l + e)}
respectively.
Hence the equation of the circle is
(x - a) (a + o) + {y-ma (1- e)} {y + ma{l + e)} = 0,
or B 1! + 2/ 2 +2ma^-a 2 -m !! 6 2 =0.
X.] .CONIC SECTIONS. 175
The circle cuts the ellipse where
y'- (a 2 - ft 2 ) - 2maeWy + m?b 4 = 0,
that is (aey-mV*)*=0.
Thus the circle cuts the ellipse in two eoinoident points,
84. The points A , B are the points of intersection of the director-circle
and a tangent to the ellipse.
The equation of the director-circle is
x 2 + y*=a?+b*.
Hence, if Ix + my — 1 =0 be the equation of the tangent AB, the equation
a?+y*-a*-& + \{te+my-l) = (i) '
is the equation of any circle through the points A and B.
The circle (i) passes through the focus (ae, 0) provided
X = 26 2 /(tae-l);
and, with this value of X, (i) may be written
/ lb 1 Y / 7ii6 a y , ,, (P+m^b* 26 2
In order that the radius of the circle may be equal to o, it is necessary
and sufficient that
(Z 2 +m 2 )Z > 2 2
(lae - 1) 2
or (Z 2 +ro 2 ) b 1 +2lae-2 + PaV-2lae + l = 0,
or Z 2 a 2 +m 2 & 2 =l (ii).
But (ii) is the condition that Ix + my = 1 should touch the ellipse
x 2 /a 2 + 2/ 2 /6 5 =l.
85. Let the equations of the parabola and of the circle be
y*-4ax=0 and a 2 +.(^-/3) 2 =c a
respectively.
The line i/=mx+— touches the parabola for all values of m; the line
m
touches the circle provided ■
^--)/(l + m 2 )= C *,
or . . c 2 m 4 + c 2 m s -(/3m-a) 2 =0.
Since the coefficient of m 3 is zero in the above equation, it follows that
the sum of the tangents of the angles the four common tangents make with
the axis of the parabola is zero.
176 CONIC SECTIONS. [CHAP.
86. Let (a cos a, a sin a) be the point from which the tangents are
drawn ; then the equation of the tangents will be
(.r 2 /a 2 + y 2 /6 2 - 1) (cos 2 a + a 2 sin 2 a/6 2 - 1) - (a cos a/a + ya sin a/6 2 - 1) 2 =0.
Hence the equation of any conic through the four points in which the
tangents cut the directrices is
(K 2 /a 2 + j/ 2 /6 2 - 1) (cos 2 a + a 2 sin 2 a/6 2 - 1) - (x cos aja + ya sin a/6 2 - 1) 2
+ X(e 2 x 2 -a 2 ) = (i).
We have to shew that, for some value of \, the left member of (i) will have
a factor of the form y - nix, and may therefore be written in the form
. f j/sin 2 a II sin a cos a msin 2 a\ „ . ,,,) . ....
(y-mx) |- iL -j* *{ p + pj— J+2asina/6 2 j-=0...(u).
Hence, equating the coefficients of a; 2 and x and the constant terms in (i)
and (ii), we must have
2m sin a cos a +m 2 sin 2 a = sin 2 a-l + e 2 +X6'e 2 (iii),
2amsina/6 2 = -2 cos a/a (iv),
and 0=cos 2 a/a 2 + sin 2 a/6 2 +X (v),
simultaneously true for some values of X and m and for all values of a.
From (iv) m=s - (1 - e 2 ) cot a ; and, when this value of ro is substituted in
(iii), we have X6 2 =e 2 cos 2 a- 1, which is equivalent to (v). Thus the values
of X and m which satisfy two of the relations (iii) (iv), and (v) will also satisfy
the third relation, and therefore the equation (i) can be written in the
form (ii).
Hence two of the four points of intersection of the tangents and direc-
trices are on a straight line through the centre, and the equation of the other
two, is, from (ii),
x (2 sin a cos a +m sin 2 a) + y sin 2 a -2a sina=0,
where m= - (1 - e 2 ) cot a ;
and this line cuts the major axis in the point
{2aseca/(l + e 2 ), 0}.
87. Let T be the point of intersection of PC and P'C, and let PCQ,
P'G'Q,' be diameters of the two conies; then, since PP' is parallel to CC,
it follows that QQ' must also be parallel to CC'.
Hence TQ . TP _ TQ> . TP'
But, from Art. 186, Cor. in., TQ . TP : XCP=:u : «„,
and TQ' . TP' : TC'*=v : v .
Hence the locus of T is given by uv =vu . The locus of T clearly passes
through the points of intersection of the given conies [as is obvious S priori].
When u = and v = are similar and similarly situated, the conic uv -u v=Q
will be similar and similarly situated to either.
X.] CONIC SECTION'S. 177
88. Let the equation of the conic be ax 1 + by 1 - 1 = 0. If a circle have
double contact with a conio the chord of contact must be parallel to an axis
of the conic, for ax s + by i -l+\(lx + my + n) 2 =0 can only represent a circle
when to=0. Also a circle will not pass through the four points of intersec-
tion of the conic ax*+by* -1=0 and the lines {x -. a) (y - p) = 0. Hence the
two chords of contact of the circles in question must be parallel. The equations
of the three circles will therefore be of the forms
ax> + by 3 -l + \(x-a)*=0 (i),
aa? + ty"-l + 0(a:-/3)»=O (ii),
and ax i + bij 2 -l + v(x-a)(x-p)=0 (iii).
But, in order that (i), (ii) and (iii) should represent circles, we must have
\=/i=v = b-a.
The squares of the tangents to the different circles from any point (x, y)
on the conic are respectively
{b -a)(x- a) 1 , (6 - a) [x - p) 2 .and (6 - a) (x - a) (x - 0).
Hence tH' 2 - T 2 =0.
89. If a circle have double contact with a conic, the chord of contact must
be parallel to one or other of the axes of the conic, and the centre of -the circle
must be on an axis of the conic. Hence if a conic have double contact with
each of two circles the chords of contact must either be parallel or perpen-
dicular.
First let the chords of contact be parallel ; then the centres of the two
circles must be on the same axis and the chords of contact must be perpen-
dicular to this axis.
The equation of the conic must therefore be of either of the forms
(x - a) 2 + y* - c 2 + X (x - a) 2 = 0,
or (x-6) 2 +2/ 2 -d 2 +yn(x-/3) 2 =0.
From this it follows that
1+X _ 1 _ c 2 -a 2 -Xa 2 _ a + Xa
1 + H~1~ &- IP -p.fi ~6+M/3'
HenceX^, a-6=X(0-a) and 6 2 + c 2 -a 2 -d 2 =X(a 2 -/3 2 );
.'. Xa={6 2 + c 2 -o 2 -d 2 + (6-a) 2 }/2(6-a).
Hence the general equation of the conic is
4(6-a) 2 X{\x 2 +(.'C-a) 2 -|-2/ 2 -c 2 }-4(6-a){26 2 -l-c 2 -d 2 -M.}Xx
+ {2i 2 + c 2 -d 2 -2a&} 2 =0.
Next suppose that the two chords of contact are at right angles, and that
their equations are
a: cos a + 2/ sin a -#=0, and £ sin a-?/ cos a -2 = 0.
S. C. K. 12
178 CONIC SECTIONS. [CHAP.
Then the equation of the conic will be of either of the forms
(x-a) 2 +^ 2 -c 2 +X (xcosa+3/sina-i>) 2 =0,
and (x - b) 2 +y 2 - d? ■{■ /j. (a; sin a-?/ cosa-g) 2 =0.
We therefore have
1 + X cos 2 a _ 1 + X sin 2 a _ X _a + \p cos a _ \p sin a _a?-c 2 + Xp 2
l+/t sin 2 a~ l+/iC0s 3 a — - /i~ b+/iq sin a - -/J.q cos o — W-dP + pa-'
Whence we have p sin a=a cos a, and then
(6 2 - d*) X + (o 2 - c 2 ) /i + Xjip 2 sec 2 a = 0,
X6 + /ua + X^p sec o = 0,
and X-f-/*+X/*=0.
Eliminating X and /4 we see that p sec a is constant ; we have also
X = (6 - a) I (p sec a - b).
Hence in this case the general equation of the conic having double contact
with the two circles is
(x-a) 2 +2/ 2 -c 2 +^^-^{coso(a;-pseco)+ysino} 2 =0,
where p sec a is known.
In the second part of the question the chords of contact are assumed to
be parallel and the locus of the extremities of the latus rectum which is
parallel to the chords of contact is required.
The general equation of the conic is found as above to be
(x + a) 2 + if - c 2 + Xx 2 - 2xa + ^ = 0,
A
or a 2 (l + X)+2/ 2 =c 2 -a 2 -^.
X
Hence, if (x, y) be an extremity of the latus rectum parallel to x=0, we
have
"-^-(*-"-9--&.
and y t =i ( < ?-^-~\(l+\).
Eliminating X we have the required result.
X.] CONIC SECTIONS, 179
90. The centre of the first conic is given by
{I'm' - J'%) x + (I - V) mm'y = lm' - I'm,
and (l-l')mm'x + (m-m')mm'y=0;
whence ; = -„- —,— ,—, — jr - (i)-
m-m I -I lm - lm
The centre of the second conic is also the point given by (i), and the com-
mon centre of the two oonics is the point of intersection of the two straight
lines Ix + my -1=0 and l'x + m'y -1=0.
Now it is clear that all conies through the points of intersection of two
given concentric conies are concentric with the given conies ; and therefore
the lines Ix+my - 1 = and l'x + m'y - 1 =0 are diameters of all the conies of
the system.
The equation of any conic through the points of intersection of the given
conies is {Pm' - l*m - Ml' (I - l')}x* + 2{{l - V) mm' -\{m- m') IV} xy
+ {(m-m')mm'-\(mH'-m'n)}y*-2(lm'-l'm)(x+\y) = (ii).
The lines Ix+my -1 = and l'x + m'y -1 = will [Art. 183] he parallel to
conjugate diameters of (ii), provided
IV { (m - m') mm' - X (mH' - m K l) }+mm' { Pm' - V*m - XIV (I - 1') }
= (lm' + I'm) {(I- V) mm' - X (m - m!) IV),
and it is easy to see that this condition is satisfied.
Hence the lines Ix + my - 1 = and Vx + m'y -1=0 are conjugate diameters
of (ii) for all values of X.
91, Let ox 2 +6y 2 -l = be the equation of the conic, and (/, g) be the
fixed point.
Then, if (f, i;) be the middle point of any chord through (/, g), the equa-
tion of the chord will be a J (x - £) + by (y - ij) = 0.
Hence, for some value of c, the circle (x - |) 2 + (y - iff - c 2 =0 will cut the
conic in points such that a( (x - £) + br/ (y - 7/) = will go through two of their
points of intersection, and the equation of the line through the other two
points will therefore be af x - bifij + n = 0.
Hence for some values of X, c, and n, the equations
asc 2 + % 2 - 1 + X { (x - 1) 2 + (?/ - 1?) 2 - c 2 } = 0,
and (a£x + bijy - at? - brf) (a £x - bqy + n) =
will represent the same pair of straight lines. Hence, comparing the
coefficients of x and y, we have
a$ (n - a£ 2 - Jtj 2 )/2X? = b-q (m + a£ 2 + 6ij 2 )/2Xij,
whence n = (a? + brj i )(a + b)l{a-b).
But the chord a£x + briy-a^-Vrp=0 goes through the fixed point (/, g) ;
12—2
180 CONIC SECTIONS. [CHAP.
And this shews that the line
passes through the fixed point given by
x _ y _ a + b
f~~g~~T-b'
92. Iiet ABC be the triangle, and let P, Q be the given points. Then, if
AP, BQ, CR out BC, CA, AB respectively in the points A', B', C, we know
that BA' . CB' . AC'=BC . AB' . CA'.
So also BA" . CB" . AC"=BC" . AB" . CA".
Hence BA' . BA" . CB' . CB" .AC. AC"=BC. BC" . AB' . AB". CA' . CA".
Hence, by the converse of Carnot's theorem, the six points A', B', C, A",
B" and C" are on a conic.
93. Let AB, BC, CD be three of the sides of the quadrilateral, and let
these lines pass respectively through the fixed points P, 0, Q which lie on a
straight line. Take the point O for origin, and POQ for the axis of x, and let
P be (a, 0) and Q be (/3, 0) ; then the equation of AB and CD will be
{y-^x-a)} {y-m tl {x-p)}=0.
Hence, if ax i + 2hxy + by 2 + 2gx + 2fy + c=0
be the equation of the conic, the equation of BC and AD will be
included in
ax i +2hxy + by ! ' + 2gx + 2fy+c+\{y-m 1 {x-a.)}{y-m 2 {x-p)} = (i).
The points where (i) cuts the axis of x are given by the equation
ax i +2gx + c + \m 1 m s (x-a) (a:-/3) = 0.
Hence if (i) pass through the origin we have
and the other point of intersection is given by
ax+2g- ~ {as- (a + /3)} =0,
so that a;=(2jra(3 + ca + c i 8)/(c-aa|3),
and this is independent of X, %, and m 2 . Thus any conic through A, B, C,D
and will cut OPQ in another fixed point.
[The theorem follows at once from Art. 320, Ex. 2.]
94. Project the conic into a circle having the point of intersection of
PQ and P'Q' for centre. [See Art. 318.] Then, if PP' pass through the
fixed point S, it is obvious that QQ' will pass through the fixed point ;S',
where S, S' are on a line through the centre of the circle and are equidistant
X.] CONIC SECTIONS. 181
from the centre. Also, since the angles SPQ' and SP'Q are right angles, it
follows that PQ' and P'Q touch a conic of which S (and similarly S') is a
focus, and of which the given circle is the auxiliary circle. This proves the
proposition, since a conic and its auxiliary circle have double contact with
one another, and this relation is unaltered by projection.
95. Let the tangents at the points whose eccentric angles are a, 3 cut
the lines x 2 -a 2 =0 in points P, Q such that PQ is parallel to one of the
equi-conjugates.
Then, if (x, y) be the point of intersection of the tangents, tan 5 and
tan ? are given by
?cos0 + ? sin 9 = 1,
a b
orby ?-l + 2?tanf-(? + l)tan 2 ? = (i).
Hence
tan£tanC = (a_a;)/(a+a:) and tan- +tan;; = 2ay/& (x + a) (ii).
Now, y = b tan - at P ; and y = b cot 5 at Q. Hence, if PQ be parallel to
one of the equi-conjugates, we have
tan^-cot 5=±2, or tan5tang-l = ±2tang.
x B
Hence, from (ii), = ±tan£.
a + x 2
Substituting for tan P in (i) we have 2xy= Jr-ab. Hence the locus of the
point of intersection is one or other of two rectangular hyperbolas.
96. Let the equation of the rectangular hyperbola be xy=c 2 , and let
L, M, N, It be (*„ y x ), (x 2 , y 2 ), (x 3 , y s ) and (x 4 , yj respectively.
Then the equations of LM and NR are respectively
*jhy, + c'ty-c»(y 1 + y s ) = and xy 3 y 4 +e 2 y-e 2 (ft + yj = 0.
Hence, if P be (£, ij), the equation of PAa is
5-Z^ = V-JI = r
W2 c* x/(2/iV + c 4 )'
Hence, Pa is the value of r given by
too/* + rvmMJ4*JWm + c*r,-c*(y s + yi ) +
But PA = {twjz+ch, - C 2 (y^+y^UWyf+c*)
=Hvi-n) (ys-vVsHy^+c*).
Hence
pa. Pa--?. {yi-v)(y Z> 2 -X 2 (1-c 2 cos 2 m)
06 C ^- Ci - "^r j 2 e 2 cos 2 W -a 2 + aVcos 2 U -
Since the value of Cg.Cl would be unaltered by changing e into - e, it
follows that Cg.Cl= Cg' . CV.
184 CONIC SECTIONS. [CHAP. X.
100. Let the equations of BO, CA, AB be
/ 1 K+m 1 j/-l = 0, Izx+mjy -1=0,
and ltfc + m0-l=O
respectively ; and let the equations of ajbfa and aj)^ be
X 1 a; + /t 1 j/-l =
and t X 2 a; + /i 2 7/-l=0
respectively.
Then the general equations of eonics circumscribing
Jj&^jCij, CjC^o, and a^Z^&a
are respectively
(ZoZ+m.^ - 1) (7 3 x+ ?%?/ - 1) H-ftj (Xja+Mi?/ - 1) (Xp + WJ - l)=0...(i),
(l^ + mtf-l) (fji + nhy - l) + i 2 (^as + ^y - 1) (XjjS+^j/- l)=0...(ii),
and (l^x+mfl-l) (l^x + m^y -l) + k 3 (\ 1 x-^ii 1 y~l) (\ i x + /j.0 -l) = 0...(iii).
Now k 3 (iji + TB^ - lJ.fZjjc+mjy - 1) - k 2 fax+mtf - 1) [l^+rn^y - 1) =
is a conic through the points of intersection of (ii) and (iii), and this is two
straight lines of which BC is one and the other is given by the equation
k 3 (l 3 x+vi s y -l)=Je 2 (l 2 x + m. 2 y -1) (iv).
Similarly the equations of the other common chords of (iii) and (i) and of
(i) and (ii) are respectively
^{IjX+m^y -l) = k 3 (l 3 x+m 3 y -1) : (v),
and k% (l 2 x + m0-l) = k 1 {l 1 x + m ] y-l) (vi).
It is obvious that each of the lines (iv), (v), (vi) passes through an angular
point of ABC, and that they all meet in a point, namely in the point given
by k 1 (l 1 x + m 1 y-l) = k 2 (I^:+m i y-l) = k 3 {l 3 x + m 3 y-l).
CHAPTER XI.
Pages 257—263.
1. Take the two straight lines for axes, and let u be the angle between
them.
Let (£, 17) be the centre and c the radius of a circle which intercepts fixed
lengths 2a, 2b respectively from the axes.
Then the equation of the circle is
(x - if + (y - vf + 2 (x - {) {y - v ) cos w - c 2 = 0.
This circle meets y = where
x=f +7i cos h>±^/(c 2 - rp sin 2 a).
Hence a = ^/(c 2 - ij" sin 2 a) ;
and similarly b — J (c 2 - £ 2 sin 2 w) .
Hence a 2 - 6 2 = (£ 2 - ij 2 ) sin 2 u.
Hence the locus of the centre of the circle is the rectangular hyperbola
whose equation is
x*-y*= (a 2 - 1 2 ) cosec 2 a.
2. Take for origin and OPP', OQQ' for axes, and let the equation of
the conic be
ax 2 + 2hxy + 2by 2 + 2gx + 2/y + c = 0.
Then the equation of any other conic through P, P', Q, Q' is
ax 2 + 2hxy + by* + 2gx + 2fy + c + ~hxy = 0,
or in polars
r 2 {acos 2 + (27iO)sin0cos0 + &sin 2 0} + 2r(0COS0+/sin 0)+c = O.
Hence - + - = - 2 (g cos +/ sin 0)/c,
and is independent of \.
Henoe cSj + sfc'SS + ds"
where S, S' are the points where ORE' cuts an^ conic through P, P', Q, Q'.
186 CONIC SECTIONS. [CHAP.
3. Take the tangent and normal at as axes, and let the equation of
the conic on which lies be
as 2 + 2hxy + by 2 + 2fy = 0.
Let oV + 2 h'xy + b'y* + 2g'x + 2f'y + c' =
be the equation of any other conic through the four given points ; then the
equation of any other conic through the four point3 is
a.r 2 + 2hxy + by* + 2fy +X (oV + 2h'xy + b'y 1 + 2g'x + 2f'y + c') = 0. . . (i).
The abscissae of the points where y=0 cuts the conic (i) are given by
ax* + X {a'x* + 2g'x + c') = 0.
Hence — h — = — ^- and is therefore independent of X. Thus
1 1 .
7>p + 0F ia constant -
4. Let the equation of the hyperbola be xy = a? and the equation of the
circle (x-a) 2 +(?/-|S) 2 -c 2 =0.
The abscissae of the points of intersection are given by
x 2 {x - a) 2 + (a 2 - §xf - c 2 z 2 = ;
.'. x 1 + x i +x a +x i =2a.
Similarly 2/i+2/ 2 +2/s+2/4= ;2 / s '
Now if the points (x 3 , y 3 ) and (x 4 , ?/.,) are extremities of a diameter of the
hyperbola x 3 + x t = and y 3 + y t = 0. Hence in this case
a; 1 + a; 2 =2o and y 1 +y 2 =2p,
which shews that the centre of the circle is the middle point of the line
joining {x lt y,) and (x 2 , yj.
5. We know from Art. 297 that the axes of the parabolas are always
parallel to conjugate diameters of any conic of the system.
We also know that in a given ellipse the acute angle between two con-
jugate diameters is least when they are the equi-conjugates; and that in
different ellipses the angle between the equi-conjugates is greatest in that
which has the least eccentricity. Hence if the directions of a pair of con-
jugate diameters are known, the conic has the least eccentricity when they
are the equi-conjugates.
6. If TQ, TQ' be the tangents, and Vhe the middle point of QQ'; then
will TV be a diameter and QQ' will be parallel to its conjugate. Hence, as
in 5, the eccentricity of the conic will be least when one of the equi-
conjugates is TV and the other is parallel to QQ'.
7. The equation of any conic touching the axes in the points (a, 0)
(0, 6) is
(H- 1 ) 2 " 2 ^ 01
XI.] CONIC SECTIONS. 187
and the condition that -= + ^ - 1 = should touch the conic is '
I m
*-e-})G-s- [ "- 216 -'
Now, if (x, y) be the middle point of the intercept made on
I m
hy the axes, we have 2x = I and 2i/ = m.
Hence the equation of the required locus is
2
The locus is therefore a rectangular hyperbola unless X = — - ; and when
ab
2
X = -j the locus reduces to a straight line, and the original conic is a para-
bola for this value of X,
8, Let the equation of the conic be
-2Xx»y = 0.
Let (a, j8) be the centre of the circle ; then its equation will be
x 2 +y* + 2xy cos u - 2x (a+/3cos w) - 2y (fi + a cos w)=0.
Hence the equation of PQ is
x 11
— J-— ;=1.
(H-0'
2(a+/3cosw) 2 (/3 + o cos u) "
In order that PQ should touch the conic we must have [Art. 216]
x=2 J! L 1 11 A I
[a 2(a+j3cosu)J [6 2(/3 + acos u)J '
Hence the locus of (x, y) is the hyperbola whose equation is
2 (x +y cos w) (j/+a;cosw) (2-Xa6)-2a(i/+xcos a) -2b (x+yoos u)+ab = 0.
9. Let OA = a and OJB = J; then, if the area of the triangle OAB be
constant, ab= constant =c 2 suppose.
The equation of any conic touching the axes at A and B is
-2Xot/ = 0;
and if the conic pass through D, (a, b), we have 1 - 2Xai = 0.
(M-)'
188 CONIC SECTIONS. [CHAP.
Hence the equation of the conic is
x 2 y* xy 2x 2y ■
a' l? (H-i)=o.
One of the conies of the system ia the pair of coincident straight lines
(H-')°=°.
and any line through (a, /3) cuts this conic in coincident points : this explains
the presence of - + ^-1 = as part of the locus of the points of contact.
11, Let the equation of the conic be
ax" + 2hxy + by 2 + igx + 2fy + c = ;
and let the equations of the straight lines be ljx + m^j - 1=0, &c.
XI.]
CONIC SECTIONS.
189
Then we have four relations of the form
Al 1 " + 2Hl i m 1 +Bm 1 a + 2Gl 1 + 2Fm l + C =0 (i),
where A, S, C, F, G, 11 are the co-factors of a, b, c, /, 17, h respectively in
the determinant
= A.
a
h
9
h
b
f
a
f
c
The equation of the polar of ({, 77) is
(aZ+hri+g)x + m + bri+f)y + gt+f v + c = 0.
Hence if the polar of ({, tj) is the fixed straight line \x + iaj + 1 = 0, we
have
a£ + hT}+g + k\ = 0,
M + br)+f + kfi.=0,
and g£+fy + c + k=0.
Hence f {\G+pF+C)=\A + /iH+G (ii)
and ii[\G+iJ.F+C)=\H+fiB+F (iii).
Eliminating A, S, C, F, G, H from the equations (i), (ii) and (iii), we have
= 0.
h*
2\my
m^
*h
2/%
1
V
2l.jii 2
™ 2
m 2 -
*h
2?^
1
V
2' s »»s
m 3 2
«,
2r> h
1
7 2
'4
2Z 4 m 4
m 4 2
2J 4
2m i
1
X
A»
1-X|
-vi
-1
X
7*
— X17
l-/"7
-7)
Multiply the last column by X and subtract from the last but two, then
multiply the last column by /i and subtract from the last but one, and we
have the equivalent determinant
= 0.
7 2
'1
2? 1 m 1
mj 2
27^-X
2m 1 - n
1
7 2
'2
27 2 m2
m 2 2
27 2 -X
2)112 -jU
1
7 2
2737)83
>V
2? 3 -X
2m 3 -/x
1
7 2
'4
2Z 4 m 4
m 4 2
2/^-X
2m i -n
1
X
7»
1
-J
X
7*
1
-fl
from which it is obvious that the locus of (f , r;) is a straight line.
[The theorem can be easily proved'by using Trilinear Co-ordinates, as in
Art. 281.]
12. Let the equation of the hyperbola be £i/=c 2 . Then the equation
of any conic touching the asymptotes of the hyperbola is of the form
(lx,+ my - l) 2 =Xin/.
190 CONIC SECTIONS, [CHAP.
Hence the conic whose equation is
(te + OTJ/-l) 2 =\c 2
passes through the points. of intersection; and this conic is the two straight
lines
Ix + my — X±c ^=0,
which are clearly parallel to, and equidistant from, the chord of contact
whose equation is
Ix + my — 1 = 0.
13. The equation of any one of the conies may be taken to be
a 2 + 6 2 '
with the condition a + b = constant = c suppose.
The equation of the polar of (x' t y') with respect to the above conic iq
xx'la 2 +yy'jb !S =l.
Hence if the polar is the fixed straight line Ax+By = l, we have
x'ja?=A and y'/i»=B,
Hence ^/ x - + ^/t=a + b=c
Hence the required locus is the parabola whose equation is
= c.
14. Take* two of the tangents for axes and let the equation of the
other given tangent be
? + f-l=0.
h k
Let the equation of the parabola be
x/fVH'
then the equation of the line through the points where the parabola touches
the axes is
5 + ^-1 = (i).
a o w
Since the parabola touches the line
h k
the equation.
XI.] CONIC SECTIONS. 191
must have equal roots, and therefore
^-1=0 (ii).
a b v '
Now (ii) shews that the chord of contact (i) passes through the fixed point
(h, k) ; this proves the theorem.
The theorem may be thus enunciated: — TA, TB are tangents to a
parabola which are met by any other tangent in the points P, Q respectively,
and the parallelogram TPOQ is completed; then the point will be on AB.
15. Take the two given lines for axes, and let the equation of the
parabola be
and let the chord of contact pass through the given point (a, b).
Now the equation of the chord of contact is
therefore ^ + ^-1 = (i).
The focus is given by the equations
x* + y , + 2xy cos w = hx=ky (ii).
Eliminating h and ft from (i) and (ii), we have the equation of the
locus of the focus, namely
ax + by = x 1 + y % +-2xy cos u.
The locus of the focus is therefore a circle.
16. The axis of the parabola must be parallel to the line ax-by = 0.
Hence if the axis pass through the fixed point (a, /3) its equation must be
a(x-a)-b(y-p) = 0.
Now the focus is given by the equations
x i + y i +2xycos■
which is the required relation between 0, X 2 and X 2 .
XI.]
CONIC SECTIONS.
From (i) we have
'-I
(l-tan 2 ?)* <^"
and this equation is
a \
clearly satisfied by tan 2 s = - l or by
I A. 2
tan 2 "- V.
2 X,
195
(ii).
Since the normals to the confocals through any point biseot the angles
between the tangents drawn from that point to the original conic, the equa-
tion of the tangents when referred to the two normals as axes will be
y= isctan ^ ; or, from (ii),
?-V^=0.
Xj x 2
25. Let d v d^ be the lengths of the semi-diameters parallel to OPP' and
OQQ' respectively.
Then [Art. 186] ^| = J-
PP' d 2
Also [Art. 229] ^, = | 2 ;
.-. OP .OP 1 . QQ'=OQ . OQ' . PP'.
Or thus
The line through 0, (a, /3) which makes an angle with the axis of x
cuts r«! 1! /a 2 +2/ 2 /6 1! -l = the points whose distances from are the roots of
-/cos 2 sin 2 0\ fa cos 8 j3sin0\ o !
\W + ~W) + r \~~^~ + ~b>~) + rf
Hence
OP . OP' a 2 ^ 6 2
t-. + ^-i=o.
PP' I If aoosB /3sin0\ 2 /a 2 /S 2 ^\f cos-8 sin£0\|
= a& (-2+-S -l) / x /{cos 2 fl(6 2 - l 8 2 )-2a/3sin9cose-l-sin 2 9(a !! -a 2 )}.
Now if OPP' touch the conic a; 2 /(a 2 + X)+t/ 2 /(6 s + X) = l we have
/3-atane= x /{(a 2 +X)tan 2 e + 6 2 + X},
.-. sin 2 8 (a 2 - a 2 ) - 2a/3 sin cos 9 + cos 2 B (/3 2 - b") = \.
13—2
196 CONIC SECTIONS. [CHAP.
„ OP. OP' db fa? 0> \
Hence __ = __ . (_ 1+p - lj .
„. ., . OQ.OQ' db /a 2 , /3 2 ,\
Simil an y ^«=_ x) .^ +r2 -ij.
Hence OP . OP' . QQ'=OQ . OQ' . PP'.
26. The tangent at (n 1 , */') to the conic a 2 /(a 2 + X) + j/ 2 /(& 2 + X) = 1 will pass
through the fixed point (a, (3) if
jc'a j/'/3
a a '+\" l "6 2 + \ '
a 2 +X 6 2 +X
To find the locus of (x', y') we have to eliminate X between the last two
equations. We have
1 1
ffl^+X ft 2 +X _ -1
I 3 !/' - 2/' 2 ~a' 2 -cu;' ~ a'l/' (a?/' - /3.t'} '
1 1
Py'-y' 2 sc'v-ax' -1
a 2 -6 2 x'y' (ay' - fix!)
Hence the locus of (a/, y') is the cubic curve whose equation is
{x*+y*-ax-py)( a y-px)-{a,*-b2){x- a )(y-l3)=0.
The cubic clearly passes through the point (a, /3), and also through the
foci.
27. The tangent to the conic a 2 /(a 2 +X)+«/ 2 /(6 2 +X) = l at the point
(x , y') will be parallel to the fixed line j/=xtana, if
tan o = - x' (6 2 + X)/y' (a 2 + X) .
_ „ aVsina+ifrVcosa
Hence X= *y-= ; >
y sino+ar cos a
„ „ (a 2 -J 2 ) a;' cos a , ,„ . (ft 2 - a 2 ) if sin a
.-. a 2 + X=-V^ — ; and 6 2 +X=-\— -*-i .
y sma + % cos a 2/ sin a + a; cos a
Substituting for a 2 +X and 6 2 +X in the equation
a;' 2 /(a 2 +X)+y' 2 /(6 2 +X) = l,
we have (x 1 aeea-y' cosec o) (a;' cos a+y' sin o) = a 2 - 6 2 .
Hence the locus of (x 1 , y') is the rectangular hyperbola whose equation
is
a; 2 -2/ 2 +xi/(tana-cota)=a 2 -6 2 (i).
XI.] CONIC SECTIONS. 197
Now the axes of the hyperbola are given by
gzl! = 2 *y (ii) .
2 tan a- cot a " '
E limin ating (a) from (i) and (ii) we have the equation of the locus of the
vertices of the rectangular hyperbolas for different values of a, namely
1,2
i.e. (a?+if) 3 =(a?-b i ){x*-if),
or, in polars, r l = (a 2 - V) cos 20.
28. Let ABC be the triangle, and let the sides BC, CA, AB touch the
confocal ellipse in A', B', C respectively.
Let EAF, FBD and DCE be the tangents at A, B, C.
Then the locus of the pole of BC with respect to all the conies of the
given confocal system is the line through A' perpendicular to BG [Art. 227].
Hence A'D is perpendicular to BC.
But, since BG and BA make equal angles with FBD [Art. 228], and BC
and CA make equal angles with DCE, it follows that D is the centre of one
of the escribed circles of the triangle ABC. Hence A', B', C are the points
where the escribed circles touch the sides of ABC.
29. If two conies have double contact the line through the intersection
of the tangents at the points of contact and through the middle point of the
chord of contact will pass through the centres of both curves. Hence if a oonic
have double contact with two concentric conies it must be concentric with
the other two; and hence both the chords of contact must be diameters, for
the conies
aa? + by 2 -l = and aa?+by 3 -l + \ {hc + my+nf=0
are not concentric unless n=0.
Let now a conic have double contact with two confocal conies, and let
PCP' and QCQ' be the two chords of contact. Let the tangents at P and
Q meet in T ; then CT will bisect PQ, since TP and TQ are tangents to the
same conic ; and since TP and TQ are tangents to two confooals, and GT
bisects PQ, it follows from Art. 226, Ex. 4, that the tangents TP and TQ
are at right angles. Hence the tangents at P, P' and Q, Q' form a rect-
angle.
30. Let T be (a, p) ; then the equation of the polar of T with respect to
the conic
a 3 /(a 2 +X)+ 3 / !! /(J 2 +X) = l
is a;o/(a 2 +X) + 2//3/(6 2 +X) = l.
Hence, if (if , ij) be the point of intersection of the normals at the ex-
tremities of the chord, the equation of the line joining the other two points
the normals at which pass through ({, r/) will be
- + 1 + 1=0. [Art. 197].
198 CONIC SECTIONS. [CHAP.
Also, for same value of n, the equations
a^ + X + 62 + X * *\&+i + 6 2 + X 1 ){a + f} + 1 )-"
and xy (a 2 - ft 2 ) + (ft 2 + X) ijs - (a 2 + X) & =
will represent the same conic.
Hence /»=1; anil comparing the coefficients of xy, x and y in the two
equations, we have
a. p a 1 p 1
(a 2 +\j~/9 + a(6»+X) a 2 + X a ft 2 + X /3
a 2 -ft 2 (6 2 +X)»; ~~-(a 2 + X)£'
. a 2 (6 2 +X)+/3 2 (q 2 + X^ _ /?(a 2 -a 2 -X) ^a( |3 2 -6 2 -X)
a 2 - ft 2 ~~ ij ""'"-{
Hence each fraction
_ q 2 ft 2 + j3 2 a 2 + j8° (a 2 - a 2 ) + a 2 (3 2 - ft 2 ) _ aff (a 2 - jS 2 - a 2 + ft 3 )
a 2 -6 2 + /3?)-a! _ <"J + /3?
.'. 2aS(<«; + S|) = (a 2 -/3 2 -a 2 + 6 2 )(a 2 -6 2 + /37;-a|).
Hence the locus required is the straight line whose equation is
{ax + py) (a, 2 + /3 2 + a 2 -6 2 ) = (a 2 -6 2 ) (a 2 -^-a 2 +6 2 ).
31. Let A, B, G be respectively the points of contact of the three
tangents B'AC, G'BA', A'GB' to the ellipse a; 2 /a 2 + !/ 2 /6 2 - 1=0; and let
u., p, y be the eccentric angles of A, B, C respectively.
Then, at the point A',
x=acos-(p+y)lcoa^(p-y) and y = b sin ^ (S + 7 )/cos ^ (P -y);
and similarly for B' and C".
Hence, if B' and C" are on the conic
x r /(a 2 + X)+ 2 / 2 /(6 2 + X) = l,
we have
co R 2 i( 7 -a)=^ x cos 2 J( 7+a ) + ^sin 2 |( 7 + fl ),
1 a 2 1 ft 2 1
and e0 s 2 -(a-/3) = 5 j- j -^cos 2 -(a + /3) + 62+x sin 2 -(a + /3).
Hence X 2 -a 2 ft 2 =(a 2 -ft 2 )Xcos( 7 +a)-(a 2 + X)(ft 2 +X)cos( 7 -a),
and X 2 - a 2 6 2 = (a 2 - ft 2 ) X cos (a + /S) - (a 2 + X) (ft 2 + X) cos (a - p) ;
whence
(a 2 !) 2 + 26 2 X + X 2 ) cos y cos a+ (a 2 6 2 + 2a-X + X 2 ) sin y sin a + X 2 - a 2 6 2 = 0,
XI.] CONIC SECTIONS. 199
and (a 2 6 2 + 26 8 X + X 2 ) oos a cos /3 + (a 2 6 2 + 2a 2 X + X 2 ) sin a sin j3 + V - a 2 6 2 = ;
. (a 2 6 2 + 26 2 X + X 2 ) cos a (a 2 6 2 + 2a 2 X + X 2 ) sin a X 2 - a 2 6 2
cos-(/3+ 7 ) Bing(/S + 7) C os|(^- 7 )
Hence, at the point A', we have
a(a 2 6 2 + 26 2 X+X 2 ) , 6(a 2 6 2 +2a 2 X+X 2 ) .
* = .
Then kS + S'=0 is the equation of any conic through A, B, C, D the
points of intersection of S=0 and the circle S'=0 which has its centre at
the point ({, ij) and is of radius r. Now if k be determined so that
kS + S' = may represent straight lines, these straight lines will be one
of the three pairs AB and CD, AG and BD, AD and BG. The equation
determining k is the cubic
(i+ 1 )(J +1 )« ,+ ^- , "-* )o (i +1 )* + (S +1 ) ft
which is equivalent to
I 2 q 2 _r 2
o 2 +ft + 6 2 + fe - "ft '
Now, if Xj, X 2 be the parameters of the two conies confocal to S=0 which
pass through ({, ij) we shall have
f , "" x- (*-xj(*-xj
a 2 +fc 6 2 +fc (a 2 +fc)(i 2 + fe)"
* The following interesting investigation is duo to Mr A. R. Forsyth, of Trinity College.
XI.] CONIC SECTIONS. 201
Hence the equation to determine ft is
.ft(ft-X,)(ft-X 2 )
(a 2 + ft) (ft 2 + ft)
.(i).
Now, if the circle touch the ellipse, two of the three pairs of lines
become coincident, and therefore the cubic in ft must in this case have
equal roots.
But when
ft(ft-X 1 )(i-X 2 )+r a (a 2 + ft)(J2+ft) =
has equal roots, the common value satisfies also
(ft-X 1 )(ft-X 3 ) + i(ft-\ 2 ) + ft(ft-A 1 )+,- 2 (a 2 +ft) + r 2 (6 2 +ft) = 0;
.-. , eliminating r, we have
1 1 1 _J_ 1 _
ft + fc-X 1 + ft-X 2 a*~+~k~V>Tk- (ll) -
Now (ii) is of the fourth degree : there are therefore four values of 4
depending on \ and X 2 , that is on the point ({, rj), but not on r, such that
the circle will touch the ellipse ; and when the values of ft are known, the
corresponding values of r are given by (i).
But when the circle touches the ellipse, r is the length of the normal.
Hence the length of any normal is given by (i), the quantity k occurring
in it being one of the roots of (ii).
Hence, if r„ r 2 , r 3 , r t be the lengths of the four normals, we have
- vvv ... n(*)-g(*-V)-n(t-A 1 )
12 3 4 ~ n (a 2 + ft) . n (& 2 + k) ■
It will now be found that
n(ft)=x 1 x 2 o 2 * 2 . n(ft-x 1 )=x 1 (\ 1 -x 2 )(a 2 +x 1 )(i 2 +x 1 ),
n(ft-X 2 )=X 2 (X 2 -X 1 )(a 2 +X 2 )(& 2 + A 2 ), n(a 2 +ft) = a 2 (a 2 + X,)(a 2 + X 2 )(& 2 -a 2 )
and n (6 2 + ft) = 6 2 (6 2 + XJ (6 2 + X 2 ) (o 2 - ft 2 ).
Hence r 1 2 r 2 V 3 V= ^%fy* ■
34. It is known [Art. 187, Ex. 1] that any rectangular hyperbola through
the three points P, Q, Ii will pass through 5, the orthocentre of the triangle
PQB. It is also known [Art. 213] that if PQ, RS meet in A, QS, PR in Ii,
and PS, QR in C; then ABC is a self-polar triangle with respect to any
conic through P, Q, R and S. This proves the theorem, since A, B, G are
the feet of the perpendiculars of the triangle PQR.
35. Take the tangents TP, TQ for axes, and let the equation of the
conic be (aa! + 62/-l) 2 + 2Xa;!/ = 0.
202 CONIC SECTIONS. [CHAP.
The equation of the bisectors of the angles between the axes is x 2 -y 2 =0;
and these bisectors cut ax + by - 1 = in the points 0, 0' where
is (-^ , -!—) and 0' is f-^- , =-?— ) .
\a + b a + bj \a-b b^aj
Hence the equation of BOB' is
x r + k(y 1 1=0,
a+b V a+bj '
and therefore the equation of TB, TB' is
{ax + by - (x + ley) (a + 6)/(l + k)} 2 + 2\xy =0,
or (x-2/) 2 (aft-Z>) 2 + 2X(l + ft) 2 x2/ = 0,
which is of the form
(x-yf + iucy = (i).
Now if two straight lines be cut by any circle whose centre is at the
intersection of the lines, two pairs of parallel lines will go through the
points of intersection, and each pair of lines will be parallel to one or other
of the bisectors of the angles between the original lines.
Hence the bisectors of the angles between the lines
(x-yf+fiay=0
are parallel to
(x - j/) 2 + pjcy + L (x 2 + 2/ 2 + 2xy cos u) = Xc 2 ,
provided L is so chosen that
(a; - y) 2 + /jjcy +L(x 2 +y 2 + 2xy cos a)
is a perfect square ; and it is obvious that if
(x - y) 2 + nxy + L (x 2 +y 2 +2xyaosa)
is a perfect square it must be (x±^) 2 . Hence the lines xJ=y = are the
bisectors of the angles between TB and TB' : thus TO and TO' are the
bisectors of BTR'.
Similarly, if SO'S' be any line through 0' cutting the conic in 8, S' ; then
TO and TO' will be the bisectors of STS'.
Or thus :
Let TO, TO' be the tangents to the confocals through T, and 0' being
on PQ. Then TO, TO' are the bisectors of the angle PTQ, and the pole of TO'
with respect to any one of the system of confocals lies on TO [Art. 227].
But the pole of TO' with respect to the original conic must be on PQ, since
TO' passes through the pole of PQ. Hence is the pole of TO' with respect
to the given conic ; and similarly 0' is the pole of TO.
Now let any line through cut the conic in B, B' and TO' in K ; then
the pencil T{BOR'K} is harmonic, and TO is perpendicular to TK; therefore
TO bisects the angle BTB'.
XI.J CONIC SECTIONS. 203
33. The axis of the first parabola bisects the interior angle between AB
and DC ; and the axis of the second parabola bisects the interior angle
between AB and CE.
Hence the angle between the axis is half the angle DOE, or one-quarter of
the angle DE subtends at the centre of the circle.
37. It is known (or easily proved) that when a circle cuts a 'parabola in
four points, the sum of the distances of the four points from the axis of the
parabola is zero : the axis of the parabola therefore passes through the centre
of mean position of the four points. Hence the point of intersection of
the axes of the two parabolas through A, B, C, I) is the centre of mean
position of A, B, C, D.
Now, since ABC is a maximum triangle in an ellipse, the eccentric angles
of A, B, C may be taken to be a, a +-=- , and a + -j- ; and therefore [Art. 186,
o o
Ex. 1] the eccentric angle of D will be 2nir - 3a.
Hence, if (x, y) be the point of intersection of the axes of the parabolas,
we have
4x -a -I cos a + cos (a + -77- ) + cos I a + -^ \ + cos {2nir - 3a) I ,
or ix — a cos 3a ;
and similarly 4y = b sin 3a.
Hence the equation of the required locus is
a 2 + 6 2 "" 16 '
38. The axis of a parabola through the four points a, ft 7, 5 will make
equal angles with the join of a, /3 and of 7, S. Hence the axes of the two
parabolas are parallel to the lines
x cos I (a + p) + y sin ^ (a + /S) = ± \x cos - (7 + S) + y sin - (7 + S)i ,
i. e. parallel to
a;cosS + 2/sinS=0 and asinS-j/cosS=0.
Also the sum of the perpendiculars on either axis from a, ft 7, 8 will be
zero. Hence, if the axes be
x cos S + y sin S -p =0
and xsiaS-y cosS- J=0,
we have - (acosacosS + asinasinS-p)=0;
.■. p— -j-Scos(S-a).
Similarly g = |ssin(S-a).
204 CONIC SECTIONS. [CHAP.
Hence the equations of the axes are
x cos 8+y sin S= -r 2 cos (S-a),
and xsinS-j/cosS= jSsin(S-a).
Since trie sum of the perpendiculars from a, /3, y, 8 on the axis of either
of the two parabolas through these points is zero, it follows that the point
of intersection of the axes is the centre of mean position of o, /3, 7, 8. And
similarly for any other four of the five points a, {3, 7, 8, e. Whence it follows
that all the five points of intersection of pairs of axes lie on the circle whose
equation is
(5*0-4^ + (52/ -42/)2=fl2,
where 5x = a (cos a + cos jS + cos 7 + cos 3 + cos e)
and 5i/ =a(sina + sin/3 + sin7 + sin 8+ sin e).
39. The first part of the question is proved in Art. 212. To prove the
second part, divide the polygon into quadrilaterals whose sides are A , B , C, X ;
X, D,E,Y; Y, F,G,Z; &c. Let the equations of the sides A, S, G, &c, be
a=0, 6=0, c=0, &c. Then for points on the conic we have ac=\bx,
xe=fidy, yg=vfz, &c. Hence a. v. e.g... varies as h : d . f. h...
40. The normals at two consecutive points on a curve intersect in the
centre of curvature. But the normals at the ends of -coso+-sin a -1=0
a b
and of h ,-. h 1 = meet in a point. Hence the equation of Qlt
a cos a 6 sin a r * ^
is of the form
x V
+ 1=0;
a cos a b sin a
and therefore, if T be (xf, y'), we have x'= -a sec a and y'= -bcosec u .
Eliminating a we see that (x', y') is on the curve whose equation is
a? & 2 ,
- +-5 = 1.
x* y i
41. The ordinates of the points of intersection of
(x-a) a +(j/-/3) 2 -c 2 =0 and y 2 -4ax=0
are given by the equation
y* + 4a (4a - 2a) j/ 2 - 32a?py + 16a 2 (a 2 + jS 2 - c 2 ) = 0.
If 4a -2a be positive, the signs in/ (y) are+, +, -, ±1 and the signs in
/(-«/) are +, +, +, ±. Hence the total number of changes of sign in f(y)
and in / ( - y) is two, and therefore by Descartes' Rule of Signs there cannot be
more than two real roots.
The equation of any conic through the four points of intersection of
y*-iax=Q and (x-a) 2 + (j/-/3) 2 -c 2 =0 is of the form
\(j/ 2 -4ax) + (x-a) 2 +(j/-/3) 2 -c 2 =0 (i).
XI.] CONIC SECTIONS. 205
In order that (i) may represent a pair of straight lines, \ must satisfy the
equation
1, 0, -a-2a\ |=0,
0, X + l, -j3
-a-2tt\, -/3, o' + ^-c 2
or 4a 2 \ 3 + 4a(a + o)X 2 + X(4aa-/3 2 +c i! ) + c !! =0.
Hence \ + \ 2 +\=-- — ^ (ii).
Now when (i) represents a pair of straight lines the equation of the
parallel straight line through the vertex is \y 1 +x* + y 2 =0, which meet the
parabola in points where x— -4a(l + X). Hence the sum of the abscissae is
constant when the sum of the values of \ is constant, and this from (ii) is the
case when a is constant.
42. Let x = 0, y = 0, Ix + my + 1 = be the equations of the given straight
lines.
Let the equation of a conio with respect to which the three straight lines
form a self -polar triangle be
ax* + 2hxy + by 2 + 2gx + 2fy + 1 = 0.
Then since Ix + my + 1=0 is the polar of (0, 0), it represents the same
straight line as gx +fy + 1 = 0, and therefore g = l, /= m.
Again the polar of ( - j , ) with respect to the conic is x=0; hence x=0
is the same as-^ - -j + lx- l+my + 1=0, whence h=lm; and this is a suffi-
cient condition that y=0 is the polar of ( 0, J ,
Hence the general equation of a conio with respect to which the given
triangle is self-polar is
ax* + 2lmxy + by* + 2kc + 2my + 1 = (i),
where a and 6 are arbitrary.
Now the centre of (i) is given by
ax + lmy+l=0, lmx+by + m=0.
And, if (i) is a rectangular hyperbola, a + b - 2lm cos u = 0. Eliminating a
and 6 from the last three equations we have the equation of the locus of the
centres of the rectangular hyperbolas, namely
hn (x* + y*+2xy cos w) + ly + mx=0.
Hence the locus is the circle circumscribing the triangle.
[The theorem can be easily proved geometrically from the fact that two
conjugate diameters of a rectangular hyperbola make equal angles with an
asymptote. See also the solution of 22, page 311.]
43, The equations of any three tangents to the circle are .
xcosa + 2/sina-c=0, x cots fi+y sin p - c=0, and x cos y+y sin y -c=0.
206 CONIC SECTIONS. [CHAP.
The vertices of the triangle formed by these tangents are
\c cos g (/3+ 7 ) sec g (fi-y), c sin ^(p + y) sec ^ p-y\ , &c.
These three points will all be on the ellipse provided values of a, ft, y can
be found which will satisfy the three simultaneous equations
^cos3-(/3+ 7 ) + - a sin ! ^(|3+7) = ci! cosVp- 7 ) (*)■
^cos 2 ^(7+o)+pSin 2 2( 7 + a)=- 2 cos I! g(7-o) (ii),
and ^cos 2 2(a+/3) + pSin a £(a+/3) = - 2 cos 2 2(a + /3) (Hi).
Now (i) can be written in the form
1 1 1 Z 1 1 1N \ f 1 x ^ • a ■ n
- 2 +p- ? + ^-6 2 - c - 2 j oo ^ cos T+^-^-- 2 j^^m7 = 0.
t, x ■ 111
But, since - = - + -. ,
cat
a 2+ 6 a c 2_ of a 2 6 2 c 2_ 6c' aD 6 2 a 2 c 2_ ac '
Hence (i) and (ii) are equivalent to
+x). f> + x>}.
48. Let the tangents at ^, B, C be Q.4.R, iJBP and PCQ respectively.
Then, since 45 and A G touch a confocal, AB and .4(7 must be equally
inclined to QAB. Similarly BC, BA must be equally inclined to BBP,
and GB, CA equally inclined to PCQ. Hence ABC is the pedal triangle of
PQB.
Hence PA is perpendicular to QAB, and therefore PA touches the con-
focal hyperbola through A.
Again, as in Example 28, PA' is perpendicular to BA'G, and therefore
PA' touches the confocal hyperbola through A'.
But, since A BG is the pedal triangle of PQB, the angle CBP is equal to
PQB, and therefore £PA'= complement of PQB=QPA. And, since the
angles BPA' and QPA are equal, and PB, PQ touch an ellipse, PA' and PA
will touch the same confocal ellipse. Hence the confocal hyperbolas through
A and through A' are the same.
Or thus :—
We have found in question 31, that if A', B', C" be on the conic
a; 2 /a 2 +2/ 2 /& 2 =l,
and A, B, C on the confocal conic ay'l{a i +\) + y !! l(i i + \) = l; then, if the co-
ordinates of A' be a cos o, 6 sin a, the co-ordinates of A will be
a(a 2 & 2 +2& 2 X+X 3 ) &(a 2 6 2 +2a 2 X+X 2 ) .
aW^X* C0S ° aW^K* ' Bm «■
The confocal hyperbola through A' will be [question 20]
z 2
cos" a sin' a
=a 2 -& 2 .
XI.] CONIC SECTIONS. 209
The condition that this hyperbola should pass through A is that
a 2 (a 2 S 2 + 26 2 \ + X 8 ) 2 - 6 2 (a 2 6 2 + 2a 2 X + \ 2 ) 2 = (a 2 - 6 2 ) (a 2 6 2 - X 2 ) 2 ,
and it is easily seen that this condition is satisfied.
49. Let the equation of the hyperbolas be
x 2 -2/ 2 = a 2 -/3 2 (i),
and xy~px-ay=0 (ii).
Then equation of any circle through the centre of (i) will be
x>+y* + 2gx+2fy = (iii).
The abscissae of the points in which (ii) and (iii) intersect are given by
x 2 (x - a) 2 + 2gx (x - a) 2 + /3 2 x 2 + 2/jSas (x - a) = 0,
or x s -2(a.-g)x i + {a? + p ! >-4:ag + 2pf)x + 2ga?-2f a p=0 (iv),
and the ordinates are given by
y 3 - + 2fp*-2g a p=0 (v).
These points being (x^ yj, (x 2 , y 2 ) and (xg, y 3 ), the conditions that they
form a conjugate triad with respect to (i) are easily found to be
w - yd/a = ^s^i - y0i = x i x n - ynii = <* s - P 2 -
Now
V 3 - y03 =2{fp-ga) (t + E\
\ x i Vi/
= 2(fp-ga), since (Xj, y x ) is on (ii).
Hence the necessary and sufficient conditions that PQR may be a self-
polar triangle with respect to (i) is that
2/jS-2sra = a 2 -/3 2 .
But this condition is not satisfied by any circle through (0, 0) but only by
circles whose centres are on the fixed line
2ax-2#/-a 2 +/3 2 =0.
50. Let the equation of the hyperbola be 2xy=c 2 , and the equation of
the first circle
x*+y* + 2gx+ 2fy=Q>.
Then if (x 1 , yj, &c. be the four points A, B, G, D respectively, the equation
of the second circle will be
a?+y* + 2x i x + 2y 4 y=0 (i).
A', the point of intersection of the tangents at (xj, y 2 ) and (x 3 , y 3 ), will be
found to be f^'-, -?—),
\x 2 + x 3 x i -Yx z j
and this point is on (i) if
c 4 (x 2 +x 3 + x 4 ) + 4x2X3X4 (x 2 x 3 +x 3 x 4 +x 4 x 2 ) = (ii).
S. C. K. 14
210 CONIC SECTIONS. [CHAP. XI.
Now the abscissae of the points of intersection of the hyperbola and the
first circle are given by the equation
4a; 4 + 8ga? + ifipx + c 4 = 0.
Hence 4a; 1 x 3 a: B x 4 =c 4 ,
and x x (x 2 +x 3 + x t ) + x^ + x^K t + x t x 2 = 0,
and these shew that (ii) is true.
CHAPTER XII.
Pages 271—275.
1. Let the equation of the parabola be y 2 - iax=0, and let P be (x', y') ;
then the equation of NM is
^ + -,-1=0, oi4ax+yy'-y'*=0.
, * y
Hence the equation of the envelope is - 16ax=y 2 .
2. Let the equation of the line be - + — — = 1, where c is the given dif-
a a — c
ference of the intercepts. The equation may be written a 2 - a (c + x + y) + cx= 0;
hence the equation of the envelope is icx—(x+y + c) 2 , which represents a
parabola.
3. Take the fixed straight lines for axes, and let the equation of the line
be- + r=L Then a5=constant=c 2 suppose. Hence we have
a b
a 2 y -an? +xc 2 =0,
and therefore the envelope is the hyperbola whose equation is 4xy = cK
4. The equation of PD is
-co S ^ + i j+jsm^ + i j = cos s ,
g + f)cos*-g-f)sin0=l,
Hence the equation of the envelope is
(H-i)(H +i ) + e-f) a = o:
?! £-1
a» + 6 8 ~2'
14—2
212 CONIC SECTIONS. [CHAP.
The co-ordinates of the middle points of NP and MD are a cos , 5 sin (j>
and - a sin 0, = cos #. Hence the equation of the line joining them is
x . . 2« , ,
-(sin 0-eos0)--^(cos0+sin0)= - 1,
- (M-)- 2 (M) tan i-e + T +i ) tana -t=°-
Hence the equation of the envelope is
(M*')G +! H + e-?)*=°'
2a 2 8« 2 ,
or — r+ 7F = 1 -
a 2 ft 2
5. Let the straight lines meet in 0, and let 0A=a, 0A' = a', AB = l,
A'B' = V ; also let AP=kl, then will A'P' = kV.
Hence the equation of the line PP' is
x , _V _i
- J.,
fl + W o'+W
or ft 2 Z2' + k (at + a7 - xV - yl) + aa' - xa' -ya=0.
Hence the equation of the envelope is
All' (aa' - xa' - ya) = (xl' + yl-al' - a'Vf.
The envelope is therefore a parabola, and it is easy to verify that the axes
are tangents.
6. Take OAP and OBQ for the axes of x and y respectively. Let 0A = a,
and 0B = b, and let AP . BQ=c 2 .
Then the equation of PQ will he
x V ,
1- 1
a + k + , c 2 '
b + I
or A 2 (6 - y) + k (c 2 + ab - bx - ay) + c 2 (a - x) = 0.
Hence the equation of the envelope is
i(a-x) (b-y) = (c 2 + ab-bx-ay)K
Thus the envelope is a conic touching the lines x=a and y — b.
7. Take the two given straight lines for axes ; then the equation of the
system of circles is
a; 2 + 1xy cos 01 + y* - lax - 2ay + a 2 = 0.
xii.] conic sections: 213
Let (a, p) be the given point ; then the polar is
aw + (ar/3 + ya) cos a + yfl - a (x.+ y + a + /3) + o 2 = 0.
Hence the equation of the envelope of the polar, for different values of a,'
is 4 {x [a+ p cos a) + y (p + a cos u)} = (x + y + a + /3) 2 .
Thus the envelope is a parabola.
8. Let ' =2kb.
Then the equation of the chord is
x v . d> + d>' '
- os^ + |sm^=cos^.
with the condition sin ^ ^ cos - ~ = k .
Hence ? B in«cos^' + f sin 2 *±* =*,,
\b } 2 a 2
Hence the envelope is the parabola whose equation is
9. Let the equation of the fixed tangent be y=m 1 x-{ — , and let
y=mx -i — be the equation of any other tangent PT. Then the co-ordinates of
T are and - H ; and therefore the equation of TO is
mm 1 m m^
f a a\ a
y j — \m+x~ =0,
\ m m^/ mmy
m?[y | +m(x - a) =0,
Hence the envelope is the parabola whose equation is
4- '-(y--) + (x-a)*=0.
10. Take the given straight line for axis of x, and let the conic be given
by the general equation.
Let P be (a, 0) ; then the equation of the polar of P is
x (aa.+ g) + y ( ha +f) + ga + c = 0.
214 CONIC SECTIONS, [CHAP.
Hence the equation of the line through P parallel to its polar is
{x - a) (aa+g) +y (ha+f) =0,
° r o s . a~a(ax+hy-g)-[gx+fy)=0.
Hence the envelope for different values of a is the parabola whose equation
is {ax+hy-g)2+4a(gx+fy) = 0.
11. If the corner G fall on C, the line of the crease will bisect CO' at
right angles, in Y suppose.
Hence the locus of 7 is a fixed line bisecting the page ; and, since the locus
of the foot of the perpendicular from G on the line of the crease is a straight
line, the line of the crease must envelope a parabola of which O is a focus.
12. Let P be one of the points of intersection of the fixed and moving
ellipses ; then OP must make equal angles with the major axes of the two
ellipses, whence it follows that a common chord which is not a diameter
must subtend a right angle at the centre. The envelope is therefore a circle
from Art. 138, Ex. 3.
13. Let S be the fixed point, and AB the fixed straight line. Draw SA
perpendicular to AB ; and through A draw AG such that the angle SAG is
equal to the given angle. Let SPQ be any position of the angle, P being on
AB, and let PQ meet AC in Y; then, since SAG=SPQ, 8, A, P, Y are all on
a circle, and therefore SYP is a right angle.
Hence PQ moves so that the foot of the perpendicular on it from the
fixed point S always lies on the fixed line AG: the envelope is therefore a
parabola whose focus is S and of which AC is the tangent at the vertex.
14. Take for axes two conjugate diameters one of which is parallel to
the given straight line. Let the equation of the ellipse be
ax 3 +by"=l,
and the equation of the straight line x—a. Then the equation of the chord
whose middle point is (a, /3) is
aa (x - a) + 6/3 (y-p)= 0.
Hence the envelope, for different values of j3, is the parabola whose equa-
tion is
by i +im(x-a)=0.
15. Suppose the conic to be an ellipse, and let 6, v fa be the eccentric
angles of 0, P, Q respectively.
Then, since OP, OQ are in given directions, 6 + v and + 2 are both
constant. IJence i + 2 )-
The above applies to the case of the hyperbola, the eccentric angle of any
point (x, y) being the imaginary angle given by
x=acos0, y = J -1 .b sin 6.
If the conic is a parabola, and y lF y„, y, be the ordinates of 0, P, Q
respectively, then y-i+y^ and y-^ + y^ are both constant since OP, OQ are
drawn in constant directions, and therefore y s -y s = constant = 2c.
Hence the equation of PQ is
y(yi+y i )-^x-y^. i =o,
or 2y(y 2 -c)^-4ax-y i (y i -2c)=0,
whence the envelope is the parabola whose equation is
y*-4ax + c*=0>
16. Let the equation of the circle be x* + y s -c 2 =0, and let the equation
of PQ be Ix+my - 1=0. Then the equation of the diameters through P and
Qis
x i +y"-c 2 {lx+my)' l =0.
Hence, if these diameters are conjugate, we have
a 2 (1 - cV) + ft 2 (1 - c 2 m 2 ) = 0,
or aW+& 3 e 2 m 2 =a 2 +i> 2 (i).
The envelope of Ix + my = 1 with the condition (i) is
jtj. y* - *
a "r t.'g.q — .
a 2 c 2T 6 2 c 2_ a 2 + 6 2 "
Thus the envelope is a similar and similarly situated ellipse. The
envelope will be the original ellipse provided c 2 =a 2 + 6 2 , that is provided the
circle is the director-circle.
17. Let the equation of the straight line be
lx+my + l=0,
and let the n fixed points be (x v ?/,), &c.
216 CONIC SECTIONS. [CHAP.
Then, by supposition,
2 (lx 1 +my 1 + l)*=oon&tant=nc 2 ;
.-. I*2x 1 !s + 2lm2x i y 1 +m 1 2y 1 !i +2l2x 1 + 2m2y 1 + n(l -c 2 ) = 0,
which from Art. 239 is the tangential equation of a conic.
18. Take AB, AG for axes of x and y respectively, and let the equation
of £(7 be bx + cy=l.
Let the moving line cut BC, GA, AB respectively in L, 31, N, and let the
equation of LMN be lx + my= 1.
Then the abscissae of N and L will be -r and — =- — =- respectively. Hence
I mb-lc
MN : ML—-i : — ^r = constant = l : k.
I mb-lc
Therefore I (m - c) = k (nib - le) ;
/. lm-(l- k) lc-kbm=0 (i).
We have to find the envelope of
lx+my-l =
with the condition (i).
The directions of the two lines through (x, y) are given by
Im- {(l-k)cl+kbm) (lx+my)=0.
The lines will therefore coincide if
4k (1 - k) bcxy = {kbx + (1 - k) ey - 1} 2 .
This is equivalent to
Jkbx+ */(l - k) cy =1;
the envelope is therefore a parabola touching the' sides of the triangle.
19. Take OA, OB for axes, and let the given fixed point be (o, /3).
The equation of any circle through is
x 2 + 2xy cos a + J/ 2 + Ix + my = 0.
Hence the equation of PQ is
| + I + 1 = (i).
I m
But, since the circle goes through (a, /3), we have
a 2 +2a/?cosw+/3 2 +Za+m/3=0 (ii).
The directions of the two lines through (x, y) are given by
Im (a 2 + 2a/3 cos w + 13 2 ) - (ly + mx) (la. + m/3) = 0.
XII.] CONIC SECTIONS. 217
Hence the equation of the envelope is
iapxy = (ax + @y - a 2 - 2a.p cos a - /3 2 ) 2 ,
or Jax + Jfy + ,/(a 2 + 2a/3 eos u + /3 2 ) = 0.
Hence the envelope is a parabola which touches the axes.
20. Let the equation of the ellipse be
ok 2 + !>!/ 2 -1 = 0;
then, from Art. 197, if the equation of PS be
Ix + my — 1 = 0,
the equation of PQ will be
^ + ^ + 1 = 0.
I m
Hence we have to find the envelope of
lx+my-l=Q
with the condition
^Ai=o,
l m
(a, p) being the fixed point through which PQ passes.
From Art. 219, Ex. 1, or Art. 239, Ex. 2, the equation of the required
envelope is
sjaax + sjbfiy =1.
Thus PS envelopes a parabola touching the axes.
21. Let the equation of the hyperbola be
The equation of the chord whose middle point is (a,,p) is
(x-a)a-{y-p)p=0.
Hence the equation of the line through the middle point of the chord and
perpendicular to it is
(x-a)p + (y-p)a = 0.
Now if the extremities of the chord are on a circle whose centre is {f y 0),
the line which bisects the chord perpendicularly must pass through (/, 0) ;
we therefore have the condition
(/-a)/3--a|8--=0,
whence /3=0 or 2a =/, If ^=0 the line
(a;-o)a-(y-/3)/3 =
218 CONIC SECTIONS. [CHAP,
is parallel to an axis ; and the envelope of
(«-|)J-dr-«/»=o.
is the parabola whose equation is
(2x~f)f=y*.
22. Take the line on which the centres lie for the axis of x, and the
conjugate diameter for the axis of y; then the equation of any ellipse of the
system is
+ 1-1 = 0,
a 2 T S 2
where a and b are known.
The polar of the fixed point (x lt yj is
^Ij-Wl-i *{x + *i) . *"_«
a? + 6 2 x a a + a 2_ " -
Hence the required envelope is the parabola whose equation is
\ a 2 + 6 2 ~ j ~ a 2 '
23. Let (7, C be the centres of the two circles of which C is fixed, and
let be the fixed point. Then the locus of C is a circle whose centre is 0.
The radical axis bisects GC at right angles, and therefore the locus of the
foot of the perpendicular from G on the radical axis is a circle whose centre
is the middle point of CO. Hence the line envelopes a conic of which C is
one focus and whose centre is the middle point of CO ; hence is the other
focus.
24. Let the equation of the ellipse be ox 2 + by*= X.
Let (xj, ?/i) be the pole of the chord joining the extremities of a pair of the
diameters ; then the equation of the chord is
ax 1 x + by 1 y=l.
Hence the equation of the corresponding pair of radii vectores is
an? + by* = (ax-p + by-^yf ;
and, since the sum of the angles these two lines make with the major axis
is a right angle, we have a - ah^ = 6 - bh/^.
Hence the locus of (a^, y x ) is the concentric and co-axial hyperbola
whose equation is oV - Wy 2 =a-b.
The envelope' of the chords is the envelope of axjX+by^y-l^O with the
condition a\'-b"y 1 a -(a-b)=0.
The envelope is therefore the rectangular hyperbola whose equation is
« 2 -« 2 =— i-, .
J a-b
XII.] CONIC SECTIONS. 219
25. Let (ka, kb) be any point on the equi-conjugate bx = ay ; then the
equation of the lines joining this point to the extremities of the axis major
is easily found to be
k?(bx-ayf-a?(y-i;b)2=0.
Hence ^ + p -1 ~ X {W(bx-ayf-a? (y- kb)*}=0
•will, for some value of X, represent PQ and y=0. The coefficient of a? must
therefore be zero, whence
l-XaW=0.
The equation of PQ is then found to be
2abk 1 x+a?y -2a?bk=0 ;
and therefore the envelope, for different values of k, is the rectangular
hyperbola whose equation is 2xy =db.
26. Let the equation of PNP 1 be 2x-a=0. Let a parabola through
G, P, P' cut the ellipse at the extremities of the chord whose equation is
Ix + m y + n = 0. Then the equation of the parabola will be
^ + ^-l-\{2x-a)(lx+my+n)=0 (i).
Since (i) goes through (0, 0), we have-l+Xa?i=0; also, since (i) is a
parabola, we have l- s -2iX)p= XW.
Hence, eliminating X, we have
■n?-2aln-bW=0.
The equation of the chord may therefore be written
rih: + 2an (my +n)-Wxm?=0.
Hence the envelope is
b*x(x+2a) + a?y 2 =0,
(x + a)' « 2 „
27. Take the line parallel to and midway between the given parallel
lines for axis of y, and the perpendicular line through the fixed point for
axis of x ; and let the equations of the parallel lines be x= ±a, and let the
fixed point be (c, 0).
Then the line y=m(x-c) cuts the parallel lines in the points
{a,m(a-c)} and {-a, -m{a+c)}.
Hence the equation of the circle on PQ as diameter is
(y+mc-am) (y+mc + am)+x i -a?=0 ;
.•. x 1 + y 2 -a? + 2cym+[c i -a?) m 2 =0.
220 CONIC SECTIONS. [CHAP.
Hence two circles pass through any particular point (x, y); and when
(x, y) is a point on the envelope the two circles will coincide. The equation
of the envelope is therefore
(cP-a 2 ) (x 2 +y 2 -a 2 )=c 2 y 2 ,
x 2 y 2 ,
28. Take the axes of x parallel to the chords and let the conic be given
by the general equation.
The chord y=\ cuts the conic in points given by
ax 2 + 2x ( h\ + g ) + SX a + 2/X + c =0.
Hence the equation of the circle of which the chord is a diameter is
a a *
Hence the equation of the envelope of the circles, for different values
of \, is
(a + b) (aa? + ay 2 + 2gx + c) = (hx - ay +f) 2 .
The required envelope is therefore a conic.
29. Let y v y 2 be the ordinates of the extremities of the chord, then the
equation of the chord is
V(Vi +Vi) -4ax- yi y 2 =0 ;
also the equation of the circle of which the chord is a diameter is
(V - 2/i) (2/ - Vi) + (« - a=i) (a - z 2 ) = 0-
Where the circle meets the parabola we have
16« 2 (V - 2/i) (y - Vz) + [y 2 - 2/i 2 ) W - yfl = ;
■•• {y - 2/i) (y - y») {16a 2 + (y + yj (y + y a ) } = 0.
Hence, in order that the circle may touch the parabola, the roots of
Ua 2 + (y+y 1 )(y+y 2 )=0
must be equal ; and the condition for this is
4(16a*+y 1 y 2 ) = (y 1 +y 11 )=,
or y 1 ~y !l =8a.
Hence the equation of the chord may be written
2y (Vi + 4a) -4ax-y 2 (y 2 + 8a) = 0,
and the equation of the envelope is therefore
ia(x-2y) = (4a-y) 2 ,
or y 2 =4a{x-4a).
The required envelope is therefore an equal parabola.
XII.] CONIC SECTIONS. 221
30. Take AP for axia of x, and a perpendicular line through A for axis
of y, and let AP—a.
Let xcosa+j/sino-j>=0
be the equation of any possible directrix ; then
( -2) cos a, -p sin a)
is the corresponding focus, and the equation of the corresponding parabola
will therefore be
(x +p cos a) 1 + (y +p Bin a) s =(xeosa+y sin a-p) e .
But the parabola cuts y=0 in the point (a, 0) ; hence
a 2 sin 2 a+4apcosa=0 (i).
The equation of the directrix may therefore be written
iax cos 2 a + lay sin o cos o + a 2 sin 2 a = 0.
Hence the envelope is the parabola whose equation is y*=ax.
31. Take.the two fixed diameters to which the biseetors are parallel for
axes, and let the equation of the conic be
-4m?a i -6al(lx+my) + {lx+my) ti =0.
Hence the equation of the envelope is
(a a + x 2 - 6ax) (j/ 2 - 4a 2 ) = {xy - 3ay)\
or 2# 2 +(a:-3a) !! =8a 2 .
35. The simplest proof of this theorem is found by reciprocating
with respect to the fixed point. The theorem then becomes : —
' The locus of the point of intersection of two tangents to a parabola
which cut at a constant angle is a conic having double contact with the
parabola.' [See Art. 107, Ex. 1.]
Or thus :
Take the tangent and the normal at the fixed point for axes, and let
the equation of the conic be
ax' + 2hxy + by a +2fy=0.
224 CONIC SECTIONS. [CHAP.
Let Ix + my- 1=0 be the equation of any one of the chords ; then the
equation of the lines joining its extremities to the origin is
ax 2 + 2tecy + by 2 + 2fy (U + my) = 0.
The angle between these lines is
tan- 1 2 N /{(ft +/Z) 2 - a (6 + 2fm)}\(a + 6 + 2/m).
Hence, if the chord subtend at the angle a, we have
4(A+/Z) 2 -4a(& + 2/m)-tau 2 a(a + &+2/m) 2 =0 (i).
It therefore follows from Art. 239 that the envelope of the chord is a conic,
except when a=^, in which case the chord always passes through the fixed
point
(»-*)■
We will now shew that the envelope has double contact with the given
conic, the chord of contact being the polar of I , — - 1 whose equation is
y a-v o J
2hx+(b-q)y + 2f=0.
The line Ix + my -1 = will touch the conic
ax 2 + 2hxy + by 1 + 2fy + X { 2hx + (6 - a) y + 2/} 2 = 0,
provided
ax 2 + 2hxy + by 2 + 2fy[lx+my)+\{2hx+{b-a)y + 2f(lx+my)} 2 =0
is a perfect square in x and y, the condition for which is
{a + i\(h+fl) 2 }{b + 2fm + \(b-a + 2fmf} = {h+fl+2\(h+fl)(b-a + 2fm)} 2 ,
or {h +fl) 2 -a(b + 2/m) - ak {4 (h, +flf + (ft - a + 2/m) 2 } = 0,
or (fc+/Z) 2 -a(& + 2/m)- 1 -^^(a + & + 2/m) 2 =0 (ii).
Now (ii) coincides with (i) provided
a\ tan 2 a , . _
, — j-r= — 3 — < or 4a\=sin z a.
1 - 4aX 4
Hence the envelope oihc+my -1=0 with condition (i) is the conic
4a {ax 3 + 2kxy + by* + 2fy} + sin 2 a {2hx + (6 - a) y + 2/} 2 = 0.
Thus the envelope is a conic having double contact with the given conic.
36. Take the fixed point for origin, and let the equation of the circle be
a?+y 2 + 2gx+c=0.
XII.] .CONIC SECTIONS. 225
Let Ix + my -1 = be a line joining extremities of a pair of the perpen-
dicular chords ; then the lines given by
X s + 2/ 2 + 2gx (Ix + my) + c ( Ix + my) 2 =
must be at right angles, the condition for which is
2 + 2gl+c(l 2 +m?) = (i).
The envelope of Ix + my - 1=0 with condition (i) is then found to be
[2c-g s )y* + 2cx i + 2cgx+cS=0 (ii).
Writing (ii) in the form
(g 2 -2c)(x 2 +y 2 ) = (gx + c) 2 ,
we see that the origin is one focus. Also the centre of (ii) is clearly midway
between the origin and the centre of the circle ; and therefore as the origin
is one focus the centre is the other.
37. Let y a -4ax=0 be the equation of the parabola, and let (a, /3) be
the point S; then the equation of the parabola referred to parallel axes
through (a, /3) is (y + /3) 2 = 4a (x + a).
The polar of (0, 0) is (3y- 2ax + /3 3 - 4aa = ; therefore the equation of
SC is 2ay + f3x=0. Hence SG meets the axis y + p=0 in the point (2a, -/S).
Let lx+my+l=0 be the equation of any chord which subtends a right
angle at 5 ; then the lines given by
{y - (3 (lx + my)} 2 + 4ax (Ix+my) - 4aa (lx+myf=0
are at right angles, and therefore
(l-/3m) 2 +/S 2 Z 2 + 4aJ-4aa(J 2 +m 2 ) = (i).
From Art. 241 the centre of the conic whose tangential equation is
(i) is (2a, -§) : thus the centre of the conic (i) is G.
Note. To find the centre of the conic whose tangential equation is given.
The centre of a conic is always on the line midway between any pair of
parallel tangents. Now the two tangents parallel to the axis of y are
l L x + l=0 and l^e+ 1 = 0, where l x , Zj'are the roots of al* + 2gl + c=0. Hence
the centre is on the line 2x+ - + j =0, or 2a:-.— =0 ; and similarly on the
line 2y - - =0. Therefore the centre is the point ( - , ■- J .
38. Let the equation of the conic be ax 2 + by 2 =l, and let O be (a, p).
Transfer the origin to O ; then the equation will be
as 2 + by 2 + 2 (aax + bfiy) + aa 2 + 6/3 5 - 1 = 0.
Let the equation of PQ, one of the chords, be Ix + my + 1 = 0; then the
equation of OP, OQ will be
ax 3 + by 1 - 2 [aax + 6/3?/) (Ix + my) +{aa? + bB 2 - 1) (Ix + my) 2 = 0.
S. C. K. 15
226 CONIC SECTIONS. [CHAP.
Since OP, OQ are at right angles we have
a+&-2aaZ-2&/3m + (aa 2 + & J 3 2 -l)(Z 2 +m 2 ) = (i).
The directions of the two chords which pass through (x, y) are given by
(o + b) (Ix + my) 3 + 2 [aal + bpm) (Ix +my) + (aa? + ft/3 2 - 1) (P + m 2 ) = 0.
Hence if (x, y) be on the envelope we must have
{(a+b)x i +2aax + aa?+.bp i -l}{(a+b)y* + 2bpy + aa?+bp*-l}
= {{a + b)xy + aa.y + bpx}*;
.: (a + b) (ao 2 + &/3 2 - 1) (a 2 + j 2 ) - (aay - bpxf
+ 2 (aax + bpy) (oa 2 + bp 1 - 1) + (aa 2 + &/3 2 - 1) 2 =0,
or {aax + bpy + aa? + &/3 2 - 1) 2 = (x 2 + 2/ 2 ) {a 2 a 2 + & 2 /3 2 - (a + b) (ao 2 + Z>/3 2 - 1)},
from which it is evident that the origin is a focus and that the equation of
the corresponding directrix is
aax + ipy + aa? + bp 2 - 1 = :
the directrix is therefore the polar of with respect to the original conic.
The centre of the conic whose tangential equation is (i) is
\a + &' a+bj L J
and therefore the other focus is
G
' - 2qq -26,3\
K a + b' a + b J '
Hence the second focus is fixed if the ratio a : b is fixed : the envelopes
corresponding to a system of concentric similar and similarly situated conies
are therefore confocal.
39. Let Ix + my - 1 = be the. equation of PQ; then the equation of RS
will be
(a* + \)V (b* + \)m + '
.*. V>mx + a?ly + a?bHm + \(mx + ly + aHm + bHm) + \Hm = 0.
Hence the envelope of RS, for different values of X, is the parabola whose
equation is
Aim (bhnx + aHy + a-bHm) = {mx +ly + a-lm + b'-lmf,
\l m J Im
Hence the envelope touches the axes, the equation of the chord of con-
tact being
I m
XII.] CONIC SECTIONS. 227
40. Let a, 6 be the radii of the circles and 2c the distance between their
centres. Take the origin midway between the centres of the circles, and let
the line joining the centres be the axis of x.
Then the lengths of the intercepts made by the circles on the line whose
equation is Ix + my - 1=0 are respectively
v r ~T+H*\'
Hence, if the intercepts be in the ratio k : 1, we have
a 2 (Z 2 +m 2 )-(Zc-l) 2 =W(Z 2 + m 2 )-/c 2 (Zc + l) 2 ,
and therefore the line will envelope a conic.
If ft=l we have
(a 2 -6 2 )(Z 2 + m 2 ) + 4cZ=0,
whence the envelope of Ix+my -1 = is at once found to be the parabola
whose equation is
4c 2 i/ 2 = (« 2 - & 2 ) (4cz+a 2 - Z< 2 ),
41. Let the equation of the hyperbola be a?- j/ 2 -a 2 =0, and let (a, /3) be
the fixed point 0. Then the equation referred to will be
{x + a)*-(y+p) 2 -a?=0.
Then, if Ix + my- 1 = be the equation of any straight line which subtends a
right angle at O, the lines represented by
x*-y 3 + 2(ax-py){lx + my) + (a*-pP-a*)(lx + m>jy>=0
will be at right angles, the condition for which is
(a 2 - p 2 - a 2 ) (Z 2 + m 2 ) + 2aZ - 2£m= 0.
Hence the directions of the two chords which pass through (x, y) are
given by
(a 2 -/3 2 -a 2 ) (Z 2 +m 2 ) + 2 (oZ-/3m) (lx+my)=0.
Hence if the two chords through (x, y), which subtend a right angle at 0,
are at right angles, we have
a 2 -/3 2 -a 2 + ox-^2/ = 0,
so that (x, y) is on the polar of 0.
42. Let the equation of PQ be Ix+my- 1 = 0; the equation of AP, AQ
will be y 2 - iax {he + my) = 0.
Since AP, AQ make an angle of j with one another we have
(l-4aZ) 2 =4{4a 2 m 2 + 4aZ},
or 16a 2 Z 2 -16a 2 m 2 -24aZ + l=0.
15—2
228 CONIC SECTIONS. [CHAP.
Hence the directions of the two chords through (x, y) are given by
16a 2 P - 16a 2 m 2 - teal (lx + my) + (lx + my) 1 = 0.
Hence the equation of the envelope is
(a; 2 - liax + 16a 2 ) (y 2 - 16a 2 ) = (xy - May) 2 ,
j, 2 (x-12f
or T5tf + 128a 2
43, Take the given point for origin, and the given straight line for axis
of x, and let the conic be given by the general equation. Then, if
Ix+my- 1=0
be the equation of the line joining a pair of the points, the lines represented
by
ax* + 2hxy + by 2 + 2 (gx +fy) (Uc + my) + c (lx + my) 2 =
must be equally inclined to the axis of x ; and therefore
h+gm+fl+clm=0 (i).
We have therefore to find the envelope of Ix+my- 1=0 with the con-
dition (i).
The equation of the two lines through (x, y) is given by
clm + (gm+ft) (Ix + my) + h(lx + my) 2 =0 (ii).
The lines given by (ii) are coincident provided
i (hx 2 +fx) (7m/ 2 +gy) = (2hxy + gx +fy + e) 2 ,
or i(fg-hc)xy = (gx+fy + c) 2 .
Hence the chords envelope a conic which touches the axes of coordinates I
the origin must therefore be on the director-circle of the conic.
44. Take the fixed point for origin, and let S x =0 and S 2 =0 be the
equations of any two of a system of conies which pass through four fixed
points ; then the equation of any other conic of the system is S 1 + XS 2 =0.
Let Ix+my + 1 = be the equation of any chord of S 1 + XS 2 =0 which
subtends a right angle at ; then the straight lines represented by the
equation
(a l + -\a 2 )x i + 2(h 1 + \h i )xy + (b 1 + X6 2 ) y 2
-2{(g 1 + \g 2 )x + (/i + X/ 2 ) y] (lx + my) + (C! + Xc,,) (Ix + my) 2 =
must be at right angles to one another. Hence we have
fll + 6j - 2lg x - 2mf x + Cj (P + m 2 )
+ X {a 2 + b 1 -2lg i -2mf i + c 2 (P + m 2 )} = (i).
Now (i) is the tangential equation of a conic, and therefore the chords
of S! + XS 2 =0 which subtend a right angle at envelope a conic. Also
£11.] .CONIC SECTIONS. 229
the system of oonics of which (i) is the tangential equation will clearly all
touch the four common tangents of the conies whose tangential equations
are respectively
u r + &! - 2lg t - 2m/i + Cj (J a + w?) = 0,
and a i +b a -2lg 2 -2mf i + c i (P + m > ) = 0.
45. Take AB, AC for axes and let B, G be (g, 0), (0, /) respectively, and
let D be {x v yj.
The general equation of a conic through A, B, C is readily found to be
ax 3 + 21ixy + by 2 - agx - bfy = 0.
The conio will also pass thr6ugh (x v j/J if the three unknown constants
(/, h and b satisfy the relation
a(x 1 1 -gx 1 )+2hx l y 1 +b(y 1 s -fy^ = (»)•
The equation of the tangent at B is
2agx + 2hgy -ag (x+g) -bfy = 0.
Hence if the equation of PQ be Ix + my + 1=0, we have
mag i + 2hg~bf=0 (ii).
We have similarly
lbf + 2hf-ag=0 (iii).
From (i), (ii), (iii) we have
x* -gx lt Xiy lt yf-fy-L
mg 3 , g, -f
-9 , f, V
=0 (iv).
Now (iv), being of the second degree in it and m, shews that lx + my + l=0
envelopes a conic.
46. Take the fixed point O for origin, and let the equation of the circle
be x"+y :l +2gx + c = 0. Let lx + my + l = be the equation of PQ; then to
find the distances of P and Q from O we have to eliminate 6 between
r 2 +2<7rcos0 + c=O and ?-Z cos + rm sin + 1 = 0.
The result is
(Ir* + cl - 2jr) 2 - VmV + m 2 (r 2 + c) 2 = 0.
Hence, if OP. OQ= constant = a 2 , we have
{c i -a i )[P + m 1 )-4cgl + 4g' > =0 (i).
Also, if OP 2 +OQ 3 = constant = 6 2 , we have
(6 2 +2c)J 2 +(i 2 + 2c-4 ? 2 )m s -4^=0 (ii).
The Cartesian equation of the envelope whose tangential equation is (i) 13
easily found to be
( C 2 _ a i + i egx + 4j/V) (c 2 - a* + 4gY ) - (2cgy + 4g*xy)*= 0,
230 CONIC SECTIONS. [CHAP,
•which may he written
Hence in the first case the locus is a conic of which the origin is one
focus.
The Cartesian equation of the conic whose tangential equation is (ii) is
easily found to be
(ft 2 + 2c - 40 s ) (i 2 + 2c - 4gx) = 4f the conic be the ge
IjX + m$ + 1 = 0, &o.
5Q. Let the equation of the conic be the general tangential equation of
the second degree, and let
be the three given tangents.
Then we have
al 1 2 +21il l in 1 + bm l s + 2tjl 1 + 2fm 1 + c = (i),
aZ 2 2 + 27j7 2 m 2 + &m 2 2 + 2gl 2 + 2fm^ + c = (ii) ,
al 3 1 +2hl 3 m 3 +bm. i 2 + 2gl 3 +2fm 3 + c=0 (iii).
Also, if (x, y) be the centre of the conic, we have [see 37]
g-cx = (iv),
f-ey = (v).
The square of the radius of the director-circle is constant, and equal to 7c 2
suppose. Hence, from Art. 241, we have
JcV=g*+f s -c(a + b);
or, using (iv) and (v),
a + b-gx-fy + Wc = (vi).
Eliminating a, h, b, g,f, c from the equations (i), (ii),..., (vi) we have the
equation of the required locus, namely
1, , 1 , -x, -y, To 2 =0.
, , , 1 , ,
0, , , , 1 ,
'i 2 > ^"hi m i 2 > 2 h< 2m i<
7 2 2 , 7 2 m 2 , m£, 27 2 , 2%,
Vi h m 3> m -4> 2 h> 2n h<
Multiply the fourth column by x and the fifth by y, and add the sum to
the sixth; we then have
-y
1
1
1
1, , 1 , -x, -y, k*-x*-y*
0,0,0,1,0,
0, , , 0, 1 ,
Jj 2 , 1^, m^, 2^, 2%, 2l 1 x+2m 1 y + l
7 2 2 , 7 2 m 3 , m 2 2 , 27 2 , 2m 2 , 27 2 x + 2m 2 i/ + 1
Z 3 2 , 7 3 m 3 , m 3 2 , 27 3 , 2mj, 27 3 x + 2m^/ + 1
=0,
234
.CONIC SECTIONS.
[CHAP. XII.
that is
1, , 1 ,
Zj 2 , z^, jjvS
Z 2 2 , Z 2 7%, m 2 3 ,
Z» , Zonio, m 3 .
W-x^-if
2l 1 x + 2m 1 y + l
2Z 2 E+2m 2 2/ + l
27 3 x+2ms# + l
= 0,
from which it is obvious that the locus of the centres of the conies is a circle
for any particular value of A, and that the circles are concentrio for different
values of k.
Or thus :—
The general tangential equation of a conic which touches three given
straight lines is
where \, X a , X 3 are arbitrary and S^O, S 2 =0, S s = are the tangential
equations of any three conies which touch the lines.
Now from Art. 241 we see that the equation of the director-circle of a
conic is of the first degree in a, h, b, &c. It therefore follows that if
(^=0, C 2 =0, C 3 =0 be the director-circles of the conies S 1 =0, S„—0, S 3 =0
respectively, the equation of the director-circle of
XjOj + XgOjj + AgOg —
will be X 1 C 1 +X 2 C„ + X s C 3 =0.
Now it is easy to prove that a circle will cut the three circles 0^=0,
C 2 =0, C 3 =0 orthogonally, and that this circle will also cut orthogonally any
one of the systems of circles given by
X 1 C 1 + X„O 2 + X 3 C 3 =0.
We therefore have the following theorem : — The director-circles of all
conies which touch three given straight lines are cut orthogonally by the
same circle.
It follows from the above that all those director-circles which are of
given radius have their centres (which are also the centres of the corre-
sponding conies) at a given distance from the centre of the orthogonal
circle : this proves the proposition.
CHAPTER XIII.
Page 285.
1. In order that a point may be on the bisector of an angle it is sufficient
and necessary that the perpendiculars from the point on the line bounding
the angle should be equal. Hence for any point on the bisector of A we
have p=y; and similarly the equations of the other bisectors are 7 = a and.
2. At A', the middle point of BG, it is obvious that bp = cy; and the
ratio /S : 7 is clearly the same for all points on AA'. Hence at any point on
AA' we have bf3=cy.
The equations of the three medians of the triangle are therefore
bp-cy=0, cy-aa=0 and aa.-bp=0.
3. At B' we have /3 = and cy-aa=0; also at C wo have 7=0 and
bp-aa=0. '
Hence the line whose equation is bp + cy-aa=0 passes through B' and
through C".
Similarly the equations of G'A' and A'B' are respectively
aa-bp+cy=0 and aa + bf3-cy = 0.
4. The trilinear co-ordinates of the centre of the in-circle are pro-
portional to 1, 1, 1 ; also the co-ordinates of the centre of the circum-circle
are proportional to cos A, coaB, cos G.
Hence the equation of the line joining these two points is
1,1,1
cos A, cos B, cos G
=0.
5. The co-ordinates of the four centres are given by
«i=Pi=7i =
a + b + c'
2 -i
-a 2 = &, = 72= — ;— - — 1
236 CONIC SECTIONS. [chap,
2A
ana a, = ft=-7,= — — r •
4 ri '* a + b-c
The co-ordinates of the middle point of the line joining (oj, ft, 7J and
, „ . 1 / 2A 2A \
(«, ft, 72 ) are - ^-— - _ a + b + c ) ,
1 /__2A_ 2A \ ^1 f 2A 2A \
2 \a+6+c -a+6+c/' 2Va+6+c-a+6+c/'
which are proportional to - a, b + c, b + c.
The co-ordinates of the middle point of the line joining (a.,, ft, y 3 ) and
K. ft. 7i) are
1 / 2A 2A \ 1 / 2A 2A \
2 \a-6 + c + a + 6-"cJ' 2 \ a-6 + c + a + b-c) '
1 / 2A 2A \
2 \a-b+c a + b-c) '
which are proportional to a, c-b, b-c.
Hence the six middle points are
(-a, b + c, b + c), (0 + 0, -b, c + a),
(a + b, a + b, -c), (a, c-b, b-c),
(c-a, b, a — c) and {b — a, a — b, c).
It is now easy to verify that these six points ore all on
aftv + bya + ca/3 = 0.
6. Let be (<*!, ft, 7J. Then the equations of AOA', BOB', COC are
respectively
i?-*=0,
ft 7i '
X_« =0 and i-£ = 0..
7i a i *i ft
Hence £' is the point /3 = 0, — = 0,
7i «i
and C" is the point7=0, £ =0;
ft a i
and both these points are clearly on the
ft 7l »l
XIII.] CONIC SECTIONS. 237
Thus the equations of S'C, G'A' and A'B' are respectively
ft 7i «■! 7i "i ft
and ^ + £_T1 =0 _
°i ft 7i
Now P is the point of intersection of o=0 and
1 + ^-^=0,
ft 7l «1
orofa = 0and £ + 7_ .
ft 7i
hence the line — (-£ + -X=0 goes through P, and the symmetry of the
a i ft 7i
equation shews that it also goes through Q and R. Thus P, Q, R are all on
the line
^ + # + ^=0.
«1 ft 7i
Again, since Q is the point of intersection of /3 = and — h — = 0,
a i 7i
the equation of BQ is h — =0 :
a i 7i
so also the equation of GR is — + £-=0;
«i ft
S *y
and the equation of AA' is • —■ = — .
ft 7i
These three lines obviously meet in the point P' given by
-— =A = X
«i ft 7i'
7. At the point P it is easy to prove that (3=a cos C + - sin C, and
7 = a cos P + s sin P. Hence the equation of AP is
P 7
2cosC+sinC 2cosP + sinP'
or ^(2oosP + sin£) = 7(2cosC + sinC) (i).
The equations of BQ and CR are similarly
7 (2 cos C + sin C) = a(2cos 4 + sin A) {ii),
and a (2 cos .4+ sin .4) =|9 (2 cos P +sinP) :...(iii),
and it is obvious that the three lines whose equations are (i), (ii), (iii) meet
in a point.
238
CONIC SECTIONS.
[CHAP.
8. Let la + m,p+ny=0 be the equation of the straight line j then the
lengths of the perpendiculars on this line from
(?.o,o),(o,?,o),^(o,o,?)
are proportional to
2AJ 2Am , 2An ., . „.„
— , -y- , and [Art. 2571.
a b c L
_ , I m n
Hence we have — = j- = — .
ap bq cr
The equation of the line may therefore he written in the form
apa + bqfl + cry=0.
9, Let ABC, A'B'C he two triangles such that AA', BB', CC meet in
some point 0. Take ABC for the triangle of reference, and let be
(f, 9, h).
Then the equation of AA'O is " = I, and therefore A' may be taken to be
(oj, g, h). So also B' and C may be taken to be (/, ft, 7j) and (/, g, 7J
respectively. The equation of B'C is therefore
= 0.
a ,
f>,
y
/.
ft.
h
f,
g <
7i
Hence B'C meets BC where
o = and —&— + -^-=0.
9-Pi ' l -Yi
Similarly the point of intersection of CA and G'A' is given by
|8=0 and
= 0,
f-o-i ft -7i
and the point of intersection of AB and A'B' is given by
= and
f-a-i g-
=0.
The three points of intersection of corresponding sides are therefore on
the straight line whose equation is
« + ^s + _L. =0 .
/-«i 0-Pi ft-7i
XIII.]
GONIC SECTIONS.
239
Pages 309—314.
1. The perpendicular from the centre of a conic on any tangent cannot
be less than the semi-minor axis.
Hence, if (a , ft, 7o) be the centre of any conic inscribed in the triangle
of reference, and p he the length of the semi-minor axis, and r be the radius
of the inscribed circle, we have
also
« a o + 6 ft+ c 7o > ( a + 6 + c ) P ;
aa + bp cy — 2 A = (o + 6 + c) r ;
/. r>p.
2 x area of triangle = sin C
2. Let (oj, ft, 7^, &c. be the angular points of the triangle; then, as
in Art. 6,
aj cosec C, u a cosec C, 1
u 2 cosec C, ft cosec C, 1
a s cosec C, ft cosec (7, X
1 adj, 6ft, 2 A
'2Aa6sinC ad 2 , 6ft, 2A
0%, 6ft, C7i
aa. it 6ft, C7 2
oaj, 6ft, cy 3
by subtracting the sum of the two first columns from the third
1
: (2A) 2
dbc
: (2A) 5
<*i> ft ■ 1\
"2> ft' 72
"3' &> 7s
3. The equations of the conies which have a common self-polar triangle
may be taken to be
« 1 a 2 + « 1 /3 a +20 1 7 s =0, &c.
The four points of intersection of the first two are given by
a p _ 7
=W(«iU> a -tV>]) ± «yK« 2 -«'2"i) ± V(¥r¥i)'
The four points of intersection of the other two are given by
a = P = 7
±»J{i>t'i>4-v 4 tt>J ^^(w^-w^) i^K^-U^,)'
The eight points will all lie on the conic whose equation is
La?+Mp i + Ny*==0,
240
CONIC SECTIONS..
[CHAP.
provided L (djW 2 - v^w^ + M (w^ - w s u x ) + N(u 1 v a - w^) = 0,
and L (« s m; 4 - v t w s ) + M (m; 3 m 4 - m^Uj) + N (u 3 v t - u t v 3 ) = 0.
Hence the eight points are all on the conic whose equation is
VM.-VM.
WM.-WM,
U,V„ - UoVi
= 0.
4. Eefer the conies to their common self-polar triangle; then their
equations will be
« 1 a 2 + B,^ 2 + M) 1 7 2 =0 and u 2 a 2 -t-i>, ! /3 2 +Hvr ! =0.
The iine la+mp+ny =
will touch both conies provided
lib lb -. -,
-+ — + — = and
P
■ + — + —
-.0.
Hence the equations of the common tangents are of the form
Za±m/3±n7=0.
The points where the four common tangents touch the two conies are
given by
Mja _ ojjjS _ w-fl
»„/3
I ±nt ±« I ±m'
w 2 y
These eight points are all on the conic whose equation is
r
n 2
V
= 0.
5. Eefer the conies to their common self -polar triangle; then their
equations will be
M 1 a 2 +« 1 /3 2 + i(; 1 7 2 =0 and iitfp+vrfP+WjpzzO.
The co-ordinates of their four common points are given by
a 7
VKW2-V1) iVlWA-WjUi) ^(ttjCjj-M^)'
which are of the form ±/, ± g, ± 7i.
Hence the equations of the eight tangents are of the form .
±«]/a± i^jr/3 ±10^7=0 and ±^^±^^±^717=0 (i).
The eight tangents will all touch the conic
La? + Mp- + Ny-=0
XIII.] CONIC SECTIONS. 241
provided
y/ 2 Bl y a V) = ^ / («i V - «gV).
Hence the eight lines (i) touch the conic
»iV - « 2 2 ">i 2 »i V - «vS 2 «i V - «aV ~
that is * + £ + _J^ =0 .
6. Take the triangle formed by the diagonals for the triangle of
reference ; then the equations of the four sides may be taken to be
Za±m/3±n7=0.
Hence the equation of the lines through A and through the extremities
of the opposite diagonal is
m 2 /3 a -nV=0.
Now the equation of any other pair of lineB through A which are harmoni-
cally conjugate to
«V-nV=0
may he put in the form
m 2 /3 2 + nV + 2W=0.
Hence the three pairs of points which satisfy the given conditions are
given by
m 2 /3 2 + nV+2Xj37=0, a=0; rcV + Z 2 a 2 + 2/t 7 a=0, /S=0
and Z 2 a 2 + m 2 /3 3 + 2ra/3 =0,7=0.
The six points are all on the conic whose equation is
Pa 2 +m 2 /3 2 +«V+ 2X/3-y + ipiya + 2»a/3=0.
7. The perpendicular distance of (a, /3, 7) from -aa + 6/3 + 07=0 is
-aa+bfi + cy _ -aa+bji + cy
V(o 2 + 6 2 + c 2 - 26c cos A + lea cos B + 2a6 cos C ~ 2a~
S. C. K. 16
242 CONIC SECTIONS. [CHAP.
The perpendicular distances of (a, (3, y) from
aa-bf} + cy=0
and aa + bf3-cy=0
. ., , aa-bp + cy __„ a a + b^-cy
are similarly =t and jjc '
Also the lengths of the sides of the triangle formed by the three lines
whose equations are
-aa + bf! + cy=0, &e.
a b e
are 2' 2' 2'
Hence, from Art. 269, the required equation is
« 2 . + h l + t =0 .
-aa + bp + cy aa-bfi + cy aa + bfi-cy
8. The equation of any circle concentric with the circumscribing circle
is
a/3y + 67a + ca/3 + \ (aa +bp + cyf= 0.
Along the line through A parallel to BG we have
aa=2Aand bf} + cy=0;
hence at the points, P, P' suppose, where this line meets the circle, we have
a 2 6/3 a +26 2 ^A-2c 2 ^A-4acXA 2 =0 ;
4A 2 c\
PA . AP' sin?C=p 1 p 1 = -
ub
„ m 4e\A 2 ,
.-. ?~ ! -Ji 2 =- . „ =-abc\.
ab sin 2 G
Hence the required equation is
afiy + bya+cap =- — (aa + bp+cyf=0.
9. The point which is at a distance p from (a , j3 m y ) on a line parallel
to .BC is
(«o » A> + P sirl C. 7o - P sin B).
Hence, if this point be on the circumscribing conic whose equation is
ipy + mya+ naj3 = 0,
we have
I (/3„ + p sin G) {y - p sin B) + mo, (y - p sin B) + no (ft + p sin G) = ;
Zft7 +TOy a +na ft
■■ PiPi- -isinBsinC -
XIII.] CONIC SECTIONS. 243
Hence, from Art. 186, Cor. I., we have
r5 r 3 ~ 2
~1 r 2 r 3
a 6 c
7 m n
The equation of the conic must therefore be
'1 '2 ? 3
10. Take ABC for the triangle of reference, and let the equation of the
conic be
I8y + my a + naB = 0.
The equations of B'C, G'A', A'B' -will then be respectively
rwy+nB=0, no + Z7 = and lB + ma=0.
The equations AA', BB', CC are easily seen to be respectively
£-1=0, ?-?=0and 5-£=0,
m n n I I m
from which it is obvious that AA', BB', CC meet in a point.
Again the point D is the point of intersection of a=0, my+nB=0.
Hence D is on the line whose equation is
? + £+?=<);
I m n
and it is obvious that E and F are also on this straight line.
11, Take ABC for the triangle of reference, and let P be (/, g, h). Then
the equations of AA', BB', CC are
8 y y a , a 8
respectively. Hence the equations of B'C, G'A', A'B' are respectively
-5 + §+2=0, ?-§+?=<>
/ 9 » / 9 »
and 7 + ^ -?=<>■
Hence the points Jf, L, M are on the line whose equation is
f 9 h
Now, (i), if P be on the fixed straight line
la+mB + ny=0,
■we have If + mg + nh = 0,
16—2
244 CONIC SECTIONS. [chap.
which shews that
f 9 h
touches the conic
Jla + »/m£ + k/tvy = 0.
Also, (ii), if P move on the fixed conic
lj3y + my a + no/3 = 0,
, I m n „
we have ■-. H h T =0,
f 9 *
whence it follows that
/ 9 &
passes through the fixed point [I, m, n).
And, (iii), if P move on the fixed conic a§=\y i , we have fg=\h?. The
equation of KLM may now be written in the form
/> + Xft» + /ft '
the envelope of which is iap=\y s .
12. If be (/, g, ft) and 0' be (/', 3', ft'), 4' will be o=0, 2 = 2 and
«
so for B' and C; also A" will be o=0, -, = ?,, and so for B" and C"-
9 ft
Hence B'C is 2 + 2_" = o,
9 h f
and B"C" is £ + 2 * 0;
9 h' f
and therefore at P s ( i _ £ ) + 7 (/ - =CW 0.
\9 9 J \h h'J
Hence the equation of AP is
P - _L_
gg'(hf'-h'f) hh'(fg'-f'g)-
The equations of BQ, CB are similarly
7 a
and
hh'U9'-f'9)~ff'(gh'-g'h)
P
ff'{gh'-g'h) gg'{hf-h'f)-
XIII.] CONIC SECTIONS. 245
The point Z is therefore given by
ff(gh'-g'h) gg'(hf-
*'/)
hh
'{fg'-
-fg)
Now,
if 0, 0' are both on the conic
a £
V
0,
wp VinvA
X /*
T =
V
VY /3' 2 + M>y 2 =0 (ii).
From (i) and (ii) we have
7' (na - Ic) - p' (lb - via)
Substitute for u, v and w in the condition
Z 2 m 2 m 2 „
- + — + - = 0;
u v w
and we obtain the equation of the required locus, namely
2Pa {a (lb-ma)-y(mc-nb)} {/3 (mc- rib) -a (na~lc)}=0.
Thus the required locus is a cubic curve which passes through the angular
points of the triangle of reference.
16. Let r lt r i be the semi-axes of an ellipse inscribed in the triangle
of reference, and let (a„, /3 ; 7 ,) be the centre of the ellipse. Then, if the
sides of the triangle make angles S v + c 7o) ( - aa o + 6 A> + c 7o)
(aa - 6/3 + c-y ) (aa + &y3 - C7 ) .
In the case before us the centre of the conic is
(R co$A,R cos B, R cos G) ;
and it will be found that
)-! 2 +?-., 2 =iJ 2 (1-4 cos A cos B cos G),
and »YV,j 2 = 4i? 4 cos 2 A cos 2 £ cos 2 C.
Now (i 2 = J R 2 (l-8cos^cosBcosC).
Hence r l 2 +r 2 2 =^( J B 2 +d 2 )
and '•iV^^-d 2 ) 2 .
so that 1r x and 2r s are equal to i? + 0,
or (2A-2a|)(2A-2&ij)(2A-2cf)>0.
Now within the triangle DEF all the factors on the left are positive ;
and within either of the angles vertically opposite to one of the angles of the
triangle DEF two of the factors are negative and the other factor is positive.
This proves the theorem.
XIII.] CONIC SECTIONS. 249
18. Theoonio ua?+vp^ + ti>7 2 =
is a parabola if it touches
aa + bp + cy=0,
the condition for which is
a 2 /u + 6 2 /» + c 2 /iu = 0.
But this condition shews that the parabola also touches the lines
-aa + bf3 + cy=0, aa-bp' + cy=0 and aa + bfi- cy=0.
The focus of the parabola is therefore on the circle circumscribing the
triangle formed by these tangents, and this circle is clearly the nine-point
circle of the triangle of reference.
19. Let (o, j3, 7) and (a', p", 7') be a pair of foci of any conic touching
the four lines Za±in/3=t 717=6, both foci being real or both imaginary; then
since the product of the perpendiculars from the foci of a given conic on
any tangent is constant, we have
la + mB + ny la' + viB' + n-/ . .
g- . g - = constant = A.
"1 "1
Hence la' + mB' + ny' =
la + mB + ny '
la' - m/3' - n-/ = , J ,
r la — mB - ny
XP 2
- la' + mB - ny'= —r— — ^ ,
- la + mB - ny
\P 2
and -la'- mB' + ny'— —
la-mB + ny
Hence, by addition,
jy pf p t * , rs
la + m(3+ny la-mB-ny -la + mp-ny -la-mB+ny
20. Let (|, i), f) be one focus; then the other will be
= 0.
(«' v' f)-
The equation of the line joining the foci is given by
». P. 7
i, V, f
111
f v' i
= 0.
250 CONIC SECTIONS. [CHAP.
By supposition the line joining the foci passes through (/, g, h). Hence
we have
/, g, h =0.
f. V, f
111
V v' t
Hence the locus of (£, i\, f ) is the cubic
/a(^_ 7 2) +?J 3( 7 2_ a 2) + fe 7 ( o 2_ j8 2) = 0.
21. Let the line on which the centre moves he
la + mf3 + n-y=0.
Let ({, i\, f ) be one focus ; then the co-ordinates of the other will be
XXX
£ v f
where X is given by
„, oX 6X cX
2A= — + — + — .
f 1 f
If (o , /}„, 7 ) be the centre of the conic, we have
2ft,=„ + 2A/„(| + J+^,
and 2 7o =i-+2A/r(| + | + '|).
Hence
^ + m , + n f+ 2A( | + - + ? )/( ? + - + ? ) = 0,
or (Z£ + mi; + nf ) (aijf + &$? + cfij) + (a| + &17 + cf ) (ftji"+ mff + njij) = 0.
22. The equation of a rectangular hyperbola with respect to which
the triangle of reference is self -polar is
ua t +vp'+wy 3 =0,
with the condition
u + v + w = (i).
The centre is given by
«a _ vfi ivy
a b c
Substituting for u, v, w in (i), we have the equation of the locus of the
centres, namely
a b c
<* P 7
XIII.] CONIC SECTIONS. 251
23. The equation of any one of the conioa is
N /\o+ N /^+ N /n;=o,
or in the rationalised form
W+ l i 1 p ll + v ;! v s -2nvPy-2v\ya-2\na.p=0 (i),
with the condition
\"+/i 2 + i' !l + 2fi.ii cos A + 2v\cob B + 2\fn.oos C = 0= (ii).
Now the centre of (i) is given by
X (Xa - ft/3 - vy) _/j.(-\a+/i.p-vy) _ x(- Xa - ft/3 + vy)
a b c '
from which it follows that
X u. v
a(aa-bp-cy) b (- aa + 6/3 - cy) c (- aa - bfi + cy)
Substituting in (ii) we have the equation of the locus of the centre,
namely
a 2 (aa - 6/3 - c 7 ) 2 + 6 2 (- aa + 6/3 - cy) 2 + c 2 (- aa - 6/3 + cyf
+ 2bc (- aa + bp - cy) (- aa - 5/3 + 07) cos A
+ 2ca (- aa - bfi+cy) (aa - 6/3 - cy) cos B
+ 2ab (aa - 5/3 - cy) (- aa + 6/3 - 07) cos G = 0,
which reduces to
a 2 sin 24 + /3 2 sin 2.B + 7 2 sin 2C = 0.
24. The equation of the inscribed circle is
J(s-a)~aa + J(s-b)Fp + J(s-c)cy=0,
or 2 (s - a) 2 a 2 a 2 - 22 (s - 6) (s - c) bcpy=0.
Hence the equation of any other circle is included in
2 (s - a) 2 a 2 a 2 - 22 (s - 5) (s - c) 60/37 + (Xa + ft/S + ><7) (aa + 6/3 + cy) = 0.
The nine-point circle goes through the points
a=0, 6/3-C7=0; /3=0, C7-aa=0; and 7=0, aa-6/3=0.
Hence X, ft, v are given by
(«-a)» + (.-c)S-2(«-&)(«-c) + 2 (j + *-)=0.
(6- C ) 2 + 2(^)=0,
and two similar equations.
252 CONIC SECTIONS. [CHAP.
Hence we have
X /t v
a(a-b)(a-c)~ b (b -c)(b-a)~ b (c - a) (c - b) '
The equation of the radical axis of the inscribed and nine-point circles
is therefore
aa bB cy _ ..;
r — + — tL - + — -r=0 (1).
b-c c-a a-b
It follows at once from Art. 278 that the radical axis of the nine-point
circle and the inscribed circle touches the inscribed circle : the two circles
must therefore touch one another.
Again the equation of the circle which touches BG externally and GA,
AB internally is found as in Art. 279 to be
V -asa+Jb{s-c)B + Jc (0-6)7=0.
The common radical axis of this inscribed circle and the nine-point
circle is found as above to be
aa bB cy _ ....
b-c c+a -a-b v '
It follows from Art. 278 that (ii) touches the escribed circle ; from which
it follows that the nine-point circle must touch the escribed circle.
25. We have found in the preceding example that the equations of the
four tangents are
" + VL + JDL (i),
b-c c-a a-b
aa bB cy
, +— r-+ L -7 = (ii),
b-c c+a -a-b '
aa bB cy „ .....
— . + — — + — ; ^-=0 (in),
-b~c c-a a+b x "
, aa bB cy
and ■=-— + — !—- + — 4=0 (iv.
b + c -c-a a-b v '
The line joining the intersections of (i), (ii) and of (iii), (iv) is easily
found to be
c -^ + A=° w-
The equations of the other diagonals are respectively
cy aa . , ..
a^H>* + W^ =0 (vl >'
, aa bB _ , ...
Now it is dear that each of the lines given by (v), (vi) and (vii) passes
through one of the angular points of the triangle of reference.
XIII.] CONIC SECTIONS. 253
The lines joining corresponding angular points of the triangle of reference
and of the triangle formed by the lines (v), (vi) and (vii) are given by
bB cy cy aa , aa &/S
c- - a- ~ a^E 2 ' a^^F'~b'-c' ' b*^ti*~ c^a?'
and it is easy to shew that these lines are all parallel to
a cos .A +/3 cos £ + 7 cos C=0,
which [Art. 286] is the radical axis of the nine-point circle and the
circumscribing circle.
26. Take ABC for the triangle of reference, and let the conio be given
by the general equation.
Then the equations of B'C, C'A' and A'B' are respectively
ua + w'B + v'y=0,
w'a + vB + u'y=0,
and v'a+u'B + wy=0.
Hence the equations of AA', BB', CC are respectively
P _ 7 y a a B
• u'v' - ww' w'u' - vv' ' v'vl - uu' ~ uV - ww' ' u'w' - vv' ~ v'w' - uu' '
Henoe AA', BB', CC meet in the point given by
o (uu' - v'w') = B (vv' - w'u') =7 (ww' - u'v').
27. The conic given by the general equation cuts a=0 where
vpP + wy* + 2u'8y = 0.
Hence the condition that the conic should pass through the middle point of
BGia
vc*+wb 2 + 2u'bc=0 (i).
Also, if 6/3 - cy be one factor of v8 2 + wy* + 2u'8y, the other factor must be
vB wy
T~T"
Hence the equation of the line through A and the other point of inter-
section is
vBjb-wy\e—0.
Hence, if the conio given by the general equation of the second degree
pass through the middle points of all three sides of the triangle of reference,
and cut those sides again in A', B', ff respectively, the lines AA', BB', CC
all meet in the point
ua _ vB _ wy
a ~ b e
This point is on the circumscribing circle provided
u+v+w = (ii).
254 CONIC SECTIONS. [chap.
Now if the conic be a rectangular hyperbola we have
u + v + w - 2u' cos A - 2v' cos B - 2w' cos C = 0,
which reducesto u + v + w=0,
since we have vc 2 + wb 2 + 2u'bc=0 and two similar equations.
28. Let the equations of the conies, referred to their common self-polar
triangle, be
w 1 a > + » 1 /9 a +w 1 7 a =0 and u 2 a i +v i p s +w 2 y^=0.
Let the equation of the given straight line be la + mp + wy=Q.
Then the polars of (/, g, h) with respect to the two conies are
Mja/+ v^g + w-flh = 0,
and u 2 af+v^g + w 2 yh=0.
But, since (/, g, fi) is on the given line, we have
lf+mg+nh=Q.
Eliminating /, g, h from the last three equations, we have the equation
of the required locus, namely
Uja, vJS, Wjy
« 2 a, vji, t» 2 -y
I , m, n
= 0,
which clearly is the equation of a conic circumscribing the triangle of
reference.
29. The equations of the two conies may be taken to be
a 2 -2X^7=0 and o 2 - 2^7=0.
The polar of (a', j3', y') with respect to the first conic is
aa'-X|87'-?i7/3' = 0.
Hence, if this polar touch the second conic, we have
Al a' 2 -2X^V=0.
Hence the equation of the required locus is
Ha?-2\*py=0,
which proves the proposition.
30. Take one of the triangles for the triangle of reference, and let
>(°i> ft> 7i)> * 0, b e *^ e angular points of the other triangle.
Let the conic on which the six points lie be
-+£+-=0 (i).
a p 7 w
XIII.] CONIC SECTIONS. 255
Then we have
-+£+-=o 7 1 7 2 7 3 /7=0 (v).
This proves the proposition, since the conies (iv) and (v) are respectively
inscribed in and circumscribed to the triangle of reference.
32. Let ABC be the triangle whose angular points are on the conio S
and which is self-polar with respect to the conic 2.
Take any other point A' on S, and let its polar with respect to 2 cut S in
the points B' and C. Let the polar of B' with respect to 2, which we know
256 CONIC SECTIONS. [CHAP.
■will pass through A', cut B'C in the point D. Then the triangles ABC and
A'B'D are both self-polar -with respect to 2 ; and therefore, by the preceding
question, the six points A, B, C, A', B', D are on the same conic. But five
of the points, namely A, B, C, A', B' are on the conic S, and only one conic
will pass through five points, therefore D must also be on S, so that C" and
D must coincide. This proves the proposition, since A' is any point on 8.
33. The asymptotes of the conic ua s + vp 2 + toy 2 = are parallel to
u (6/3 + cyf + va?{P + w«V = 0.
The angle between these lines is [Art. 44]
_j 2 ^{mWU (utf+va?) (uc* + wa?)\ sin A
ub 1 +va 2 +uc i +wa"-2ubc cos A
_. _j %J{- vwa? - wub' 1 - wye*} sin A
a (u + v + w)
Hence, if the conic be one of a system of similar conies, we have
W (u+v +W) 1 + 2vwa'+2wub* +2tMJc a = (i),
where k is some constant.
Now the centre of the conic is given by
Ma v/3 wy
a b c
Substituting for it, v, w in (i) we have the equation of the locus of the
centre, namely
\a y] \py 7 a a? J
or ft 2 (afty + bya+ cap) 2 + 2abca.py (aa + bp + cy) = 0.
The locus is therefore a curve of the fourth degree &o.
34. This is the converse of Brianchon's Theorem, and may be proved
thus : — A conic will touch the ./foe lines AB', B'G, GA', A'B and BC ; let the
other tangent to this conic from the point 0' cut BA' in the point X. Then
by Brianchon's Theorem, BB', CC and A'X meet in a point ; but BB', CC
and A'A meet in a point ; hence X must coincide with A, and therefore the
conic touches G'A.
35. Let the equation of the conic be
J\a + J]If3 + Jry= 0.
The points of contact are a=0, fi.p=vy; &c.
Hence the normals at the points of contact are
(/i cos G - v cos B) a + /j.p - vy = 0,
(» cos A - \ cos G) p + vy - \a = 0,
(\cos B - fiCOB A)y + Xo- /j.p=0.
XIII.] CONIC SECTIONS. 257
If these lines meet in a point, the locus of the point will be found by
eliminating X, /i, v from the three equations : the equation is
(acosC + /3) {§cosA + y) (7COsP+a) = (acosP + 7) (/3cosC + a) (7 cos .4 + /3).
The locus is therefore a cubic curve ; and by considering the three conies
which have two of the sides of the triangle for asymptotes we see that the
points at infinity on the locus are in directions perpendicular to the sides
of the triangle.
36. Take the diagonal-triangle of the quadrilateral for the triangle of
reference, and let the equations of the four lines be
la±mp±ny=0.
The equation of the conic will be
with the condition
P m 2 n 2 . ...
- + — + - = (i .
U V w
Let \a + /ji.p + i>y=0
be any other tangent to the conic ; then we have
— + — + - = n).
U V w
Now the three pairs of opposite angular points of the quadrilateral are
a=0, ?nj3±)i7 = 0; &c.
The actual co-ordinates of one pair of points are therefore
2An 2Am , . 2An 2A»t
0, -i , -; ; and 0, - , ,- - ;
bn+cm bn + cm on -cm cm -on
and so for the others.
Hence the product of the perpendiculars from these three pairs of
points on Xo + ^ + >'7=0 are
4A 2 nW-yW 4A 2 »V - W j , 4A 2 W-/i?P
P 2 " SV-A»' P 2 " cH*-aW P 2 ' a 2 m 2 -W
But, from (i) and (ii),
1 1 _ 1
u (mV - »V) ~ vJnW - P**) ~~ w (*V - m2 * 2 ) '
whence it follows that the ratios of the three products of perpendiculars is
independent of X, /i and v.
37. Take ABC for the triangle of reference, and let the conic be given
by the general equation. Then the equation of the polar of A is
ua+w'f} + v'y=0.
S. C. K. 17
258 CONIC SECTIONS. [chap.
Hence A' is given by
a=0=w'f3+v'y,
■whence it follows that A', B' and C are on a straight line, namely on the
line whose equation is
-, + 4 + ^ = 0.
U V IB
Since A'B'C ia a straight line, it follows that AA', BB', CC are
the diagonals of the quadrilateral formed by the sides of the triangle and
the line A'B'C The proposition is now reduced to a particular case of
Art. 299 (5); for AA' is a limiting form of a conic touching the four
lines, and the circle on AA' as diameter is the director-circle of this
limiting conic ; and so for the other diagonals.
38. The equation of any conic which touches BC at its middle point is
Jla + JOp+s/cy^O.
The condition that the conic should touch
; + 2 =° CO-
is [Art. 278]
The condition that the line
\a + (ip+iry=Q
should touch is
|+-+-=o : (u).
But, if jp, q, r be the perpendiculars from the angular points of the
triangle on the line
\a+/tp+i>y=0,
we have
£-!-!
\ /i v'
a b c
Hence, from (ii),
I b c „
— + ■.- + - =0;
up bq cr
/. , from (i)
- + 1 - 2
q r^p'
39. The condition that
apa+ bqp + cry a@
should touch
apy+bya+-eap=@
XIII.] CONIC SECTIONS. 259
is [Art. 276] Ja^p + JWq + N /c"V=0,
that is ajp + b,jq + cjr=0.
It A', B', C" be respectively the middle points of the sides BC, CA, AB of
the triangle of reference, and Up', q', r' be the perpendiculars from A', B', G'
on any tangent to the nine-point circle, it follows from the above that
a >HP + i/3 2 +M> 1 y ! ) (u 1 $ 3 +e 1 iJ*+ u, if) - (Uifa+twJj+Wufy)'^
and
(« 2 a= + »j|3» + wtf) (u a £ 2 + vtf + m> 2 H - (u£ a + v s vP + *>,.&)* = 0.
If the two pairs of tangents are ■ conjugate pairs of a harmonic pencil,
the lines from A to the points where they out BC will be conjugate pairs of
a harmonic pencil, and the equations of these pairs are
and (iyS 2 + t&tf 8 ) (u£> + vrf + io a f 2 ) - (v a ij/9 + w 2 fr) 2 = 0.
The condition that these should be conjugate pairs of a harmonic
pencil is
v 1 w 2 Kl 2 + lOjf 2 ) (u 2 £ 2 + vtf) + WW (u 2 | 2 + »,f 2 ) KJ 2 + vrf) - 2» 1 M! 1 i; !1 M) 2 )) 2 i- 2 = 0,
whence
Mjl/jj folflj + VjU)^ f 2 + «!» a (WjU^ + W^) 1J 2 + WjlO j (Uj» s + l/jl^) f 2 = 0.
The locus of (f, ij, f) is therefore a conic.
From the point of contact of either of the conies with a common
tangent, three out of the four tangents are coincident and the four will
therefore form a harmonic pencil. Hence the locus passes through the
eight points of contact of the common tangents of the conies.
46. The equations of the circles can be taken to be
x 2 + 7/ 2 -2oa;-c 2 =0 and x*+y i +2ax-c< l =0.
The tangents from (x', if) to the first circle cut x=0 where
(j/ 2 - c 2 ) (x" + ?/ 2 - lax 1 - c 2 ) - (yy' - ax' - c 2 ) 2 = 0,
or l/ 2 (x' 2 -2ax'- c") + %jy' (ax' + c 2 ) - (c 2 + a 2 ) x" - cY" = 0.
The tangents from (x', y') to the second circle cut x=0 where
y 1 (x 1 + 2ax'-c') + iyy' (c 2 -ax')- (c 2 + o 2 ) a;' 2 - C y 2 = 0.
Hence if the pairs of tangents are conjugate rays of a harmonic pencil
we have
(a;' 2 - c 2 ) {(c 2 + a 2 ) x'°- + c 2 i/' 2 } + y '» (c* - aV 2 ) = 0,
XIII.] CONIC SECTIONS. 261
■whence the equation of the required loeus is
(e 2 + a 2 ) x s + (c 2 - a 2 ) 2/ 2 - c 2 (c 2 + a 2 ) = 0.
Hence the locus is an ellipse of c*>a?, and two parallel straight lines if
e 1 =a-. This proves the proposition.
47. Let the equations of the two conies be
?i 1 a 2 + ?.' 1 i8 2 + K> i y ! =0 and « 2 a 2 + u 2 /3 2 + «! 2 -y 2 =0.
The co-ordinates of any point on the first conic can be taken to be
/ «>i / Wi .
j./ - cos 8, . / * sin B, - 1.
V «i V v i
Hence the equations of the tangent at the three points 8, i V V»i «i V
or X cos ^ cos 5 + M sin ^ sin 8 - N= 0,
and also i cos cos 8 + M ein sin 8 - N= 0.
Whence we have
L cos $ M sin 9 2V
1 1 1 '
■■■ p 0O8 2 g (0 + + $ - ^- 2 cos 2 - (0 - # = 0.
Hence at the point of intersection of (i) and (ii) we have
262
CONIC SECTIONS.
[CHAP.
■where
L=* ■
M=-
and N**?*-
48. Refer the conic to their common self-conjugate triangle ; then their
equations will be
u 1 a 2 + i' 1 /3 1! -|-i» 1 7 2 =0 and « 2 a 8 +u a (3 !! + «) 2 7 !! =0.
Let (/, g, h) be any point P on the first conic; then the equation of the
tangents PQ, PR from P to the second conic are
(u a a? + v$> + W£f) {uj' + v^ + wji*} - [Uzfa + VtfP + wJirfio^O... (i).
But the equation of PQ, PR is included in
X (Mja 2 + VjP> + wrf) - (uja + v$p + wjiy) (La + Mp + Ny) = 0... (ii).
Comparing coefficients of py, ya and ap in (i) and (ii) we have
Nu 1 f+ Lwji = nw^ujif,
and Lvtf + Nu 1 f= iM^fg,
M N
1_ _
"iff
Hence
Wjh
=H
V -^,&c;
... A = „ I _ *& + ^h vA &c .
«i/ l "i^i «vi «i"i)
But
Hence
which shews that
touches the conic
v 1 f i +v 1 g i +w 1 h 1 =Q,
#
1 1 v ifi "Vi u i v i\
La+Mp+Ny-0
■,=o.
1 ( «!»! M^Mj Mjt).
l S a==C
«i»ij
49. Take the triangle on which the angular points lie for the triangle of
reference ; and let the two fixed points P, Q be
(fv 0i. *i) and (/ 2 , 2 , ft a ).
Let D, 25, F be points on P(7, CA, AB respectively. Suppose that P is on
DF and Q on DE.
Then if Za + m/S+ 717=0 be the equation of FE, it is easily seen that the
equations of FD, EQ will be respectively
and
h 1 (la + mp)=y(lf l +mg 1 ),
g 3 (la+ny)=p{lf !i +nhj.
XIII.] CONIC SECTIONS. 263
But, since F P and EQ meet on BC, we have
and the envelope of la + m[3 + ny=0 with the above conditions is clearly a
conic.
50. The equation of the two tangents from A to the given conic is
u (ux 2 + vy 2 + wz* + 2u' yz + 2v'zx + 2w'xy) - (ux + w'y + v'z) 2 = 0.
These meet BC in points given by
x = 0, (uv - w'-) y 1 + ( uw - v' s ) 2 2 + 2 (mi' - v'w') yz = 0,
that is x=0, Wy* + Vz>- 2 U'yz= 0,
where IT, V, W, &c. are the cd-factors of u, v, w, &c. in
u , i«', »'
w', v, u'
Hence these points, and by symmetry the corresponding points on CA
and AB, are on the conic
FW* 2 * WUy*+ UVz"-- 2UU'yz-2YV'z3i ~2WW'^=a.
This conic intersects 2 V /«U' = 0, that is
VW+ T"V + WV-2V'W'yz - 2W'V'zx-2U'V'xy i =iQ
in the same four points as the conic
S (VW- V s ) x* + 22 (V'W- UU') i/«=0.
But the latter conic is the original conic since
VW- U'°-vA, &o.
CHAPTER XIV.
Pages 336—338.
1. See Art. 148 (7).
2. Conies through four given points have a common self-polar triangle ;
and, from Art. 306, if the eonics be reciprocated -with respect to a vertex of this
self-polar triangle, the reciprocal eonics will be concentric.
3. Reciprocate with respect to any point the following known theorem :—
' Four circles can be drawn so as to touch three given straight lines, and the
reciprocal of the radius of one of the circles is equal to the sum of the reci-
procals of the radii of the other three ; also the centres of the circles lie two
and two on lines through the angular points of the triangle formed by the
three given straight lines.'
4. Let the two conies be
« 1 a a + »ijS»-ne 1 7 !, =0 (i),
M 2 o' ! +« 2 (3 2 + w 2 7 3 =0 (ii)
the polar of the point (a', /3', y) with respect to the first conic is
u 1 a'a + vj}'p + w 1 yy=0 (iii).
If (a', j3', 7') is on (ii), we have
u 2 a' 2 + tf 2 /3' !! + w 2 y 2 = (iv).
Now (iv) Bhews that (iii) touches the conic
u
-a 2 + ^-^ + - 1 - 7 ' = (v).
Hence (v) is the equation of the reciprocal of (ii) with respect to (i).
Similarly the equation of the reciprocal of (i) with respect to (ii) is
«V „ . V „„ . w.
"JLa? + "JL(jii + t* S Y. = ( vi ).
It is clear that the conies given by (i), (ii), (v) and (vi) have a common
self-polar triangle.
CHAP. XIV.] CONIC SECTIONS. 203
5. Let the conies L, and V be referred to their common self-polar triangle,
and let their equations be respectively
J l a a + m 1 j3 2 +n 1 -y 2 =0 (i)
and ji 1 o a +u,/S 3 +W]7 3 =0 (ii).
Then, by the preceding question, the equation of L., will be
v,* „„ . w
•X a * + H/fi + ™A 7 = = (iii).
Af l is the reciprocal of (i) with respect to (iii), and its equation is therefore
'£ «■' + -; iP + —> W = ° (")•
M„ is the reciprocal of (iii) with respect to (i), and its equation is there-
fore
5- • , +3o , +s?.*-° «■
It is now clear that (iv) and (v) are reciprocals with respect to (ii).
6. Let the pencil be cut by any straight line in the points A, A' ; B, B' ;
C, C ; &c. Let P be the vertex of the pencil, and let APA' and BPB' be
right angles. Then, if PO be perpendicular to ABC; it is clear that
AO . OA' = BO . OB' = PO\ so that is centre of the involution. It
therefore follows that, if C, C are any other pair of conjugate points,
CO . OC'=P0 2 , and therefore CP, CP are at right angles.
7. Let Vbe the middle point of ^J' and also of />/?'; and let O be the centre
of the involution. Let AV=VA'=a; BV=VB'=b; and \'0 = x.
Then x? - a?=x 2 - 6 2 , from which it follows that x is infinite, since a and b
are unequal.
Now let c, c' be any other pair of conjugate points, and let CV=c,
VG'=c' ; then we have
as»-a s = (as + c) [x-c'),
which gives a finite value for x, unless c=c'.
8. Eeciprocate with respect to any point ; then we have to prove the
following theorem : — ■
The points in which any straight line cuts a system of conies through four
given points are in involution.
This is proved by projection in Art. 320, Kx. 2. It may however be
proved thus : —
Take the line for axis of x, and for origin the centre of the involution
determined by the two points in which the line cuts any two conies of the
system S x =0 and S 2 =0. Then any other conic is 5 I + XS 2 =0, or
c^x 1 + 1\xy + bjj 2 + 2g x x + 2/0 + c x
+ \ (a 2 .r 2 + 2h;xy + b 2 f + 2g
THE GALLIC WAR. . SCENES FROM BOQKS Y. and VI.
Edited by C. Colbeck, M.A., Assistant- Master at Harrow;'
formerly Fellow of Trinity College; Cambridge 1 .
THE GALLIC WAR. BOOKS V. and VI. (separately). By
the same Editor. Book V. ready. Book VI. in preparation.
THE GALLIC WAR. BOOK VII. Edited by John Bond,
M.A., and A. S. Walpole, M;A.
Cicef d.-HpE SENECTUTE. Edited by E. S. Shuckburgh,
M.A., late Fellow of Emmanuel College, Cambridge.
DEAMICITIA. By the same Editor.
STORIES, OF ROMAN HISTORY. Adapted for the .Use of
Beginners. With Notes, Vocabulary, . and Exercises, by the Rev.
G. £. Jeans, M.A.^ Fellow of Hertford! College, Oxford, and
A^ V. Jones, M.A. ; Assistant-Masters at HaSleyDUry College.'!
Etitfbpius-.— Adapted for the Use of Beginners. With Notes;
Vocabulary, and Exercises, by William Welch; M.A., and C.
Gi DiUEFiELDi M.A., Assistant-Masters at Surrey County School,
, . C.ranleigh. ., . , ,
Homer. — ILIAD. BOOK I. Edited by Rev. John Bond, M.A.,;
and A. S. Walpole, M.A.
^LEME^TARY CLASSICS. 5
Homer.— iLTAp. bookxviii, the armsofachilles.
Edited by S. R. James, M. A., Assistant-Master at E^on College'.
ODYSSEY. BOOK I. Edited by Rev. John Bond, M.A. and
A. S. Walpole. M.A, ,
Horace. — ODES. BOOKS I.— IV. Edited by T. E. Page, M. A.,
late Fellow of St. John's College, Cambridge; Assistant-Master
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DIVINITY, 75
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DIVINITY. 77
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DIVINITY. . 19
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