604 
PROBLEMS AND SOLUTIONS. 
[485 
8. Supposing that the points {x 0 , y 0 ), {x x , yft are on the conic + y, — 1> and that 
we have consequently ^ ^ = 1, = 1, the equations of the tangents at these 
0/ 0 CL" 1 0 
points respectively are æ ~° + = 1 
xo^ yy x 
a? b 2 
1 ; and hence, writing for shortness 
a = 2/o-2d, (3 = x 1 — x 0 , y = x 0 y x - x x y 0 , we find x = 
a 2 a 5 2 /3 
y = — as the coordinates of 
7 
the point of intersection T, of the two tangents; and in order to transform to this 
point as origin, we must in place of x, y write x — —, y — —- respectively. Or what 
is more convenient, we may in the equation at the end of (6), in which it is to be 
OL ¡3 
now assumed that x 0 2 + y 0 2 = 1, x x 2 + y x 2 = 1, write x—, y for x, y, and then restore 
y u y u 
the original coordinates by writing j-, —, &c., for x, y, x, 
CL 0 CL 
a, ft, y, these quantities throughout signifying OL = y 0 — y x , ft = x 1 —x 0 , y = x 0 y x — x x y 0 . 
I however obtained the equation referred to the point T as origin by a different 
process, as follows: 
9. Starting from the equation at the commencement of (5), I found that the 
points (x n , y 0 ), {x x , y x ) being on the conic %+r^ = 1, the equation could be transformed 
CL D - 
into the form 
&c., and y, — , \ for 
b a ab 
, 2/2/o _ 
a 2 + b 2 
, VVi _ i 
a 2 + b 2 
(x - oc 0 ) 2 + (y- y 0 ) 2 (x - x x ) 2 + (y- y x ) 2 ’ 
an equation which (not, as the original one, for all values of {x 0 , y 0 ), (x 1} yft, but) for 
X 2 2 X ^ /^/2 
values of {x„ Vo), (®i, yd such that -^- + ^=1, -^ + ^ = 1, breaks up into the line AB 
CL 0" CL v 
and a cubic curve. 
10. To simplify the transformation, write as before ax, by, ax 0 , &c., for x, y, x 0 , &c. 
We have thus to consider the equation 
a 2 (x - x 0 ) 2 + b 2 (y - yft 2 _ a 2 {x - x x f + b 2 (y - y x ) 2 
(xx 0 + yy 0 - l) 2 (xx x + yy x - l) 2 
where x 0 2 + y 0 2 = 1, x x 2 + y x 2 = 1, and which equation, I say, breaks up into the line L = 0, 
and into a cubic. 
Write for shortness a = y 0 -y x , ft = x x -x 0 , y = x 0 y x -x x y 0 , so that the equation of 
the last-mentioned line is ax + fty + y = 0. Then it may be verified that, in virtue of 
the relations between (x 0) y 0 ), {x x , y x ), we have identically 
(x - x 0 ) (xx x + yy x -1) + (x - x x ) {xx 0 + yy 0 - 1) = {ax + fty + 7) X ~- + Xl {yx + a), 
ay 
{x - x 0 ) {xx x + yy x -\)-{x- X x ) {xXq + yy 0 - 1 ) = ftx 2 -axy-yy- ft;