602 
PROBLEMS AND SOLUTIONS. 
[485 
Taking the upper signs, the equation is 
{x 2 + y 2 ) [m (x cos /x + y sin /x) — l (x cos A 4- y sin A)] 
— 2Im {(x 2 — y 2 ) sin (A + fi) — xy cos (A + /¿)} = 0, 
which is the locus for equal angles, a circular cubic as in the case of the four distinct 
points. 
3. The question is connected with Question 1819, which is given above. In fact, 
taking A, B for the fixed points on the given conic, and P for the intersection of 
any two of the tangents, if in the conic (foci A, B) which passes through P, the 
tangent or normal at P passes through a fixed point T, then it is clear that at P 
the points A, T and B, T subtend equal angles, and consequently the locus of P 
should be a circular cubic as above. The theorem will therefore be proved if it be 
shown that the locus of P considered as the intersection of tangents from A, B to 
the variable confocal conic is in fact the foregoing circular cubic. I remark that the 
fixed point T is in fact the intersection of the tangents AT, BT to the given conic 
at the points A, B respectively. 
4. Consider the points A, B, (which we may in the first instance take to be 
arbitrary points, but we shall afterwards suppose them to be situate on the conic 
x 2 y 2 
a~ 
G '?/ 
- + = 1,) and from each of them draw a pair of tangents to the confocal conic 
a 2 q. }i + k = r -^ a k e ( x o> 2/o) for the coordinates of A, and (x 1} y L ) for those of P; 
then the equation of the pair of tangents from A is 
a 2 + h b' 2 + h 
+ 
y 
a 2 + h b 2 + h 
- 1 
XX n 
+ 
yy o 
a? + h b 2 + h 
-1=0, 
or, what is the same thing, 
(xy 0 - x 0 y) 2 _ (x - x o y _ (y - y o y = 
(a 2 + h) (ft 2 + h) a 2 + h b 2 + h ’ 
that is 
(xy 0 - x 0 y) 2 - (b 2 + h)(x — x 0 ) 2 - (a 2 + h)(y- y 0 ) 2 = 0, 
or as this may also be written 
(xy 0 - x,yf -b 2 {x- x 0 ) 2 — a 2 (y — y 0 ) 2 = h [(x - x,) 2 + (y- y 0 ) 2 J 5 
and similarly for the tangents from B we have 
{xy x - x x yf - b 2 (x - x L ) 2 - a 2 (y - y x f = h [(x - x x ) 2 + (y- y x ) 2 ] ; 
in which equations the points (x 0} y 0 ), (x 1} yQ are in fact any two points whatever. 
5. Eliminating h, we have as the locus of the intersection of the tangents 
{xy, - x 0 y) 2 -b 2 (x- x 0 ) 2 - a 2 (y - y 0 ) 2 _ (xy x - x x y) 2 - b 2 (x - x x ) 2 - a 2 (y - y x ) 2 
(x - x 0 ) 2 + {y- y 0 f (x - x,Y + (y - y x f