266 
HYPERBOLA. 
The equation to the circle, described about the exterior 
focus, is 2 o 
x + y + 2 aex + a = 0. 
Let (A, k) be any point in the hyperbola: then, tangents being- 
drawn from this point to the circle, the equation to the chord 
of contact will be 
(h + ae) x + ky + a (eh + a) = 0 . 
The equation to the hyperbola gives the equation 
(1). 
V = (e 2 - 1) (A 2 - a 2 ) 
(2)- 
The distance of the centre of the hyperbola from (1) is equal 
to 
a (eh + a) 
{(4 + aef + **)» 
a (eh + a) 
{(h + aeY + (e 2 - 1) (h* - a 2 )}* ’ 
= a. 
This result establishes the proposition. 
2. To find the locus of the centre of a circle inscribed in the 
triangle SPII, S and H being the foci and P any point of an 
hyperbola. 
The equation to the hyperbola being 
the required locus is a straight line 
x = a. 
which is therefore a tangent at the vertex. 
Lardner : Algebraic Geometry, p. 128. 
3. To find the locus of the centre of a circle, which, the 
diagram remaining the same as in the preceding problem, 
touches SP, HP produced, and IPS produced.