Consider the hyperbola x2a2 - y2b2 = 1 having one of its foci at P(-3, 0). If the latus rectum through its other focus subtends a right angle at P and a2 b2 = 2 - , , N, then + is .
Answer
1944
Solution
Foci ( ae, 0) with ae = 3; other focus S(3, 0).
Ends of latus rectum through S: L (3, b2a ), L' (3, - b2a )
LPL' = 90 and PS bisects it LPS = 45
b2/aPS = 45 b2a = 6 b2 = 6a
Also a2 + b2 = a2e2 = 9 a2 + 6a - 9 = 0
a = -3 + 3 2 (a > 0)
a2 = 27 - 18 2, b2 = 6a = 18 2 - 18
a2 b2 = (27 - 18 2)(18 2 - 18)
= 486 2 - 486 - 648 + 324 2
= 810 2 - 1134
= 810, = 1134 + = 1944
Answer: 1944