Let the product of the focal distances of the point P(4, 2 3) on the hyperbola H: x2a2 - y2b2 = 1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p2 + q2 is equal to
Answer
120
Solution
Since P lies on H, 16a2 - 12b2 = 1 ...(i)
The focal distances are |ex1 - a|, |ex1 + a| with x1 = 4, which gives 16e2 - a2 = 32.
e2 = 1 + b2a2, this becomes 16 + 16b2a2 - a2 = 32 ...(ii)
From (i), b2 = 12a216 - a2, and putting this in (ii):
19216 - a2 = 16 + a2
256 - a4 = 192 a4 = 64 a2 = 8
b2 = 968 = 12
Now p = 2b, so p2 = 4b2 = 48, and q = 2b2a = 242 2 = 6 2, so q2 = 72.
p2 + q2 = 48 + 72 = 120
Answer: 120