Let H : -x2a2 + y2b2 = 1 be the hyperbola, whose eccentricity is 3 and the length of the latus rectum is 4 3. Suppose the point ( , 6), > 0 lies on H. If is the product of the focal distances of the point ( , 6), then 2 + is equal to
- 170
- 171
- 169
- 172
Answer
(2) 171
Solution
Transverse axis along y-axis: e2 = 1 + a2b2 = 3 a2 = 2b2
Latus rectum = 2a2b = 4b = 4 3 b = 3, a2 = 6
( , 6) on H: - 26 + 363 = 1 2 = 66
Foci: (0, be) = (0, 3)
Focal distances: 66 + (6 - 3)2 = 5 3, 66 + (6 + 3)2 = 7 3
= 5 3 7 3 = 105
2 + = 66 + 105 = 171
Answer: (2)