Let the foci of a hyperbola H coincide with the foci of the ellipse E : (x - 1)2100 + (y - 1)275 = 1 and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is and the length of its conjugate axis is , then 3 2 + 2 2 is equal to
- 225
- 237
- 242
- 205
Answer
(1) 225
Solution
For the ellipse, a2 = 100, b2 = 75, so eE = 1 - 75100 = 12.
Its foci are (1 10 12, 1) = (6, 1) and (-4, 1), so the distance between the foci = 10.
For H, eH = 2 and the same foci give 2aH eH = 10, so aH = 52.
bH2 = aH2(eH2 - 1) = 254 3 = 754
= 2aH = 5 and = 2bH = 75
3 2 + 2 2 = 75 + 150 = 225
Answer: (1)