Let the ellipse E : x2144 + y2169 = 1 and the hyperbola H : x216 - y2 2 = -1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24(e + L) is :
- 296
- 126
- 67
- 148
Answer
(1) 296
Solution
Ellipse: 169 > 144, so the major axis is along the y-axis.
Foci of E: (0, 169 - 144) = (0, 5)
Hyperbola: y2 2 - x216 = 1 (transverse axis along the y-axis)
Foci of H: (0, 2 + 16)
2 + 16 = 25 2 = 9
e = 1 + 169 = 53
L = 2 163 = 323 (semi-transverse axis = 3, semi-conjugate axis = 4)
24(e + L) = 24 ( 53 + 323 ) = 24 373 = 296
Answer: (1)