Let each of the two ellipses E1: x2a2 + y2b2 = 1, (a > b) and E2: x2A2 + y2B2 = 1, (A < B) have eccentricity 45. Let the lengths of the latus recta of E1 and E2 be l1 and l2, respectively, such that 2l12 = 9l2. If the distance between the foci of E1 is 8, then the distance between the foci of E2 is
- 965
- 85
- 165
- 325
Answer
(4) 325
Solution
For E1 (major axis along x-axis): 2ae = 8 2a ( 45 ) = 8 a = 5
b2 = a2(1 - e2) = 25 (1 - 1625 ) = 9
l1 = 2b2a = 185
2l12 = 9l2 2 32425 = 9l2 l2 = 7225
For E2 (A < B, major axis along y-axis): A2 = B2(1 - e2) = 925B2
l2 = 2A2B = 18B25 = 7225 B = 4
distance between the foci of E2 = 2Be = 2 4 45 = 325
Answer: (4)