For some (0, 2 ), let the eccentricity and the length of the latus rectum of the hyperbola x2 - y2 2 = 8 be e1 and l1, respectively, and let the eccentricity and the length of the latus rectum of the ellipse x2 2 + y2 = 6 be e2 and l2, respectively. If e12 = e22 ( 2 + 1 ), then ( l1 l2e1 e2 ) 2 is equal to .
Answer
8
Solution
Hyperbola: x28 - y28 2 = 1
e12 = 1 + 2 , l1 = 2(8 2 )2 2 = 4 2 2
Ellipse: x26 2 + y26 = 1 (major axis along y-axis)
e22 = 1 - 2 = 2 , l2 = 2(6 2 ) 6 = 2 6 2
e12 = e22( 2 + 1) 1 + 2 = 2 (1 + 2 ) 2
2 = 2 = 4, 2 = 12, 2 = 1
e1 = 32, e2 = 1 2, l1 = 2 2, l2 = 6
l1 l2e1 e2 = 2 2 6 32 = 4 3 32 = 8
( l1 l2e1 e2 ) 2 = 8 1 = 8
Answer: 8