Let the foci of a hyperbola coincide with the foci of the ellipse x236 + y216 = 1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :
- 96 5
- 24 5
- 12
- 16
Answer
(1) 96 5
Solution
For the ellipse, a2 = 36 and b2 = 16, so the focal distance is ae = a2 - b2 = 20 = 2 5 and the foci are ( 2 5, 0).
The hyperbola x2a12 - y2b12 = 1 has eccentricity e1 = 5 and a1e1 = 2 5, so
a1 = 2 55 = 2 5, a12 = 45
b12 = a12(e12 - 1) = 45 24 = 965
Length of latus rectum = 2b12a1 = 2 965 2 5
= 96 55 = 96 5
Answer: (1)