Let x2a2+ y2b2=1, a>b be an ellipse, whose eccentricity is 1 2 and the length of the latus rectum is 14. Then the square of the eccentricity of x2a2- y2b2=1 is :
- 3
- 72
- 32
- 52
Answer
(3) 32
Solution
Ellipse: e2=1- b2a2= 12 b2= a22
Latus rectum = 2b2a= a2a=a= 14 a2=14, b2=7
Hyperbola: e12=1+ b2a2=1+ 714= 32
Answer: (3)