JEE MainMathematicsHyperbola
Let a hyperbola share its foci with the ellipse x^2 25 + y^2 16 = 1 . If the length of the latus rectum of the hyperbola is equal to the length of the latus rectum of the given ellipse, then the eccentricity of the hyperbola is:
Options
- A5 3
- B3 5
- C5 4
- D4 5
Correct answer
A. 5 3
Step-by-step solution
For the given ellipse x^2 25 + y^2 16 = 1 , we have a_E^2 = 25 and b_E^2 = 16 . The eccentricity of the ellipse is e_E = 1 - b_E^2 a_E^2 = 1 - 16 25 = 3 5 . The foci of the ellipse are at ( a_E e_E, 0) = ( 3, 0) . The length of the latus rectum of the ellipse is 2b_E^2 a_E = 2(16) 5 = 32 5 . Let the hyperbola have parameters a_H and b_H , and eccentricity e_H . Since it shares the foci with the ellipse, its foci are also at ( 3, 0) . Thus, a_H e_H = 3 . For the hyperbola, b_H^2 = a_H^2(e_H^2 - 1) = (a_H e_H)^2 - a_