JEE MainMathematicsHyperbola
Let the hyperbola H : x^2 16 - y^2 9 = 1 and the ellipse E : x^2 a^2 + y^2 b^2 = 1 ( a > b ) be such that the length of the latus rectum of E is equal to the length of the latus rectum of H . If the eccentricity of E is 2 5 times the eccentricity of H , then the value of 4(a^2 + b^2) is equal to ____.
Correct answer
63
Step-by-step solution
For the hyperbola H : x^2 16 - y^2 9 = 1 , we have a_H^2 = 16 and b_H^2 = 9 . The eccentricity of H is e_H = 1 + b_H^2 a_H^2 = 1 + 9 16 = 5 4 . The length of the latus rectum of H is 2b_H^2 a_H = 2(9) 4 = 9 2 . Given that the eccentricity of E is e_E = 2 5 e_H = 2 5 5 4 = 1 2 . For the ellipse E , e_E^2 = 1 - b^2 a^2 1 4 = 1 - b^2 a^2 b^2 = 3a^2 4 . The length of the latus rectum of E is 2b^2 a = 2 a ( 3a^2 4 ) = 3a 2 . Since the latus rectums are equal, 3a 2 = 9 2 a = 3 . Then a^2 = 9 and b^2 = 3(9) 4 = 27 4 . The