Let l and l' be the lengths of the latus rectums of a hyperbola H and its conjugate hyperbola H' respectively. Let e and e' be their respective eccentricities. If l l' = 48 , e^2 + (e')^2 = 625 144 , and e < e' , then the area of the rectangle formed by the four extremities of the latus rectums of H is

Step-by-step solution

Let the hyperbola H be x^2 a^2 - y^2 b^2 = 1 . The lengths of the latus rectums are l = 2b^2 a and l' = 2a^2 b . Given l l' = 48 , we have ( 2b^2 a ) ( 2a^2 b ) = 4ab = 48 ab = 12 . The eccentricities are e^2 = 1 + b^2 a^2 and (e')^2 = 1 + a^2 b^2 . Given e^2 + (e')^2 = 625 144 , we get: 2 + b^2 a^2 + a^2 b^2 = 625 144 2 + a^4 + b^4 a^2b^2 = 625 144 Since ab = 12 , a^2b^2 = 144 . Substituting this gives: 2 + a^4 + b^4 144 = 625 144 a^4 + b^4 144 = 337 144 a^4 + b^4 = 337 . We know (a^2 + b^2)^2 = a^4 + b^4 + 2a^2b^