Let E : x^2 a^2 + y^2 b^2 = 1 ( a > b ) and H : x^2 A^2 - y^2 B^2 = 1 . The length of the latus rectum of E is equal to the length of the latus rectum of H , and both are equal to 6 . If the length of the major axis of E is 4 times the length of the transverse axis of H , and the product of their eccentricities is 1 , then the square of the distance between the foci of E is equal to

Step-by-step solution

Let the eccentricities of E and H be e and e' respectively. Given that the lengths of the latus rectums are equal to 6 : 2b^2 a = 6 b^2 = 3a a^2(1-e^2) = 3a a(1-e^2) = 3 2B^2 A = 6 B^2 = 3A A^2(e'^2-1) = 3A A(e'^2-1) = 3 We are given that the length of the major axis of E is 4 times the length of the transverse axis of H : 2a = 4(2A) a = 4A Substituting a = 4A into the first equation: 4A(1-e^2) = 3 Dividing this by the hyperbola equation A(e'^2-1) = 3 : 4(1-e^2) e'^2-1 = 1 4(1-e^2) = e'^2 - 1 Given that the product