Let E: x^2 a^2 + y^2 b^2 = 1 ( a > b ) be an ellipse such that the distance between its foci is 4 2 and the length of its latus rectum is 4 . If H is the hyperbola x^2 a^2 - y^2 b^2 = 1 , then the square of the distance between the foci of H is:

Step-by-step solution

For the ellipse E , the distance between the foci is 2ae = 4 2 . Squaring both sides, we get 4a^2e^2 = 32 a^2e^2 = 8 . Since b^2 = a^2(1 - e^2) = a^2 - a^2e^2 , we have a^2 - b^2 = 8 . The length of the latus rectum is given by 2b^2 a = 4 b^2 = 2a . Substituting b^2 = 2a into the first equation: a^2 - 2a - 8 = 0 (a - 4)(a + 2) = 0 Since a > 0 , we get a = 4 . Thus, a^2 = 16 and b^2 = 2(4) = 8 . For the hyperbola H: x^2 a^2 - y^2 b^2 = 1 , the distance between the foci is 2ae_H , where e_H is its eccentricity. The s