Let E: x^2 a^2 + y^2 b^2 = 1 ( a > b ) be an ellipse and H: x^2 a^2 - y^2 b^2 = 1 be a hyperbola. If the eccentricity of H is 2 times the eccentricity of E , and the length of the latus rectum of E is 6 , then the value of a^2 + b^2 is:

Step-by-step solution

Let e_E and e_H be the eccentricities of the ellipse and the hyperbola, respectively. Given that e_H = 2 e_E , we square both sides to get: e_H^2 = 2e_E^2 Using the standard formulas for eccentricity: 1 + b^2 a^2 = 2 (1 - b^2 a^2 ) 1 + b^2 a^2 = 2 - 2b^2 a^2 3b^2 a^2 = 1 a^2 = 3b^2 The length of the latus rectum of the ellipse E is given as 6 : 2b^2 a = 6 b^2 = 3a Substituting b^2 = 3a into the relation a^2 = 3b^2 : a^2 = 3(3a) = 9a Since a > 0 , we have a = 9 . Thus, a^2 = 81 and b^2 = 3(9) = 27 . The value of a^2