Let the ellipse E : x² 144 + y² 169 =1 and the hyperbola H : x² 16 - y² ² =-1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24( e + L ) is :

Step-by-step solution

Ellipse E: x^2 144 + y^2 169 = 1 has a^2 = 169, b^2 = 144 , so c_E = 25 = 5 with foci at (0, 5) . Hyperbola H: y^2 x^2 - x^2 16 = 1 has the same foci, so c_H^2 = x^2 + 16 = 25 , giving x^2 = 9 . Thus H is y^2 9 - x^2 16 = 1 with a = 3, b = 4, c = 5 . Eccentricity: e = 5 3 . Latus rectum: L = 2b^2 a = 32 3 . Therefore 24(e + L) = 24 ( 5 3 + 32 3 ) = 24 37 3 = 296