Let an ellipse E: x^2 a^2 + y^2 b^2 = 1 (a > b) and a hyperbola H: x^2 A^2 - y^2 B^2 = 1 be confocal. The length of the minor axis of E is 4 and the length of its latus rectum is 2 . If e₁ and e₂ are the eccentricities of H and E respectively, and e₁ = 2e₂ , then the perpendicular distance between the tangents to H which are parallel to the line y = 2x is :

Step-by-step solution

For the ellipse E , the length of the minor axis is 2b = 4 b = 2 . The length of the latus rectum is 2b^2 a = 2 8 a = 2 a = 4 . The eccentricity of E is e₂ = 1 - b^2 a^2 = 1 - 4 16 = 3 2 . The foci of E are ( ae₂, 0) = ( 4 3 2 , 0 ) = ( 2 3 , 0) . Given e₁ = 2e₂ , the eccentricity of the hyperbola H is e₁ = 2 ( 3 2 ) = 3 . Since E and H are confocal, the foci of H are also ( 2 3 , 0) . For H , the foci are ( Ae₁, 0) , so Ae₁ = 2 3 . A( 3 ) = 2 3 A = 2 . For the hyperbola, B^2 = A^2(e₁^2 - 1) = 4(3 - 1) = 8 . The eq