For the hyperbola H : x 2 - y 2 = 1 and the ellipse E : x 2 a 2 + y 2 b 2 = 1 , a > b > 0 , let the (1) eccentricity of E be reciprocal of the eccentricity of H , and (2) the line y = 5 2 x + K be a common tangent of E and H . Then 4 a 2 + b 2 is equal to

Step-by-step solution

Given H : x 2 - y 2 = 1 , E : x 2 a 2 + y 2 b 2 = 1 e H = 2     &   e E = 1 e H = 1 2 For hyperbola e 2 = 1 - b 2 a 2 = 1 2 ⇒ b 2 a 2 = 1 2 Also given that the common tangent of H   &   E is y = 5 2 x + k i . e . m = 5 2 We know that the condition for common tangent of ellipse x 2 a 2 + y 2 b 2 = 1 and hyperbola x 2 A 2 - y 2 B 2 = 1 is a 2 m 2 + b 2 = A 2 m 2 - B 2 Now, for common tangency: 5 2 a 2 + b 2 = 5 2 - 1 5 2 + b 2 a 2 = 3 2 a 2 ⇒ a 2 = 1 2 ∴   &#160