JEE Math Practice Question
Let the eccentricity of the ellipse $${x^2} + {a^2}{y^2} = 25{a^2}$$ be b times the eccentricity of the hyperbola $${x^2} - {a^2}{y^2} = 5$$, where a is the minimum distance between the curves y = ex and y = logex. Then $${a^2} + {1 \over {{b^2}}}$$ is equal to :
- A.$${3 \over 2}$$
- B.$${5 \over 2}$$
- C.3
- D.5
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