Question Id : 13015 |
Context : JEE Main 2022 (30 June Morning Shift)
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 :
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