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Ellipse

JEE Main 2022 (Online) 30th June Morning Shift

MCQ+4 / -12022

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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