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JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1
If the eccentricity of the hyperbola x2βˆ’y2coces⁑2Ξ±=25x^2-y^2 \operatorname{coces}^2 \alpha=25 is 5\sqrt{5} times the eccentricity of the ellipse x2cosec⁑2Ξ±+y2=5x^2 \operatorname{cosec}^2 \alpha+y^2=5, then Ξ±\alpha is equal to :
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Correct option: A
tanβ‘βˆ’12\tan ^{-1} \sqrt{2}
Option A (tanβ‘βˆ’12\tan^{-1} \sqrt{2}) is correct because by equating the eccentricities of the hyperbola and the ellipse as given in the question, we can solve for Ξ±\alpha and find that Ξ±=tanβ‘βˆ’12\alpha = \tan^{-1} \sqrt{2}.

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