Let H_1: frac{x^2}{a^2}-frac{y^2}{b^2}=1 and | vedclass

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(D) For $H_1: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,the length of latus rectum is $\frac{2b^2}{a} = 15\sqrt{2}$ and $e_1^2 = 1 + \frac{b^2}{a^2} = \frac{

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$55$

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(D) For $H_1: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,the length of latus rectum is $\frac{2b^2}{a} = 15\sqrt{2}$ and $e_1^2 = 1 + \frac{b^2}{a^2} = \frac{5}{2}$.From $e_1^2 = 1 + \frac{b^2}{a^2} = \frac{5}{2}$,we get $\frac{b^2}{a^2} = \frac{3}{2}$,so $b^2 = \frac{3}{2}a^2$.Substituting into the latus rectum equation: $\frac{2(\frac{3}{2}a^2)}{a} = 15\sqrt{2} \implies 3a = 15\sqrt{2} \implies a = 5\sqrt{2}$.Then $b^2 = \frac{3}{2}(50) = 75$,so $b = 5\sqrt{3}$.The transverse axis length of $H_1$ is $2a = 10\sqrt{2}$.For $H_2: \frac{y^2}{B^2}-\frac{x^2}{A^2}=1$,the length of latus rectum is $\frac{2A^2}{B} = 12\sqrt{5}$.The product of transverse axes is $(2a)(2B) = 100\sqrt{10}$.$(10\sqrt{2})(2B) = 100\sqrt{10} \implies 20\sqrt{2}B = 100\sqrt{10} \implies B = 5\sqrt{5}$.Using $\frac{2A^2}{B} = 12\sqrt{5}$,we get $\frac{2A^2}{5\sqrt{5}} = 12\sqrt{5} \implies 2A^2 = 60(5) = 300 \implies A^2 = 150$.For $H_2$,$e_2^2 = 1 + \frac{A^2}{B^2} = 1 + \frac{150}{125} = 1 + \frac{6}{5} = \frac{11}{5}$.Therefore,$25e_2^2 = 25 	imes \frac{11}{5} = 55$.

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