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For some  \theta \epsilon \left ( 0, \frac{\pi}{2} \right ) if the eccentricity of the hyperbola x^{2}-y^{2}\sec ^{2}\theta =10 is \sqrt{5} times the eccentricity of the ellipse , x^{2}\sec^{2}\theta +y^{2}=5, then the length of the latus rectum of the ellipse is 
Option: 1 \sqrt{30}
Option: 2 \frac{4\sqrt{5}}{3}
Option: 3 \frac{2\sqrt{5}}{7}
Option: 4 \frac{2\sqrt{5}}{6}

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\\\text{equation of hyperbola}\\x^{2}-\frac{y^{2}}{\cos ^{2} \theta}=10 \\\text{equation of ellipse}\\\frac{x^{2}}{\cos ^{2} \theta}+y^{2}=5

\\e_{h}=\sqrt{1+\frac{\cos ^{2} \theta}{1}},\;\;\;e_{e}=\sqrt{1-\frac{\cos ^{2} \theta}{1}}\\1+\cos ^{2} \theta=5 \sin ^{2} \theta\\ 2=6 \sin ^{2} \theta\\\sin^2\theta=1/3\\\text{LR of ellipse}=\frac{2 \times \frac{2}{3} \times \cdot 5}{\sqrt{5}}=\frac{4}{3}\sqrt5

 

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Suraj Bhandari

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