# 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 Option: 2 Option: 3 Option: 4

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

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

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