કોઈ theta inleft(0, frac{pi}{2}right) માટે, જ | vedclass

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Question

Given $	heta \in\left(0, \frac{\pi}{2}ight)$

equation of hyperbola $\Rightarrow x^{2}-y^{2} \sec ^{2} 	heta=10$

$\Rightarrow \frac{x^{2}}{10

Vedclass · Accepted Answer

$\frac{4 \sqrt{5}}{3}$

Vedclass · Answer

Given $	heta \in\left(0, \frac{\pi}{2}ight)$

equation of hyperbola $\Rightarrow x^{2}-y^{2} \sec ^{2} 	heta=10$

$\Rightarrow \frac{x^{2}}{10}-\frac{y^{2}}{10 \cos ^{2} 	heta}=1$

Hence eccentricity of hyperbola

$\left(\mathrm{e}_{\mathrm{H}}ight)=\sqrt{1+\frac{10 \cos ^{2} 	heta}{10}} \ldots$.

$\left\{e=\sqrt{1+\frac{b^{2}}{a^{2}}}ight\}$

Now equation of ellipse $\Rightarrow x^{2} \sec ^{2} 	heta+y^{2}=5$

$\Rightarrow \frac{\mathrm{x}^{2}}{5 \cos ^{2} 	heta}+\frac{\mathrm{y}^{2}}{5}=1 \quad\left\{\mathrm{e}=\sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{b}^{2}}}ight\}$

Hence eccenticity of ellipse

$\left(e_{E}ight)=\sqrt{1-\frac{5 \cos ^{2} 	heta}{5}}$

$\left(e_{E}ight)=\sqrt{1-\cos ^{2} 	heta}=\sin 	heta=\sin 	heta \quad \cdots(2)$

$\left\{\because 	heta \in\left(0, \frac{\pi}{2}ight)ight\}$

given $\Rightarrow \mathrm{e}_{\mathrm{H}}=\sqrt{5} \mathrm{e}_{\mathrm{e}}$

Hence $1+\cos ^{2} 	heta=5 \sin ^{2} 	heta$

$1+\cos ^{2} 	heta=5\left(1-\cos ^{2} 	hetaight)$

$1+\cos ^{2} 	heta=5-5 \cos ^{2} 	heta$

$6 \cos ^{2} 	heta=4$

$\cos ^{2} 	heta=\frac{2}{3}$

Now length of latus rectum of ellipse

$=\frac{2 \mathrm{a}^{2}}{\mathrm{b}}=\frac{10 \cos ^{2} 	heta}{\sqrt{5}}=\frac{20}{3 \sqrt{5}}=\frac{4 \sqrt{5}}{3}$

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