Consider a hyperbola mathrm{H} having centre | vedclass

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Question

$ 	ext { Let } \mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad\left(\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{

Vedclass · Accepted Answer

$\frac{28}{3}$

Vedclass · Answer

$ 	ext { Let } \mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad\left(\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1ight)ight) $

$ 	herefore \mathrm{eq}^{\mathrm{n}} 	ext { of } \mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2 $

$ 	ext { Ar. }=36 \pi $

$ \pi \mathrm{a}^2=36 \pi $

$ \mathrm{a}=6 $

$ 	ext { Now radius of } \mathrm{C}_2 	ext { can be } \mathrm{a}(\mathrm{e}-1) 	ext { or } \mathrm{a}(\mathrm{e}+1) $

$ 	ext { for } \mathrm{r}=\mathrm{a}(\mathrm{e}-1) \quad 	ext { for } \mathrm{r}=\mathrm{a}(\mathrm{e}+1) $

$ 	ext { Ar. }=4 \pi \quad \pi \mathrm{r}^2=4 \pi $

$ \pi \mathrm{a}^2(\mathrm{e}-1)^2=4 \pi \quad \mathrm{a}^2(\mathrm{e}+1)^2=4 $

$ 36 \pi(\mathrm{e}-1)^2=4 \pi \quad 36(\mathrm{e}+1)^2=4 $

$ \mathrm{e}-1=\frac{1}{3} \quad \mathrm{e}+1=\frac{1}{3} $

$ \mathrm{e}=\frac{4}{3} \quad-\frac{2}{3} $

$ 	ext { Not possible } $

$ 	herefore \mathrm{b}^2=36\left(\frac{16}{9}-1ight)=28 $

$ 	herefore \mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 	imes 28}{6}=\frac{28}{3} $

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