ધારો ક P એ અતિવલય H: frac{x^2}{9}-frac{y^2}{4 | vedclass

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Question

$\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$

 $\mathrm{a}^2=9, \mathrm{~b}^2=4$

 $\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e

Vedclass · Accepted Answer

$22$

Vedclass · Answer

$\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$

 $\mathrm{a}^2=9, \mathrm{~b}^2=4$

 $\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1ight) \Rightarrow \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}$

$\mathrm{e}^2=1+\frac{4}{9}=\frac{13}{9} $

$\mathrm{e}=\frac{\sqrt{13}}{3} \Rightarrow \mathrm{s}_1 \mathrm{~s}_2=2 \mathrm{ae}=2 	imes 3 	imes \sqrt{\frac{13}{3}}=2 \sqrt{13}$

Area of $\Delta \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} 	imes \beta 	imes \mathrm{s}_1 \mathrm{~S}_2=2 \sqrt{13}$

$\Rightarrow \frac{1}{2} 	imes \beta 	imes(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 $

$\frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2}$

Distance of $\mathrm{P}$ from origin $=\sqrt{\alpha^2+\beta^2}$

$=\sqrt{18+4}=\sqrt{22}$

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