If the foci of a hyperbola are same as that o | vedclass

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Question

$\frac{x^2}{9}+\frac{y^2}{25}=1$
$a=3, b=5 $
$e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} 	herefore 	ext { foci }=(0, \pm b e)=(0, \pm 4)$
$	herefore e_

Vedclass · Accepted Answer

$7 \sqrt{\frac{2}{5}}-\frac{8}{3}$

Vedclass · Answer

$\frac{x^2}{9}+\frac{y^2}{25}=1$
$a=3, b=5 $
$e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} 	herefore 	ext { foci }=(0, \pm b e)=(0, \pm 4)$
$	herefore e_H=\frac{4}{5} 	imes \frac{15}{8}=\frac{3}{2}$

Let equation hyperbola

$\frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1$

$	herefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad 	herefore \mathrm{B}=\frac{8}{3}$

$	herefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1ight)=\frac{64}{9}\left(\frac{9}{4}-1ight) 	herefore \mathrm{A}^2=\frac{80}{9}$

$	herefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1$

Directrix : $y= \pm \frac{B}{e_H}= \pm \frac{16}{9}$

$\mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}ight|$

$=7 \sqrt{\frac{2}{5}}-\frac{8}{3}$

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