t{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm 4) \\ & \quad \therefore \mathrm{e}_{\mathrm{H}}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2} \end{aligned}$$

Let equation hyperbola

$$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1 \\ & \therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3} \\ & \therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9} \\ & \therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1 \\ & \text { Directrix }: \mathrm{y}= \pm \frac{\mathrm{B}}{\mathrm{e}_{\mathrm{H}}}= \pm \frac{16}{9} \\ & \mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right| \\ & =7 \sqrt{\frac{2}{5}}-\frac{8}{3} \end{aligned}$$

19:T473,

$$\begin{aligned} & \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \\ & \mathrm{a}=3, \mathrm{~b}=5 \\ & \mathrm{e}=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { नाभियाँ }=(0, \pm \mathrm{be})=(0, \pm 4) \\ & \quad \therefore \mathrm{e}_{\mathrm{H}}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2} \end{aligned}$$

माना अतिपरवलय का समीकरण

$$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1 \\ & \therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3} \\ & \therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9} \\ & \therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1 \\ & \text { नियता }: \mathrm{y}= \pm \frac{\mathrm{B}}{\mathrm{e}_{\mathrm{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}\right| \\ & =7 \

Hyperbola

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