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If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to
$7 \sqrt{\frac{2}{5}}-\frac{8}{3}$
$14 \sqrt{\frac{2}{5}}-\frac{4}{3}$
$14 \sqrt{\frac{2}{5}}-\frac{16}{3}$
$7 \sqrt{\frac{2}{5}}+\frac{8}{3}$
Solution
$\frac{x^2}{9}+\frac{y^2}{25}=1$
$a=3, b=5 $
$e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm b e)=(0, \pm 4)$
$\therefore e_H=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2}$
Let equation hyperbola
$\frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1$
$\therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3}$
$\therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9}$
$\therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1$
Directrix : $y= \pm \frac{B}{e_H}= \pm \frac{16}{9}$
$\mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right|$
$=7 \sqrt{\frac{2}{5}}-\frac{8}{3}$
