If the foci of a hyperbola are same as that of the ellipse x ^ 2 9...

Given equation of ellipse is, $\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{25} = 1$

$⇒ a = 3, b = 5$

We know that, $e = \sqrt{1 - \dfrac{{a}^{2}}{{b}^{2}}}$

$⇒ e = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$

Now, $\text{foci} = \left(0, \pm be\right)$

$= \left(0, \pm 4\right)$

Since, eccentricity of hyperbola is given as $\dfrac{15}{8}$ same to that of ellipse, $∴ {e}_{H} = \dfrac{4}{5} \times \dfrac{15}{8} = \dfrac{3}{2}$

Let equation of the hyperbola be $\dfrac{{x}^{2}}{{A}^{2}} - \dfrac{{y}^{2}}{{B}^{2}} = - 1$.

$⇒ B . {e}_{H} = 4$

$⇒ B = \dfrac{8}{3}$

$⇒ {A}^{2} = {B}^{2}\left({{e}_{H}}^{2} - 1\right) = \dfrac{64}{9}\left(\dfrac{9}{4} - 1\right)$

$⇒ {A}^{2} = \dfrac{80}{9}$

$⇒ \dfrac{{x}^{2}}{\dfrac{80}{9}} - \dfrac{{y}^{2}}{\dfrac{64}{9}} = - 1$

Directrix: $y = \pm \dfrac{B}{{e}_{H}} = \pm \dfrac{16}{9}$

$⇒ PS = e \cdot PM$

$⇒ PS = \dfrac{3}{2}\left|\dfrac{14}{3} \cdot \sqrt{\dfrac{2}{5}} - \dfrac{16}{9}\right|$

$⇒ PS = 7\sqrt{\dfrac{2}{5}} - \dfrac{8}{3}$