If the foci of a hyperbola are same as that of the ellipse x^2 9+ y^2 25=1 and the eccentricity of the hyperbola is 15 8 times the eccentricity of…

Given equation of ellipse is, $\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1$9x2​+25y2​=1

$$\Rightarrow a=3,b=5$$⇒a=3,b=5

We know that, $e=\sqrt{1-\frac{{a}^{2}}{{b}^{2}}}$e=1−b2a2​​

$$\Rightarrow e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$$⇒e=1−259​​=54​

Now, $\text{foci}=(0,\pm \mathrm{be})$foci=(0,±be)

$$=(0,\pm 4)$$=(0,±4)

Since, eccentricity of hyperbola is given as $\frac{15}{8}$815​ same to that of ellipse, $\therefore {e}_{H}=\frac{4}{5}\times \frac{15}{8}=\frac{3}{2}$∴eH​=54​×815​=23​

Let equation of the hyperbola be $\frac{{x}^{2}}{{A}^{2}}-\frac{{y}^{2}}{{B}^{2}}=-1$A2x2​−B2y2​=−1.

$$\Rightarrow B.{e}_{H}=4$$⇒B.eH​=4

$$\Rightarrow B=\frac{8}{3}$$⇒B=38​

$$\Rightarrow {A}^{2}={B}^{2}({{e}_{H}}^{2}-1)=\frac{64}{9}(\frac{9}{4}-1)$$⇒A2=B2(eH​2−1)=964​(49​−1)

$$\Rightarrow {A}^{2}=\frac{80}{9}$$⇒A2=980​

$$\Rightarrow \frac{{x}^{2}}{\frac{80}{9}}-\frac{{y}^{2}}{\frac{64}{9}}=-1$$⇒980​x2​−964​y2​=−1

Directrix: $y=\pm \frac{B}{{e}_{H}}=\pm \frac{16}{9}$y=±eH​B​=±916​

$$\Rightarrow PS=e\cdot PM$$⇒PS=e⋅PM

$$\Rightarrow PS=\frac{3}{2}|\frac{14}{3}\cdot \sqrt{\frac{2}{5}}-\frac{16}{9}|$$⇒PS=23​∣314​⋅52​​−916​∣

$$\Rightarrow PS=7\sqrt{\frac{2}{5}}-\frac{8}{3}$$⇒PS=752​​−38​