If the foci of a hyperbola are same as that of the ellipse [(x^2)/(9)+(y^2)/(25)=1] and th…

Question

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 $$\frac{15}{8}$$

Accepted Answer

$\begin{gathered} \frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1 \ a=3,b=5 \ e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}	herefore =(0,\pm be)=(0,\pm 4) \ \, 	herefore {e}_{H}=\frac{4}{5}	imes \frac{15}{8}=\frac{3}{2} \end{gathered}$

Let equation hyperbola

$\begin{gathered} \frac{{x}^{2}}{{A}^{2}}-\frac{{y}^{2}}{{B}^{2}}=-1 \ 	herefore B\cdot {e}_{H}=4\, 	herefore B=\frac{8}{3} \ 	herefore {A}^{2}={B}^{2}({e}_{2}^{H}-1)=\frac{64}{9}(\frac{9}{4}-1)	herefore {A}^{2}=\frac{80}{9} \ 	herefore \frac{{x}^{2}}{\frac{80}{9}}-\frac{{y}^{2}}{\frac{64}{9}}=-1 \ :y=\pm \frac{B}{{e}_{H}}=\pm \frac{16}{9} \ PS=e\cdot PM=\frac{3}{2}|\frac{14}{3}\cdot \sqrt{\frac{2}{5}}-\frac{16}{9}| \ =7\sqrt{\frac{2}{5}}-\frac{8}{3} \end{gathered}$

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