If the foci of a hyperbola are the same as th | vedclass

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(A) For the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$,we have $a^2=9$ and $b^2=25$. Since $b > a$,the foci are on the $y$-axis.Eccentricity $e = \sqrt{

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$7 \sqrt{\frac{2}{5}}-\frac{8}{3}$

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(A) For the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$,we have $a^2=9$ and $b^2=25$. Since $b > a$,the foci are on the $y$-axis.Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.The foci are $(0, \pm be) = (0, \pm 5 	imes \frac{4}{5}) = (0, \pm 4)$.For the hyperbola,the eccentricity $e_H = \frac{15}{8} 	imes e = \frac{15}{8} 	imes \frac{4}{5} = \frac{3}{2}$.Since the foci are $(0, \pm 4)$,the hyperbola is of the form $\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$.Here,$Be_H = 4 \implies B 	imes \frac{3}{2} = 4 \implies B = \frac{8}{3}$.Also,$A^2 = B^2(e_H^2 - 1) = \frac{64}{9} 	imes (\frac{9}{4} - 1) = \frac{64}{9} 	imes \frac{5}{4} = \frac{80}{9}$.The hyperbola equation is $\frac{y^2}{64/9} - \frac{x^2}{80/9} = 1$.The focal distance of a point $P(x, y)$ is $|ey \pm B|$.For $P\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}ight)$,the focal distances are $e_H y \pm B = \frac{3}{2} 	imes \frac{14}{3} \sqrt{\frac{2}{5}} \pm \frac{8}{3} = 7 \sqrt{\frac{2}{5}} \pm \frac{8}{3}$.The smaller focal distance is $7 \sqrt{\frac{2}{5}} - \frac{8}{3}$.

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