If e_1 is the eccentricity of the ellipse fra | vedclass

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(B) For the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$,we have $a^2=16$ and $b^2=25$. Since $b^2 > a^2$,the eccentricity $e_1$ is given by $e_1 = \sqrt

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$\frac{y^2}{9}-\frac{x^2}{16}=1$

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(B) For the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$,we have $a^2=16$ and $b^2=25$. Since $b^2 > a^2$,the eccentricity $e_1$ is given by $e_1 = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.Given $e_1 e_2 = 1$,we have $\frac{3}{5} e_2 = 1$,which implies $e_2 = \frac{5}{3}$.The foci of the ellipse are $(0, \pm be_1) = (0, \pm 5 	imes \frac{3}{5}) = (0, \pm 3)$.The hyperbola passes through $(0, \pm 3)$,so its transverse axis is along the $y$-axis. The equation is of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.Since it passes through $(0, 3)$,we have $b^2 = 3^2 = 9$.For a hyperbola,$e_2^2 = 1 + \frac{a^2}{b^2}$. Substituting $e_2 = \frac{5}{3}$ and $b^2 = 9$:$(\frac{5}{3})^2 = 1 + \frac{a^2}{9}$ $\Rightarrow \frac{25}{9} = 1 + \frac{a^2}{9}$ $\Rightarrow \frac{16}{9} = \frac{a^2}{9}$ $\Rightarrow a^2 = 16$.Thus,the equation of the hyperbola is $\frac{y^2}{9} - \frac{x^2}{16} = 1$.

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of a hyperbola passing through the foci of the given ellipse and $e_1 e_2=1$,then the equation of such a hyperbola among the following is

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