Let e_1 and e_2 be the eccentricities of the | vedclass

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(B) For the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$,since $b < 5$,the major axis is along the $y$-axis. Thus,$e_1^2 = 1 - \frac{b^2}{25}$.For t

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$\frac{3}{5}$

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(B) For the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$,since $b For the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$,$e_2^2 = 1 + \frac{b^2}{16}$.Given $e_1 e_2 = 1$,so $e_1^2 e_2^2 = 1$.$(1 - \frac{b^2}{25})(1 + \frac{b^2}{16}) = 1$.$1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1$.$\frac{9b^2}{400} = \frac{b^4}{400} \Rightarrow b^2 = 9$.The foci of the ellipse are $(0, \pm ae_1) = (0, \pm \sqrt{25-9}) = (0, \pm 4)$.The foci of the hyperbola are $(\pm ae_2, 0) = (\pm \sqrt{16+9}, 0) = (\pm 5, 0)$.The ellipse passing through $(\pm 5, 0)$ and $(0, \pm 4)$ is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Its eccentricity $e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

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