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(B) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$,we have $a^{2}=25$ and $b^{2}=16$.The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \

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

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(B) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$,we have $a^{2}=25$ and $b^{2}=16$.The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The foci of the ellipse are $(\pm ae_{1}, 0) = (\pm 5 	imes \frac{3}{5}, 0) = (\pm 3, 0)$.Given that the product of the eccentricities of the ellipse and the hyperbola is $1$,we have $e_{1} 	imes e_{2} = 1$ $\Rightarrow \frac{3}{5} 	imes e_{2} = 1$ $\Rightarrow e_{2} = \frac{5}{3}$.Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Since the hyperbola passes through the foci of the ellipse $(\pm 3, 0)$,we have $\frac{3^{2}}{a^{2}} = 1 \Rightarrow a^{2} = 9$.For a hyperbola,$b^{2} = a^{2}(e_{2}^{2}-1) = 9((\frac{5}{3})^{2}-1) = 9(\frac{25}{9}-1) = 9(\frac{16}{9}) = 16$.Thus,the equation of the hyperbola is $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

$A$ hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse,respectively. If the product of their eccentricities is $1$,then the equation of the hyperbola is ...... .

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