The foci of a hyperbola coincide with the foc | vedclass

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(B) For the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$,we have $a^2 = 25$ and $b^2 = 9$.The eccentricity of the ellipse is $e_e = \sqrt{1 - \frac{b^2}{a

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$\frac{x^2}{4}-\frac{y^2}{12}=1$

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(B) For the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$,we have $a^2 = 25$ and $b^2 = 9$.The eccentricity of the ellipse is $e_e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.The foci of the ellipse are $(\pm ae_e, 0) = (\pm 5 	imes \frac{4}{5}, 0) = (\pm 4, 0)$.For the hyperbola,the foci are $(\pm ae_h, 0) = (\pm 4, 0)$,so $ae_h = 4$.Given the eccentricity of the hyperbola $e_h = 2$,we have $a(2) = 4$,which implies $a = 2$.For a hyperbola,$b^2 = a^2(e_h^2 - 1) = 2^2(2^2 - 1) = 4(4 - 1) = 4(3) = 12$.Thus,the equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. The equation of the hyperbola with eccentricity $e = 2$ is

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