The equation of the hyperbola whose foci are | vedclass

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

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

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(B) For the given ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$,we have $a^2 = 25$ and $b^2 = 9$.The eccentricity $e_e$ of the ellipse is given by $b^2 = a^2(1 - e_e^2)$,so $9 = 25(1 - e_e^2)$,which gives $1 - e_e^2 = \frac{9}{25}$,so $e_e^2 = \frac{16}{25}$,hence $e_e = \frac{4}{5}$.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 $(\pm 4, 0)$,so $ae = 4$. Given the eccentricity $e = 2$,we have $a(2) = 4$,which implies $a = 2$.For a hyperbola,$b^2 = a^2(e^2 - 1) = 2^2(2^2 - 1) = 4(3) = 12$.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 equation of the hyperbola whose foci are the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and the eccentricity is $2$,is

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