If the foci of the ellipse frac{x^2}{16} + fr | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$. Here,$a^2 = 16$,so $a = 4$. The foci of the ellipse are $(\pm ae, 0)$,where

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$7$

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$. Here,$a^2 = 16$,so $a = 4$. The foci of the ellipse are $(\pm ae, 0)$,where $e = \sqrt{1 - \frac{b^2}{16}}$. Thus,the foci are $(\pm 4 \sqrt{1 - \frac{b^2}{16}}) = (\pm \sqrt{16 - b^2}, 0)$. The equation of the hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$. Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$. The eccentricity $e_h$ of the hyperbola is $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81/25}{144/25}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$. The foci of the hyperbola are $(\pm a_h e_h, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$. Since the foci coincide,$\sqrt{16 - b^2} = 3$. Squaring both sides,$16 - b^2 = 9$,which gives $b^2 = 7$.

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide with the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2$ is equal to

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