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

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(D) For the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,we rewrite it as $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.Here,$a^2 = \fra

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$7$

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(D) For the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,we rewrite it as $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.The eccentricity $e_h$ is given by $e_h^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = \frac{225}{144}$.So,$e_h = \frac{15}{12} = \frac{5}{4}$.The foci of the hyperbola are $(\pm a_h e_h, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$.For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae, 0)$ where $a^2 = 16$.Since the foci are the same,$ae = 3$,so $e = \frac{3}{4}$.For an ellipse,$e^2 = 1 - \frac{b^2}{a^2}$,so $(\frac{3}{4})^2 = 1 - \frac{b^2}{16}$.$\frac{9}{16} = 1 - \frac{b^2}{16} \implies \frac{b^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}$.Therefore,$b^2 = 7$.

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ are the same,then the value of $b^2$ is:

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