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

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(D) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.

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(D) The given 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 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 $(\pm ae, 0) = (\pm \sqrt{16 - b^2}, 0)$.Since the foci coincide,$\sqrt{16 - b^2} = 3$.Squaring both sides,$16 - b^2 = 9$.Therefore,$b^2 = 16 - 9 = 7$.

The foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ coincide. Then the value of $b^2$ is -

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