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

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(C) Given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$. Here,$a^{2}=16$,so $a=4$. The eccentricity $e$ is given by $e=\sqrt{1-\

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(C) Given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$. Here,$a^{2}=16$,so $a=4$. The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{16}}=\frac{\sqrt{16-b^{2}}}{4}$. Thus,the foci of the ellipse are $(\pm ae, 0) = (\pm \sqrt{16-b^{2}}, 0)$. Given 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 $e$ is $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}$. The foci of the hyperbola are $(\pm ae, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$. Since the foci coincide,we have $\sqrt{16-b^{2}}=3$. Squaring both sides,$16-b^{2}=9$,which gives $b^{2}=7$.

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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