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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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The correct answer is:
7
7
Given equation of ellipse is
$$
\frac{x^2}{16}+\frac{y^2}{b^2}=1
$$
$$
\begin{aligned}
& \text { eccentricity }=e=\sqrt{1-\frac{b^2}{16}} \\
& \text { foci: } \pm a e=\pm 4 \sqrt{1-\frac{b^2}{16}}
\end{aligned}
$$
Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$$
\begin{aligned}
& \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1 \\
& \text { eccentricity }=e=\sqrt{1+\frac{81}{25} \times \frac{25}{144}}=\sqrt{1+\frac{81}{144}} \\
& =\sqrt{\frac{225}{144}}=\frac{15}{12} \\
& \text { foci: } \pm a e=\pm \frac{12}{5} \times \frac{15}{12}=\pm 3 \\
&
\end{aligned}
$$
Since, foci of ellipse and hyperbola coincide
& \therefore \pm 4 \sqrt{1-\frac{b^2}{16}}=\pm 3 \Rightarrow b^2=7 \\
$$
$$
\frac{x^2}{16}+\frac{y^2}{b^2}=1
$$
$$
\begin{aligned}
& \text { eccentricity }=e=\sqrt{1-\frac{b^2}{16}} \\
& \text { foci: } \pm a e=\pm 4 \sqrt{1-\frac{b^2}{16}}
\end{aligned}
$$
Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$$
\begin{aligned}
& \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1 \\
& \text { eccentricity }=e=\sqrt{1+\frac{81}{25} \times \frac{25}{144}}=\sqrt{1+\frac{81}{144}} \\
& =\sqrt{\frac{225}{144}}=\frac{15}{12} \\
& \text { foci: } \pm a e=\pm \frac{12}{5} \times \frac{15}{12}=\pm 3 \\
&
\end{aligned}
$$
Since, foci of ellipse and hyperbola coincide
& \therefore \pm 4 \sqrt{1-\frac{b^2}{16}}=\pm 3 \Rightarrow b^2=7 \\
$$
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