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

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Question

For hyperbola, $\mathrm{e}^{2}=\frac{1+\mathrm{b}^{2}}{\mathrm{a}^{2}}$

$\mathrm{e}^{2}=1+\frac{81}{144}=\frac{225}{144} \Rightarrow \mathrm{e}=\fr

Vedclass · Accepted Answer

$7$

Vedclass · Answer

For hyperbola, $\mathrm{e}^{2}=\frac{1+\mathrm{b}^{2}}{\mathrm{a}^{2}}$

$\mathrm{e}^{2}=1+\frac{81}{144}=\frac{225}{144} \Rightarrow \mathrm{e}=\frac{15}{12}=\frac{5}{4}$

and $a^{2}=\frac{144}{25}$

Foci are $(\pm 	ext { ae }, 0) \equiv(\pm 3,0)$

Now for ellipse ae $=3 \Rightarrow \mathrm{a}^{2} \mathrm{e}^{2}=9$

and $b^{2}=a^{2}\left(1-e^{2}ight)$

$\quad b^{2}=a^{2}-a^{2} e^{2}$

$\Rightarrow b^{2}=16-9=7$

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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