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#### The foci of the ellipse $\mathrm{\frac{x^2}{16}+\frac{y^2}{b^2}=1}$ and the hyperbola $\mathrm{\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}}$ coincide, then the value of $\mathrm{b^2}$ is Option: 1 1Option: 2 5Option: 3 7Option: 4 9

For the given ellipse,$\mathrm{a^2=16, b^2=b^2 .}$

$\mathrm{\therefore \quad e=1-\frac{b^2}{a^2} \quad \Rightarrow \quad e=\frac{\sqrt{16-b^2}}{4}}$

Thus the foci of the ellipse are (± ae, 0) i.e. $\mathrm{\left( \pm \sqrt{\left(16-b^2\right)}, 0\right)}$

Re-arranging the equation of the hyperbola, we get

$\mathrm{\left[x^2\left(\frac{12}{5}\right)^2\right]-\left[y^2\left(\frac{9}{5}\right)^2\right]=1}$

Comparing this with standard equation, we have

$\mathrm{a=\frac{12}{5}, b=\frac{9}{5} \text { and the centre }(0,0)}$

$\mathrm{\therefore \quad e^2=1+\left(\frac{b^2}{a^2}\right) \Rightarrow e=\frac{5}{4}}$

Thus the foci of the hyperbola are (± ae, 0) i.e. (± 3, 0)

Since the foci of ellipse and hyperbola coincide, therefore

$\mathrm{\sqrt{\left(16-b^2\right)}=3 \quad \Rightarrow \quad b^2=7}$