NIMCET 2015 Question Paper with solutions

The foci of the ellipse $\frac{{\mathit{x}}^2}{16}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1$
and the hyperbola $\frac{{\mathit{x}}^2}{144}-\frac{{\mathit{y}}^2}{81}=\frac{1}{25}$ coincide
∴ Foci of the ellipse and hyperbola lie on X-axis
Foci of the ellipse $=\sqrt{16}-{\mathit{b}}^2$
and foci of the hyperbola $=\sqrt{\frac{144}{25}+\frac{81}{25}}$
Foci of the ellipse and hyperbola coincide
$\therefore \sqrt{16-{\mathit{b}}^2}=\sqrt{\frac{144}{25}+\frac{81}{25}}$
Squaring both sides,
$\Rightarrow 16-{\mathit{b}}^2=\frac{225}{25}$
$\Rightarrow {\mathit{b}}^2=16-9=7$