NIMCET 2016 Question Paper with solutions

Given equation of ellipse is $\frac{{\mathit{x}}^2}{16}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1$......(i)
and equation of hyperbola $\frac{{\mathit{x}}^2}{144}-\frac{{\mathit{y}}^2}{81}=\frac{1}{25}$
$\Rightarrow \frac{\frac{{\mathit{x}}^2}{144}}{25}-\frac{\frac{{\mathit{y}}^2}{81}}{25}$.........(ii)
By comparing Eq (ii) with $\frac{{\mathit{x}}^2}{{\mathit{a}}^2}-\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1$ we get
${\mathit{a}}^2=\frac{144}{25}{\mathit{b}}^2=\frac{81}{25}$
So. foci of the ellipse are $\left(\pm \sqrt{16-{\mathit{b}}^2},0\right)$ and hyperbola are $\left(\pm \sqrt{{\left(\frac{12}{5}\right)}^2+{\left(\frac{9}{5}\right)}^2},0\right)$
If they coincide then
$16-{\mathit{b}}^2=\frac{144}{25}+\frac{81}{25}$
$\Rightarrow 16-{\mathit{b}}^2=\frac{225}{25}$
$\Rightarrow 16-{\mathit{b}}^2=9$
$\Rightarrow {\mathit{b}}^2=16-9=7$