NIMCET 2016 Question Paper with solutions

Given equation of line is $3\mathit{x}+5\mathit{y}=\mathit{k}$...(i)
comparing Eq. (i) with $\mathit{l}\mathit{x}+\mathit{m}\mathit{y}+\mathit{n}=0$, we get$\mathit{I}=3,\mathit{m}=5$ and $\mathit{n}=-\mathit{k}$
Equation of ellipse is $16{\mathit{x}}^2+25{\mathit{y}}^2=400$...(ii)
On divide Eq (ii) by 400, we get
$\frac{16{\mathit{x}}^2}{400}+\frac{25{\mathit{y}}^2}{400}=\frac{400}{400}$
$\Rightarrow \frac{{\mathit{x}}^2}{25}+\frac{{\mathit{y}}^2}{16}=1$
By comparing with standard form of ellipse $\frac{{\mathit{x}}^2}{{\mathit{a}}^2}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1$
we get $\mathit{a}=5,\mathit{b}=4$
∵ Condition of tangent
${\mathit{n}}^2={\mathit{a}}^2{\mathit{l}}^2+{\mathit{b}}^2{\mathit{m}}^2$
$\left(-{\mathit{k}}^2\right)={\left(5\right)}^2{\left(3\right)}^2+{\left(4\right)}^2{\left(5\right)}^2$
$\Rightarrow {\mathit{k}}^2=225+400$
$\Rightarrow {\mathit{k}}^2=625$
$\Rightarrow \mathit{K}=\pm 25$