NIMCET 2021 Question Paper with solutions

Let $\left(\mathit{h},\mathit{k}\right)$ be the point of intersection of two
Perpendicular tangents equations of pair tangents is ${\mathit{S}\mathit{S}}_1={\mathit{T}}^2$
$\left(\frac{{\mathit{x}}^2}{{\mathit{a}}^2}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}-1\right)\left(\frac{{\mathit{h}}^2}{{\mathit{a}}^2}+\frac{{\mathit{k}}^2}{{\mathit{b}}^2}-1\right)$
$\Rightarrow \left(\frac{\mathit{h}\mathit{x}}{{\mathit{a}}^2}+\frac{\mathit{k}\mathit{y}}{{\mathit{b}}^2}-1\right)$ $\Rightarrow \frac{{\mathit{x}}^2}{{\mathit{a}}^2}\left(\frac{{\mathit{k}}^2}{{\mathit{b}}^2}-1\right)+\frac{{\mathit{y}}^2}{{\mathit{a}}^2}\left(\frac{{\mathit{h}}^2}{{\mathit{a}}^2}-1\right)+.......=0$ .....(i)
Since equation (i) represents two perpendicular lines
$\therefore \frac{1}{{\mathit{a}}^2}\left(\frac{{\mathit{k}}^2}{{\mathit{b}}^2}-1\right)+\frac{1}{{\mathit{b}}^2}\left(\frac{{\mathit{h}}^2}{{\mathit{a}}^2}-1\right)=0$
$\Rightarrow {\mathit{k}}^2-{\mathit{b}}^2+{\mathit{h}}^2-{\mathit{a}}^2=0$
$\Rightarrow$ Replace h by by $\mathit{y}$, then locus will be
$\Rightarrow {\mathit{x}}^2+{\mathit{y}}^2={\mathit{a}}^2+{\mathit{b}}^2$ (Director Circle)