The locus of the foot of perpendicular drawn from the centre of the ellipse ${\mathrm{x}}^2+3{\mathrm{y}}^2=6$ on any tangent to it is

Equation of the ellipse is $$\begin{gathered} {\mathrm{x}}^2+3{\mathrm{y}}^2=6\text{ ⇒}\frac{{\mathrm{x}}^2}{6}+\frac{{\mathrm{y}}^2}{2}=1. \\ where a^2=6,b^2=2 \end{gathered}$$
Equation of the tangent is  $\frac{\mathrm{xcos} \mathrm{\theta }}{\mathrm{a}}+\frac{\mathrm{ysin} \mathrm{\theta }}{\mathrm{b}}=1$
Let (h, k) be any point on the locus.
$\therefore \frac{\mathrm{h}}{\mathrm{a}}\cos \mathrm{\theta }+\frac{\mathrm{k}}{\mathrm{b}}\sin \mathrm{\theta }=1 .....\left(1\right)$
Slope of the tangent line is $\frac{-\mathrm{b}}{\mathrm{a}}\cot \mathrm{\theta }$
Slope  of  perpendicular  drawn  from  centre (0, 0) to (h, k) is k/h.
Since, both the lines are perpendicular.
$\begin{array}{l}\therefore \left(\frac{\mathrm{k}}{\mathrm{h}}\right)\times \left(-\frac{\mathrm{b}}{\mathrm{a}}\cot \mathrm{\theta }\right)=-1\\ \Rightarrow \cos \mathrm{\theta }=\mathrm{\alpha ha}\\ \sin \mathrm{\theta }=\mathrm{\alpha kb}\\ \text{ From Eq. }\left(1\right),\frac{\mathrm{h}}{\mathrm{a}}\left(\mathrm{\alpha ha}\right)+\frac{\mathrm{k}}{\mathrm{b}}\left(\mathrm{\alpha kb}\right)=1\Rightarrow \mathrm{\alpha }=\frac{1}{{\mathrm{h}}^2+{\mathrm{k}}^2}\\ \text{ Also, }{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\\ \Rightarrow (\mathrm{\alpha kb}{)}^2+(\mathrm{\alpha ha}{)}^2=1\Rightarrow {\mathrm{\alpha }}^2{\mathrm{k}}^2{\mathrm{b}}^2+{\mathrm{\alpha }}^2{\mathrm{h}}^2{\mathrm{a}}^2=1\\ \Rightarrow \frac{2{\mathrm{k}}^2}{{\left({\mathrm{h}}^2+{\mathrm{k}}^2\right)}^2}+\frac{6{\mathrm{h}}^2}{{\left({\mathrm{h}}^2+{\mathrm{k}}^2\right)}^2}=1\\ \left[\because {\mathrm{a}}^2=6,{\mathrm{b}}^2=2\right]\\ Now locus of (h,k) is 6{\mathrm{x}}^2+2{\mathrm{y}}^2={\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2\end{array}$
[replacing k by y and h by x]