Equation of ellipse is $x^{2}+3 y^{2}=6$ or $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$.
Equation of the tangent is $\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$
Let $(h, k)$ be any point on the locus.
$\therefore \frac{h}{a} \cos \theta+\frac{k}{b} \sin \theta=1 ....$ (i)
Slope of the tangent line is $\frac{-b}{a} \cot \theta$.
Slope of perpendicular drawn from centre $(0,0)$ to $(h, k)$ is $k / h$.
Since, both the lines are perpendicular.
$\therefore \left(\frac{k}{h}\right) \times\left(-\frac{b}{a} \cot \theta\right)=-1$
$\Rightarrow \frac{\cos \theta}{h a}=\frac{\sin \theta}{k b}=\alpha $ [say]
$\Rightarrow \cos \theta=\alpha h a$
$\sin \theta=\alpha k b$
From Eq. (i), $\frac{h}{a}(\alpha h a)+\frac{k}{b}(\alpha k b)=1$
$\Rightarrow h^{2} \alpha+k^{2} \alpha=1$
$\Rightarrow \alpha=\frac{1}{h^{2}+k^{2}}$
Also,$ \sin ^{2} \theta+\cos ^{2} \theta=1$
$\Rightarrow (\alpha k b)^{2}+(\alpha h a)^{2}=1$
$\Rightarrow \alpha^{2} k^{2} b^{2}+\alpha^{2} h^{2} a^{2}=1$
$\Rightarrow \frac{k^{2} b^{2}}{\left(h^{2}+k^{2}\right)^{2}}+\frac{h^{2} a^{2}}{\left(h^{2}+k^{2}\right)^{2}}=1$
$\Rightarrow \frac{2 k^{2}}{\left(h^{2}+k^{2}\right)^{2}}+\frac{6 h^{2}}{\left(h^{2}+k^{2}\right)^{2}}=1$
$\left[\because a^{2}=6, b^{2}=2\right]$
$\Rightarrow 6 x^{2}+2 y^{2}=\left(x^{2}+y^{2}\right)^{2}$
[replacing $k$ by $y$ and $h$ by $x$]