The locus of the mid-point of the chord of contact of the perpendicular tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$is 

Equation of the chord of contact to the tangents at $(h, k)$ is 

$\frac{hx}{a^2}+\frac{ky}{b^2}=1$                                 $....\left(i\right)$

The equation of the chord of the ellipse whose mid-point $(\alpha , \beta )$ is

$\frac{\alpha x}{a^2}+\frac{\beta y}{b^2}=\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}$                            $....\left(ii\right)$

Since Eqs $\left(i\right)$ and $\left(ii\right)$ are the same, therefore

$\begin{array}{l}\frac{h}{\alpha }=\frac{k}{\beta }=\frac{1}{\left(\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}\right)}\\ \Rightarrow h=\frac{\alpha }{\left(\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}\right)}\end{array}$

and $k=\frac{\beta }{\left(\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}\right)}$

Also, $(h, k)$ lies on the director circle of the given ellipse $x^2+y^2=a^2+b^2$.

Thus, $h^2+k^2=a^2+b^2$

$\Rightarrow \left(\frac{{\alpha }^2+{\beta }^2}{a^2+b^2}\right)={\left(\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}\right)}^2$

Hence, the locus of $(\alpha , \beta )$ is $\left(\frac{x^2+y^2}{a^2+b^2}\right)={\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}^2$.