Given,

The equation of the given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \cdots \left(1\right)$

Now let $\left(h,k\right)$ be the pole of a chord $PQ$ of the hyperbola

Then the straight line $PQ$ is the polar of the point $\left(h,k\right)$ w.r.t. the hyperbola and so its equation is

$\frac{xh}{a^2}-\frac{yk}{b^2}=1 \cdots \left(2\right)$

Now the center of the hyperbola is the origin, let it be point $C$,

Now making $\left(1\right)$ homogeneous with the help of $\left(2\right)$, the combined equation of $CP$ and $CQ$ is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}={\left(\frac{xh}{a^2}-\frac{yk}{b^2}\right)}^2 \cdots \left(3\right)$

Now the chord $PQ$ subtends a right angle at $\left(0,0\right)$

Therefore the lines $CP$ and $CQ$ given by $\left(3\right)$ are perpendicular and so in the equation $\left(3\right)$, we have the coefficient of $x^2+$coefficient of $y^2=0$

$\Rightarrow \left(\frac{1}{a^2}-\frac{h^2}{a^4}\right)+\left(\frac{-1}{b^2}-\frac{k^2}{b^4}\right)=0$

$\Rightarrow \frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}$

Thus locus is

$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}$