The locus of the mid-points of the normal chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

Let the point $P$ be $(a\cos \theta , b\sin \theta )$.

$\therefore$ The equation of normal at $P$ to the ellipse is

$a\cdot x\cdot \sec \theta -b\cdot y\cdot \mathrm{cosec}\theta =a^2-b^2$    …(i)

Let its mid-point be $(h, k)$.

$\therefore$ The equation of the chord bisected at $(h, k)$ is

$\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$   …(ii)

From Eqs (i) and (ii), we get

$\begin{array}{l}\frac{a\sec \theta }{\left(h/a^2\right)}=\frac{b\mathrm{cosec}\theta }{\left(-k/b^2\right)}=\frac{a^2-b^2}{\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}\right)}\\ \Rightarrow \cos \theta =\frac{a^3}{h}\left(\frac{({\displaystyle \frac{h^2}{a^2}}+{\displaystyle \frac{k^2}{b^2}})}{a^2-b^2}\right) and \sin \theta =\frac{-b^3}{k}\left(\frac{({\displaystyle \frac{h^2}{a^2}}+{\displaystyle \frac{k^2}{b^2}})}{a^2-b^2}\right)\end{array}$

Squaring and adding, we get

$\left(\frac{a^6}{x^2}+\frac{b^6}{y^2}\right){\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}^2={\left(a^2-b^2\right)}^2$.