The locus of middle points of normal chords of the rectangular hyperbola ${\mathrm{x}}^2–{\mathrm{y}}^2={\mathrm{a}}^2$ is

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} {\mathrm{x}}^2-{\mathrm{y}}^2={\mathrm{a}}^2 \\ \mathrm{If} (\mathrm{h}, \mathrm{k}) \mathrm{be} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{any} \mathrm{chord} \mathrm{of} \mathrm{hyperbola} \mathrm{then} \mathrm{its} \mathrm{equation} \mathrm{is} \\ \mathrm{hx}-\mathrm{ky}={\mathrm{h}}^2-{\mathrm{k}}^2 — ① (\because {\mathrm{S}}_1={\mathrm{S}}_{11}) \\ \mathrm{But} \left(1\right) \mathrm{is} \mathrm{a} \mathrm{normal} \mathrm{to} \mathrm{hyperbola}, \mathrm{then} \mathrm{its} \mathrm{equation} \mathrm{is} \\ \frac{\mathrm{ax}}{\mathrm{sec\theta }}+\frac{\mathrm{ay}}{\mathrm{tan\theta }}=2{\mathrm{a}}^2 \\ \Rightarrow \mathrm{x} \mathrm{cos\theta }+\mathrm{y} \mathrm{cot\theta }=2\mathrm{a} — ② \\ \sin ce ① \& ② represents same line then \\ \frac{\mathrm{h}}{\mathrm{cos\theta }}=\frac{-\mathrm{k}}{\mathrm{cot\theta }} =\frac{{\mathrm{h}}^2-{\mathrm{k}}^2}{2\mathrm{a}} \\ \Rightarrow \mathrm{sec\theta }=\frac{{\mathrm{h}}^2-{\mathrm{k}}^2}{2\mathrm{ah}}, \mathrm{tan\theta }=-\frac{({\mathrm{h}}^2-{\mathrm{k}}^2)}{2\mathrm{ak}} \\ \mathrm{Now} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow \frac{{({\mathrm{h}}^2-{\mathrm{k}}^2)}^2}{4{\mathrm{a}}^2{\mathrm{h}}^2}-\frac{{({\mathrm{h}}^2-{\mathrm{k}}^2)}^2}{4{\mathrm{a}}^2{\mathrm{k}}^2}=1 \Rightarrow \frac{{({\mathrm{h}}^2-{\mathrm{k}}^2)}^2}{4{\mathrm{a}}^2} \left(\frac{1}{{\mathrm{h}}^2}-\frac{1}{{\mathrm{k}}^2}\right)=1 \\ \Rightarrow \frac{{({\mathrm{h}}^2-{\mathrm{k}}^2)}^2}{4{\mathrm{a}}^2} \left(\frac{{\mathrm{k}}^2-{\mathrm{h}}^2}{{\mathrm{h}}^2 {\mathrm{k}}^2}\right)=1 \\ \Rightarrow {({\mathrm{h}}^2-{\mathrm{k}}^2)}^3=-4{\mathrm{a}}^2{\mathrm{h}}^2{\mathrm{k}}^2 \\ \therefore \mathrm{locus} \mathrm{is} {({\mathrm{x}}^2-{\mathrm{y}}^2)}^3+4{\mathrm{a}}^2{\mathrm{x}}^2{\mathrm{y}}^2=0 \end{gathered}$$