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

Given hyperbola ${\mathrm{x}}^2-{\mathrm{y}}^2=a^2$

$let p(x_1,y_1) be locus of mid point of normal chord$

 Now chord equation is ${\mathrm{S}}_1={\mathrm{S}}_{11}$

$\Rightarrow {\mathrm{xx}}_1-{\mathrm{yy}}_1={{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2 \to \left(1\right)$

Equation of normal  at  $\mathrm{P}\left(\mathrm{\theta }\right)$ to the hyperbola is  

$$\begin{gathered} \frac{\mathrm{ax}}{\mathrm{sec\theta }}-\frac{\mathrm{ay}}{\mathrm{tan\theta }}=2{\mathrm{a}}^2 \\ \Rightarrow \mathrm{x} \mathrm{cos\theta }-\mathrm{y} \mathrm{cot\theta }=2\mathrm{a} \to \left(2\right) \end{gathered}$$

 $$\begin{gathered} \sin ce \left(1\right) and \left(2\right) represents same line then \\ \frac{\mathrm{cos\theta }}{{\mathrm{x}}_1}=\frac{\cancel{-}\mathrm{cot\theta }}{\cancel{-} {\mathrm{y}}_1}=\frac{2\mathrm{a}}{{{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2} \\ \Rightarrow \mathrm{cos\theta }=\frac{2{\mathrm{ax}}_1}{{{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2}, \mathrm{cot\theta }=\frac{2{\mathrm{ay}}_1}{{{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2} \\ \mathrm{Now} \mathrm{sin\theta }=\frac{\mathrm{cos\theta }}{\mathrm{cot\theta }}=\frac{{\mathrm{x}}_1}{{\mathrm{y}}_1} \\ and {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1 \\ \Rightarrow \frac{{x^2}_1}{{y^2}_1}+\frac{4a^2{x^2}_1}{{({x^2}_1-{y^2}_1)}^2}=1 \\ \Rightarrow {\left({{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2\right)}^2{{x^2}_1}^{ }+4{\mathrm{a}}^2 {{\mathrm{x}}_1}^2 {{\mathrm{y}}_1}^2={({x^2}_1-{y^2}_1)}^2{y^2}_1 \\ \Rightarrow {\left({{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2\right)}^2({x^2}_1- {{\mathrm{y}}_1}^2)+4{\mathrm{a}}^2 {{\mathrm{x}}_1}^2 {{\mathrm{y}}_1}^2=0 \\ \therefore \mathrm{locus} \mathrm{is} {({\mathrm{x}}^2-{\mathrm{y}}^2)}^3+4{\mathrm{a}}^2{\mathrm{x}}^2{\mathrm{y}}^2=0 \end{gathered}$$