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

$Given hyperbola x^2-y^2=a^2$

Equation of  Normal at P($\theta )$is $\frac{\mathrm{x}}{\sec \mathrm{\theta }}+\frac{\mathrm{y}}{\tan \mathrm{\theta }}=2\mathrm{a}$  ……(1)

Let $Q(x_1,y_1) be locus of mid point of normal$

Now chord equation whose mid point $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is ${\mathrm{xx}}_1-{\mathrm{yy}}_1={\mathrm{x}}_1^2-{\mathrm{y}}_1^2$  ……(2)

Since  (1) & (2) represents same line then 

$$\begin{gathered} x_{1}sec\theta =-y_1\tan \theta =\frac{{x_1}^2-{y_1}^2}{2a} \\ \Rightarrow sec\theta =\frac{{x_1}^2-{y_1}^2}{2a{x}_1},\tan \theta =\frac{{x_1}^2-{y_1}^2}{-2a{y}_1} \\ Since sc{e}^2\theta -{\tan }^2\theta =1 then \\ {\left(\frac{{x_1}^2-{y_1}^2}{2a{x}_1}\right)}^2-{\left(\frac{{x_1}^2-{y_1}^2}{-2a{y}_1}\right)}^2=1 \\ \Rightarrow \frac{{{(x_1}^2-{y_1}^2)}^2}{4a^2{x_1}^2}-\frac{{{(x_1}^2-{y_1}^2)}^2}{4a^2{y_1}^2}=1 \\ \Rightarrow \frac{{{(x_1}^2-{y_1}^2)}^2}{4a^2}({y_1}^2-{x_1}^2)={x_1}^2{y_1}^2 \\ So locus is {(x^2-y^2)}^3+4a^2{x}^2{y}^2=0 \end{gathered}$$