A normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the axes in $M$ and $N$ and the lines $MP$ and $NP$ are drawn perpendiculars to the axes meeting at $P$. the locus of $P$ is the hyperbola a is.

The equation of any normal to the hyperbola

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

$ax\cos \phi +by\cot \phi =\left(a^2+b^2\right)$         …(i)

Since the normal (i) meets the x-axis at $M$ and y-axis at $N$ respectively. Then,

$M=\left(\frac{a^2+b^2}{a\cos \phi }, 0\right)$ and $N=\left(0, \left(\frac{a^2+b^2}{b}\right)\tan \phi \right)$

Let the co-ordinates of the point $P$ be $\left(\alpha , \beta \right)$ Since $PM$ and $PN$ are perpendiculars to the axes, so the co-ordinates of $P$ are

   $\left(\left(\frac{a^2+b^2}{a}\right)\sec \phi , \left(\frac{a^2+b^2}{b}\right)\tan \phi \right)$

Therefore,

    $\alpha =\left(\frac{a^2+b^2}{a}\right)\sec \phi$ and $\beta =\left(\frac{a^2+b^2}{b}\right)\tan \phi$

$\Rightarrow \alpha \left(\frac{a}{a^2+b^2}\right)=\sec \phi$ and $\beta \left(\frac{b}{a^2+b^2}\right)=\tan \phi$

As we know that, 

  $\begin{array}{l}{\sec }^2\phi -{\tan }^2\phi =1\\ {\alpha }^2{\left(\frac{a}{a^2+b^2}\right)}^2-{\beta }^2{\left(\frac{b}{a^2+b^2}\right)}^2=1\\ \Rightarrow {\alpha }^2{a}^2-{\beta }^2{b}^2={\left(a^2+b^2\right)}^2\end{array}$

Hence, the locus of $\left(\alpha , \beta \right)$ is

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