If the normal at P to the rectangular hyperbola $x^2-y^2=4$ meets the axes in G, g and C is the centre of the hyperbola, then    

Hyperbola $x^2-y^2=4$

$\frac{x^2}{4}-\frac{y^2}{4}=1$

Let $P\left(2\sec \theta ,2\tan \theta \right)$ by any point on hyperbola.

Equation of normal

$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

$\frac{4x}{2\cos \theta }+\frac{4y}{2\tan \theta }=4+4$

$2x\cos \theta +2y\cot \theta =8$

$x\cos \theta +y\cot \theta =4$

For g put  $y=0,x=4\sec \theta ,g\left(4\sec \theta ,0\right)$

For g put  $x=0,y=4\tan \theta ,G\left(0,4\tan \theta \right)$

$PC=2\sqrt{{\sec }^2\theta +{\tan }^2\theta }$

$Gg=4\sqrt{{\sec }^2\theta +{\tan }^2\theta }$

$pg=\sqrt{{\left(4\sec \theta -2\sec \theta \right)}^2+{\left(2\tan \theta \right)}^2}$

$=2\sqrt{{\sec }^2\theta +{\tan }^2\theta }$

$PG=\sqrt{{\left(2\sec \theta \right)}^2+{\left(2\tan \theta \right)}^2}$

$=2\sqrt{{\sec }^2\theta +{\tan }^2\theta }$

$\because PG=Pg=PC$