If the normal at any point Pon the ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the axis of x at G and the axis of y at g, then  $PG:pg$ is. 

Equation of the normal at  $P(a\cos \theta ,b\sin \theta )$ on the 

ellipse is  $ax\sec \theta -by\mathrm{cosec}\theta =a^2-b^2$ 

So coordinates of G are $\left(\frac{a^2-b^2}{a\sec \theta },0\right)$  and of $g$  are 

$\begin{array}{l}\left(0,-\frac{a^2-b^2}{a\mathrm{cosec}\theta }\right)\\ (PG{)}^2={\left(a\cos \theta -\frac{a^2-b^2}{a}\cos \theta \right)}^2+b^2{\sin }^2\theta \end{array}$

$=\frac{b^2}{a^2}\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)$ 

and $(Pg{)}^2=\frac{a^2}{b^2}\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)$

So $PG:Pg=b^2:a^2$