Product of length of segments in Ellipse. 
If the normal at any point $P$  on the Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2} {b^2}=1$ meets the axis in $G$ and $g$ respectively, then find $PG\cdot Pg$, in terms of $a$ and $b$.

I tried considering the parametric point as $(a\cos\theta,b\sin\theta)$ on the ellipse, then constructed both tangent and normal at that point then found the coordinated where normal cut the axes, then found the length of the perpendicular from these coordinates on the tangent (at $(a\cos\theta,b\sin\theta)$), but I end up with following: $$PG\cdot Pg=a^2b^2(a^2\sin^2\theta+b^2\cos^2\theta)$$
please help.

Hint:
Using this, the equation of normal
$$\dfrac x{\dfrac{(a^2-b^2)\cos\theta}a}-\dfrac x{\dfrac{(a^2-b^2)\sin\theta}b}=1$$
$$|PG|=\sqrt{\left(\dfrac{(a^2-b^2)\cos\theta}a-a\cos\theta\right)^2+(0-b\sin\theta)^2}=?$$
