Product of length of segments in Ellipse.

Question

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.

Answer 1

Let $P(u,v)$.

Thus, $$\frac{xu}{a^2}+\frac{yv}{b^2}=1$$ is an equation of the tangent to ellipse.

Thus the slope of the normal it's $\frac{va^2}{b^2u}$ and the equation of the normal it's $$y-v=\frac{va^2}{b^2u}(x-u),$$ which gives $$g\left(0,v\left(1-\frac{a^2}{b^2}\right)\right)$$ and $$G\left(\left(1-\frac{b^2}{a^2}\right)u,0\right),$$ which gives $$PG=\sqrt{\frac{b^4u^2}{a^4}+v^2}$$ and $$Pg=\sqrt{u^2+\frac{a^4v^2}{b^4}}.$$ Thus, $$PG\cdot Pg=\frac{a^4v^2+b^4u^2}{a^2b^2}$$ and since $$\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$$ only, we see that $PG\cdot Pg$ depend on $P$.

Answer 2

Hint:

Using this, the equation of normal

$$\dfrac x{\dfrac{(a^2-b^2)\cos\theta}a}-\dfrac x{\dfrac{(a^2-b^2)\sin\theta}b}=1$$

$$|PG|=\sqrt{\left(\dfrac{(a^2-b^2)\cos\theta}a-a\cos\theta\right)^2+(0-b\sin\theta)^2}=?$$