Question

If $CF$ be the perpendicular from the centre $C$ of the ellipse $\frac{x^2}{12}+\frac{y^2}{8}=1,$ on the tangent at any point $P$ and $G$ is the point where the normal at $P$ meets the major axis, then the value of $CF\cdot PG=$

Answer 1

Let $P$ be $(2\sqrt{3}\cos \theta ,2\sqrt{2}\sin \theta )$
Tangent at $P:$
$\frac{x\cos \theta }{2\sqrt{3}}+\frac{y\sin \theta }{2\sqrt{2}}=1$

Normal at $P:$
$\frac{2\sqrt{3}x}{\cos \theta }-\frac{2\sqrt{2}y}{\sin \theta }=4$

The point where normal meets major axis is,
$x=\frac{4\cos \theta }{2\sqrt{3}}=\frac{2\cos \theta }{\sqrt{3}}$
$\therefore G\equiv (\frac{2\cos \theta }{\sqrt{3}},0)$

Now, $CF=$ Distance of $(0,0)$ from the tangent line
$=\frac{|0+0-1|}{\sqrt{{\left(\frac{\cos \theta }{2\sqrt{3}}\right)}^2+{\left(\frac{\sin \theta }{2\sqrt{2}}\right)}^2}}$
$\Rightarrow CF=\frac{2\sqrt{6}}{\sqrt{2+{\sin }^2\theta }}$

Also, $PG=\sqrt{{\left(2\cos \theta (\sqrt{3}-\frac{1}{\sqrt{3}})\right)}^2+(2\sqrt{2}\sin \theta {)}^2}$
$=\frac{2\sqrt{2}\left(\sqrt{2+{\sin }^2\theta }\right)}{\sqrt{3}}$

$\therefore CF\cdot PG=8$