Conics - Locus of points

Question

The ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has one of its foci at the point $F$. The perpendicular from the origin to the tangent at a point $P(a\cos\theta, b\sin\theta)$ on the ellipse intersects the line $FP$ at a point $G$. Find the locus of $G$ as $\theta$ varies.

Answer 1

HINT - modified:

I would say, the locus will circle:

$FP:\,y=\frac{b\sin \theta}{a\cos \theta - e}\,(x+e)$

$SR:\,y=\frac{a}{b}\,\tan \theta\, x=\frac{a}{b}\,\tan \theta\,(x+e)-\frac{a\,e}{b}\,\tan \theta$

$\Rightarrow x+e = a\,\frac{a \cos \theta-e}{e\cos \theta -a}, \quad y = a\,\frac{b \sin \theta}{e\cos \theta -a}$

$e^2=a^2-b^2$

$(x+e^2)+y^2=a^2\left(\frac{(a\cos \theta -e)^2+b^2\sin^2 \theta}{(e\cos \theta -a)^2}\right)=a^2\left(\frac{a^2\cos^2 \theta -2ae\cos \theta +e^2+(a^2 -e^2)\sin^2 \theta}{(e\cos \theta -a)^2}\right)=$

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

$\Rightarrow\,$the equation of the locus: $\,(x+e^2)+y^2=a^2$