Question

$PG$ is the normal to a standard ellipse at $P$. $G$ being on the major axis. $GP$ is produced outwards to $Q$ so that $PQ=GP$. Show that the locus of $Q$ is an ellipse whose eccentricity is $\frac{a^2-b^2}{a^2+b^2}$

Answer 1

Let $P$ be $(a\cos \theta ,b\sin \theta )$

$⟹$ normal equation is $\frac{x}{\frac{\cos \theta }{a}}-\frac{y}{\frac{\sin \theta }{b}}=a^2-b^2$

$⟹G=(a{e}^2\cos \theta ,0)$

$PQ=GP⟹P$ is the mid point of $Q$ and $G$

$Q=\left(\right(2-e^2)a\cos \theta ,2b\sin \theta )$

Locus of $Q$ is $\frac{x^2}{a^2(2-e^2{)}^2}+\frac{y^2}{4b^2}=1$

eccentricity $=\sqrt{1-\frac{4b^2}{a^2(\frac{a^2+b^2}{a^2}{)}^2}}=\frac{a^2-b^2}{a^2+b^2}$