Question

If normal at a variable point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of eccentricity $e$ meets the axes of the ellipse at $Q$ and $R,$ then the locus of the midpoint of $QR$ is a conic with an eccentricity $e^{\prime }$ such that

• A. $e^{\prime }$ is independent of $e$
• B. $e^{\prime }=1$
• C. $e^{\prime }=e$
• D. $e^{\prime }=\frac{e}{2}$

Answer 1

The correct option is C $e^{\prime }=e$

Assuming $a>b$ for the given ellipse
Normal at point $P(a\cos \theta ,b\sin \theta )$ is given by :
$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2\dots \left(i\right)$
It meets coordinate axis at $Q(\frac{(a^2-b^2)\cos \theta }{a},0)$ and $R(0,-\frac{(a^2-b^2)\sin \theta }{b})$
Let $T(h,k)$ be the midpoint of $QR,$ then
$2h=\frac{(a^2-b^2)\cos \theta }{a}$ and
$2k=-\frac{(a^2-b^2)\sin \theta }{b}$
$\Rightarrow {\cos }^2\theta +{\sin }^2\theta =\frac{4h^2{a}^2}{(a^2-b^2{)}^2}+\frac{4k^2{b}^2}{(a^2-b^2{)}^2}=1$
Hence, required locus is
$\frac{x^2}{\frac{(a^2-b^2{)}^2}{4a^2}}+\frac{y^2}{\frac{(a^2-b^2{)}^2}{4b^2}}=1$
which is an ellipse, having eccentricity $e^{\prime }$, then
$e^{\prime }=\sqrt{1-\frac{\left(\frac{(a^2-b^2{)}^2}{4a^2}\right)}{\left(\frac{(a^2-b^2{)}^2}{4b^2}\right)}}=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e^{\prime }=e$