Question
The normal at a variable point $P$ on an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ of eccentricity $e$ meets the axes of the ellipse in $Q$ and $R$ then the locus of the mid-point of $QR$ is a conic with an eccentricity $e^{'}$ such that:
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Solution
Sol.(c) $2h = ae^{2}cos\theta$
$$2k = \frac{- a^{2}e^{2}sin\theta}{b}$$
$$\begin{matrix} & \ \Rightarrow \frac{4h^{2}}{\left( ae^{2} \right)^{2}} + \frac{4k^{2}}{\left( a^{2}\frac{e^{2}}{b} \right)^{2}} = 1 \\ & \frac{x^{2}}{\left( \frac{ae^{2}}{2} \right)^{2}} + \frac{y^{2}}{\left( \frac{a^{2}e^{2}}{2b} \right)^{2}} = 1 \\ & \frac{a^{2}e^{4}}{4} = \frac{a^{4}e^{4}}{4b^{2}}\left( 1 - \left( e^{'} \right)^{2} \right) \\ & \ \Rightarrow \ b^{2} = a^{2}\left( 1 - \left( e^{'} \right)^{2} \right) \\ & \ \Rightarrow \ e = e^{'} \end{matrix}$$
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