Question

The 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. $e^{\prime }=1/e$

Answer 1

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

Normal $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$

$Q((a^2-b^2)\frac{\cos \theta }{a},0)$

$R(0,\frac{(b^2-a^2)\sin \theta }{b})$

Mid point of $QR(x,y)=((a^2-b^2)\frac{\cos \theta }{2a},\frac{(b^2-a^2)\sin \theta }{2b})$

${\cos }^2\theta +{\sin }^2\theta =\frac{4a^2}{{(a^2-b^2)}^2}{x}^2+\frac{4b^2}{{(a^2-b^2)}^2}{y}^2=1$

$e^{\prime }=\sqrt{1-\frac{\frac{{(a^2-b^2)}^2}{a^2}}{\frac{{(a^2-b^2)}^2}{b^2}}}=\sqrt{1-\frac{b^2}{a^2}}=e$