Question

Consider a tangent to the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ at any point. The locus of the midpoint of the portion intercepted between the axes is

• A. $\frac{x^2}{2}+\frac{y^2}{4}=1$
• B. $\frac{x^2}{4}+\frac{y^2}{2}=1$
• C. $\frac{1}{3x^2}+\frac{1}{4y^2}=1$
• D. $\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Answer 1

The correct option is D $\frac{1}{2x^2}+\frac{1}{4y^2}=1$


Tangent at $R$ $(\sqrt{2}\sin \theta ,\sin \theta )$
$\Rightarrow \frac{x\sqrt{2}\cos \theta }{2}+\frac{y\sin \theta }{1}=1$
$A=(\frac{\sqrt{2}}{\cos \theta },0),B=(0,\frac{1}{\sin \theta })$
Let $p(h,k)$ be the locus of the midpoint.
$\therefore (h,k)=(\frac{\sqrt{2}}{2\cos \theta },\frac{1}{2\sin \theta })\therefore h=\frac{1}{\sqrt{2}\cos \theta },k=\frac{1}{2\sin \theta }\Rightarrow \cos \theta =\frac{1}{\sqrt{2}h},\sin \theta =\frac{1}{2k}$
$\Rightarrow {\cos }^2\theta +{\sin }^2\theta =\frac{1}{2h^2}+\frac{1}{4k^2}\therefore \text{Locus of}(h,k)\text{is}\frac{1}{2x^2}+\frac{1}{4y^2}=1$