Question

$P\left(\theta \right),D(\theta +\frac{\pi }{2})$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the locus of mid point of chord $PD$ is:

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

Answer 1

Answer: (D)

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$

The points $P\left(a\cos \theta ,b\sin \theta \right)$ and $D\left(a\cos (\theta +\frac{\pi }{2}),b\sin (\theta +\frac{\pi }{2})\right)$

$\Rightarrow D(-a\sin \theta ,b\cos \theta )$

Let the midpoints be $(h,k)$

$\therefore h=\frac{a(\cos \theta -\sin \theta )}{2}k=\frac{b(\cos \theta +\sin \theta )}{2}$

On squaring and adding both the equations we get:

$\frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{{\left((\cos \theta +\sin \theta )\right)}^2+{\left((\cos \theta +\sin \theta )\right)}^2}{4}$

$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{2({\left(\cos \theta \right)}^2+{\left(\sin \theta \right)}^2)}{4}$

$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{1}{2}$

$\therefore$ The locus is: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$