Question

Let $P,D$ be two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, whose eccentric angles differ by $\frac{\pi }{2}$. Then the locus of mid point of chord $PD$ is

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

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

Given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $P\equiv (a\cos \theta ,b\sin \theta )$, then
$D\equiv (a\cos (\theta +\frac{\pi }{2}),b\sin (\theta +\frac{\pi }{2}))$
$\Rightarrow D\equiv (-a\sin \theta ,b\cos \theta )$
Let mid point of $PD$ be $M(h,k)$
So, $h=\frac{a\cos \theta -a\sin \theta }{2}$
$\Rightarrow \cos \theta -\sin \theta =\frac{2h}{a}\dots \left(1\right)$
And
$k=\frac{b\sin \theta +b\cos \theta }{2}$
$\Rightarrow \sin \theta +\cos \theta =\frac{2k}{b}\dots \left(2\right)$
Squaring and adding equation $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 2{\sin }^2\theta +2{\cos }^2\theta +2\sin \theta \cos \theta -2\sin \theta \cos \theta ={\left(\frac{2h}{a}\right)}^2+{\left(\frac{2k}{b}\right)}^2$
$\Rightarrow 2={\left(\frac{2h}{a}\right)}^2+{\left(\frac{2k}{b}\right)}^2$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$