Question

$P\left(\theta \right)$ and $Q(\theta +\frac{\pi }{2})$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ The locus of midpoint of the chord PQ is

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{a}$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{b}$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$
• D. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{6}$

Answer 1

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

$P=(a\cos \theta ,b\sin \theta ),Q=(–a\sin \theta ,b\cos \theta )$
If $(x_1,y_1)$ is the midpoint of PQ then $(x_1,y_1)$ = $(\frac{a(\cos \theta -\sin \theta )}{2},\frac{b(\cos \theta +\sin \theta )}{2})$
$\Rightarrow \frac{2x_1}{a}=\cos \theta -\sin \theta ,\frac{2y_1}{b}=\cos \theta +\sin \theta$
But $(\cos \theta -\sin \theta {)}^2+(\cos \theta +\sin \theta {)}^2=2\Rightarrow \frac{4x_1^2}{a^2}+\frac{4y_1^2}{b^2}=2$
Locus is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$