Question

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

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=4$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
• D. None of these

Answer 1

Answer: (A)

Clearly, $P$ is $\left(acos\theta ,bsin\theta \right)$
$Q$ is $\left(acos\left(\frac{\pi }{2}+\theta \right),bsin\left(\frac{\pi }{2}+\theta \right)\right)$
i.e., $\left(-asin\theta ,bsin\theta \right)$
If $\left(x_1,y_1\right)$ is the mid-point of $PQ$, then
$x_1=\frac{a}{2}\left(cos\theta -sin\theta \right)$
$y_1=\frac{b}{2}\left(sin\theta +cos\theta \right)$
$\therefore \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=\frac{1}{4}\left[2\left(co{s}^2\theta +si{n}^2\theta \right)\right]=\frac{1}{2}$
$\therefore$ locus of $\left(x_1,y_1\right)$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$