Question

If P($\theta$) and Q($\pi$/2 + $\theta$) are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Locus of the mid-point 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

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

Given,$P\left(\theta \right),Q(\frac{\pi }{2}+\theta )$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Any point on the ellipse will be $(acos\theta ,bsin\theta )$
$\Rightarrow P=(acos\theta ,bsin\theta ),Q=\left(acos\right(\frac{\pi }{2}+\theta ),bsin(\frac{\pi }{2}+\theta \left)\right)$
$\Rightarrow P=(acos\theta ,bsin\theta ),Q=(-asin\theta ,bcos\theta )$
Let required point be $C(x,y)$
Given, $C=mid-pointofPQ$
$\Rightarrow (x,y)=(\frac{(acos\theta -asin\theta )}{2},\frac{(bsin\theta +bcos\theta )}{2})$
$\Rightarrow \frac{x}{a}=(\frac{(cos\theta -sin\theta )}{2}),\frac{y}{b}=(\frac{(sin\theta +cos\theta )}{2})$
on squaring $\frac{x}{a},\frac{y}{b}$ and adding both
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=(\frac{(cos\theta -sin\theta )}{2}{)}^2+(\frac{(sin\theta +cos\theta )}{2}{)}^2$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{2(co{s}^2\theta +si{n}^2\theta )}{4}$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$(since $co{s}^2\theta +si{n}^2\theta =1$)