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If $P(\theta)$ and $Q(\pi / 2+\theta)$ are two points on the ellipse $\frac{z^2}{a^2}+\frac{y^2}{b^2}=1$. Locus of the mid-point of $P Q$ is
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Verified Answer
The correct answer is:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$
The coordinates of $P$ and $Q$ are $(a \cos \theta, b \sin \theta)$ and $(-a \sin \theta, b \cos \theta)$ respectively.
Let $(h, k)$ be the co-ordinates of the mid-point of $P Q$. Then,
$\begin{aligned}
& 2 h=a(\cos \theta-\sin \theta) \text { and } 2 k=b(\sin \theta+\cos \theta) \\
& \Rightarrow \frac{4 h^2}{a^2}+\frac{4 k^2}{b^2}=2
\end{aligned}$
Hence, the locus of $(h, k)$ is
$\frac{4 x^2}{a^2}+\frac{4 y^2}{b^2}=2 \text { or, } \frac{2 x^2}{a^2}+\frac{2 y^2}{b^2}=1$
Let $(h, k)$ be the co-ordinates of the mid-point of $P Q$. Then,
$\begin{aligned}
& 2 h=a(\cos \theta-\sin \theta) \text { and } 2 k=b(\sin \theta+\cos \theta) \\
& \Rightarrow \frac{4 h^2}{a^2}+\frac{4 k^2}{b^2}=2
\end{aligned}$
Hence, the locus of $(h, k)$ is
$\frac{4 x^2}{a^2}+\frac{4 y^2}{b^2}=2 \text { or, } \frac{2 x^2}{a^2}+\frac{2 y^2}{b^2}=1$
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