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Q.
$P(\theta)$ 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
Conic Sections
Solution:
Clearly, $P$ is $\left( a \,cos\,\theta, b\, sin\, \theta\right)$
$Q$ is $\left( a\, cos \left(\frac{\pi}{2}+\theta\right), b\, sin\left(\frac{\pi}{2}+\theta\right)\right) $
i.e., $\left(-a\,sin\,\theta, b\, sin\, \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(cos^{2}\theta +sin^{2}\theta\right)\right] = \frac{1}{2}$
$ \therefore $ locus of $\left(x_{1}, y_{1}\right)$ is $\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} = \frac{1}{2}$