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10-2. Parabola, Ellipse, Hyperbola
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Minimum distance between two points $P$ and $Q$ on the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{4} = 1$ , if difference between eccentric angles of $P$ and $Q$ is $\frac{{3\pi }}{2}$ , is
A
$2\sqrt 2 $
B
$2\sqrt 5 $
C
$\sqrt {29} $
D
$\sqrt {62} $
Solution
$\mathrm{P}(5 \cos \theta, 2 \sin \theta), \mathrm{Q}(5 \sin \theta,-2 \cos \theta)$
$\mathrm{PQ}=\sqrt{(5 \cos \theta-5 \sin \theta)^{2}+(2 \sin \theta+2 \cos \theta)^{2}}$
$\mathrm{PQ}=\sqrt{29-21 \sin 2 \theta}$
Minmum value of $\mathrm{PQ}=\sqrt{8}=2 \sqrt{2}$
Standard 11
Mathematics
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