Minimum distance between two points P and Q on ellipse frac{{{x^2}}}{{25}} + frac{{{y^2}}}{4}

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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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