The minimum distance between two points P and | vedclass

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(A) Let the eccentric angles of $P$ and $Q$ be $	heta_1$ and $	heta_2$ respectively. Given $	heta_1 - 	heta_2 = \frac{3\pi}{2}$,so $	heta_1 = 	h

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$2\sqrt{2}$

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(A) Let the eccentric angles of $P$ and $Q$ be $	heta_1$ and $	heta_2$ respectively. Given $	heta_1 - 	heta_2 = \frac{3\pi}{2}$,so $	heta_1 = 	heta_2 + \frac{3\pi}{2}$.Coordinates are $P(5 \cos 	heta_1, 2 \sin 	heta_1)$ and $Q(5 \cos 	heta_2, 2 \sin 	heta_2)$.Substituting $	heta_1 = 	heta_2 + \frac{3\pi}{2}$,we get $P(5 \sin 	heta_2, -2 \cos 	heta_2)$.The distance $PQ^2 = (5 \cos 	heta_2 - 5 \sin 	heta_2)^2 + (2 \sin 	heta_2 + 2 \cos 	heta_2)^2$.$PQ^2 = 25(\cos^2 	heta_2 + \sin^2 	heta_2 - 2 \sin 	heta_2 \cos 	heta_2) + 4(\sin^2 	heta_2 + \cos^2 	heta_2 + 2 \sin 	heta_2 \cos 	heta_2)$.$PQ^2 = 25(1 - \sin 2	heta_2) + 4(1 + \sin 2	heta_2) = 29 - 21 \sin 2	heta_2$.To minimize $PQ$,we maximize $\sin 2	heta_2 = 1$.$PQ^2_{min} = 29 - 21 = 8$.$PQ_{min} = \sqrt{8} = 2\sqrt{2}$.

The minimum distance between two points $P$ and $Q$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$,if the difference between the eccentric angles of $P$ and $Q$ is $\frac{3\pi}{2}$,is

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