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Question: The distance of the point \(\left( \sqrt{6}\cos\theta,\sqrt{2}\sin\theta \right)\)on the ellipse \(\...

The distance of the point (6cos⁡θ,2sin⁡θ)\left( \sqrt{6}\cos\theta,\sqrt{2}\sin\theta \right)on the ellipse x26+y22=1\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1from the centre is 2 if

A

θ=π2\theta = \frac{\pi}{2}

B

θ=3π2\theta = \frac{3\pi}{2}

C

θ=5π2\theta = \frac{5\pi}{2}

D

θ=7π2\theta = \frac{7\pi}{2}

Answer

θ=7π2\theta = \frac{7\pi}{2}

Explanation

Solution

Since (6cos⁡θ,2sin⁡θ)\left( \sqrt{6}\cos\theta,\sqrt{2}\sin\theta \right)lie on the ellipse

x26+y22=1\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1

∴∣CP∣=2\therefore|CP| = 2; 6cos⁡2θ+2sin⁡2θ=26\cos^{2}\theta + 2\sin^{2}\theta = 2

⇒ 3(1+cos⁡2θ)+1−cos⁡2θ=23\left( 1 + \cos 2\theta \right) + 1 - \cos 2\theta = 2

⇒ 2cos2θ\theta +2=0

⇒ cos⁡2θ=−1\cos 2\theta = - 1

2θ=π,3π,5π,7π2\theta = \pi,3\pi,5\pi,7\pi

∴θ=π2,3π2,5π2,7π2\therefore\theta = \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}