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# Eccentric angle of a point on the ellipse ${{x}^{2}}+3{{y}^{2}}=6$at a distance $2$units from the center of the ellipse is :(A) $\dfrac{\pi }{4}$(B) $\dfrac{\pi }{3}$(C) $\dfrac{3\pi }{4}$(D) $\dfrac{2\pi }{3}$

Last updated date: 13th Jul 2024
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Hint: Find out the center of the given ellipse and consider a parametric point. Later, Equate the distance between the parametric point and center of the ellipse to 2 units.

For, any given ellipse, $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, the eccentric angle $\theta$ is related as:
$x=a\cos \theta ;$
$y=b\sin \theta ;$
The given ellipse equation is: $\dfrac{{{x}^{2}}}{6}+\dfrac{{{y}^{2}}}{2}=1.$
So, here we will have $a=\sqrt{6}$ and $b=\sqrt{2}$;
And, $x=\sqrt{6}\cos \theta ;y=\sqrt{2}\sin \theta$
Now, the distance between the center$\left( 0,0 \right)$ and the point $\left( \sqrt{6}\cos \theta ,\sqrt{2}\sin \theta \right)$ on ellipse is given as 2 units.
Therefore, $2=\sqrt{{{\left( \sqrt{6}\cos \theta -0 \right)}^{2}}+{{\left( \sqrt{2}\sin \theta -0 \right)}^{2}}}$
Squaring on both sides, we will get:
${{2}^{2}}=6{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta$
Now, substituting ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$, we will have:
$4=4{{\cos }^{2}}\theta +2$
We get, ${{\cos }^{2}}\theta =\dfrac{1}{2}$
$\to \cos \theta =\pm \dfrac{1}{\sqrt{2}}$.
Therefore, $\theta =\left( 2n+1 \right)\dfrac{\pi }{4},n\in z$
So, the value of $\theta$ is either $\dfrac{\pi }{4}or\dfrac{3\pi }{4}$.
Hence, option A and C are correct

Note: We have to make sure that you consider the both positive and negative values of $\cos \theta$after applying the square root every time. Remember that the general solution of Cosine trigonometric function is $\theta =\left( 2n+1 \right)\dfrac{\pi }{4},n\in z$.