Question

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) \[\dfrac{2\pi }{3}\]

Answer 1

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 \[\theta =\left( 2n+1 \right)\dfrac{\pi }{4},n\in z\].