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  • The distance of a point on the ellipse   <img alt="\mathrm {\frac{x^2}{6}+\frac{y^2}{2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7B%5Cfrac%7Bx%5E2%7D%7B6%7Da

The distance of a point on the ellipse   \mathrm {\frac{x^2}{6}+\frac{y^2}{2}=1}   from the centre is 2 . Then the eccentric angle of the point is
 

Option: 1

\frac{\pi}{4}


Option: 2

\frac{3 \pi}{4}


Option: 3

Both (1) and (2)
 


Option: 4

\frac{\pi}{6}


Answers (1)

best_answer

Any point on this ellipse is (\sqrt{6} \cos \phi, \sqrt{2} \sin \phi).  Since, centre is (0,0),  so 6 \cos ^2 \phi+2 \sin ^2 \phi=4 \Rightarrow 2 \cos ^2 \phi=1
\Rightarrow \cos ^2 \phi=\left(\frac{1}{\sqrt{2}}\right)^2=\cos ^2 \frac{\pi}{4} \Rightarrow \phi=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}

 

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