How do we find the eccentric angle of ellipse #x^2/16+y^2/4=1#?

1 Answer
Shwetank Mauria
Nov 27, 2016

Eccentric angle is #pi/3#

Explanation:

The equation of ellipse is #x^2/16+y^2/4=1# or #x^2+4y^2=16# and point #(2,sqrt3)# lies in first quadrant.

Equation of auxiliary circle will be #x^2+y^2=16#.

Now the point #P(2,sqrt3)# and corresponding point on auxiliary circle is

#Q(2,sqrt(16-2^2))# or #Q(2,2sqrt3)# and

eccentric angle #theta=tan^(-1)((2sqrt3)/2)=tan^(-1)sqrt3=pi/3#
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