If the distance of a point on the ellipse x^2/6+y^2/2=1 from the centre is 2, then the eccentric angle is _______?

1 Answer
Feb 27, 2018

# (2n+1)pi/4,qquad n=0,1,2,3#

Explanation:

The eccentric angle #theta# is related to the coordinates of a point on the ellipse #x^2/a^2+y^2/b^2=1# by

#x = a cos theta, qquad y = b sin theta#

For this particular ellipse, #a = sqrt{6}, b = sqrt{2}#, so that

#x = sqrt{6} cos theta, qquad y = sqrt{2} sin theta#

The distance between this point and the origin is #d = sqrt{x^2+y^2}#, so that

#2^2 = 6cos^2theta+2sin^2theta = 4cos^2theta +2 implies#
# 4cos^2theta = 4-2=2 implies cos^2theta = 1/2 implies#
#cos theta = pm 1/sqrt{2}#

This implies #theta = (2n+1)pi/4,qquad n\in ZZ#