How do you convert #x^2 + y^2 = 4y# into polar form?

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Answer 1 · 2018-04-15T03:47:29

#color(crimson)(r = 4 sin theta#

#x = r cos theta, y = r sin theta#

Given : #x^2 + y^2 = 4y#

Substituting for x & y in terms of r and theta,

#r^2 cos^2 theta + r^2 sin^2 theta = 4 r sin theta#

#r^2 (cos^2 theta + sin^2 theta) = 4 r sin theta#

#r^2 = 4 r sin theta, " as " cos^2 theta + sin^2 theta = 1#

#color(crimson)(r = 4 sin theta#