How do you convert #2y= -2x^2+3xy # into a polar equation?

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Answer 1 · 2016-09-25T19:18:49

Substitute #rsin(theta)# for y and #rcos(theta)# for x and then do some algebra.

#2rsin(theta) = -2(rcos(theta))^2 + 3(rcos(theta))(rsin(theta))#

There is a common factor of #r^2# on the right:

#2rsin(theta) = r^2{-2cos^2(theta) + 3cos(theta)sin(theta))}#

Flip things a bit:

#r^2{3cos(theta)sin(theta))-2cos^2(theta)} = 2rsin(theta)#

Remove a common r:

#r{3cos(theta)sin(theta))-2cos^2(theta)} = 2sin(theta)#

Divide by everything in the {}s:

#r = (2sin(theta))/(3cos(theta)sin(theta)-2cos^2(theta))#

There you have #r(theta)#