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

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Answer 1 · 2016-12-27T14:32:50

#r=sintheta/(1-sinthetacostheta)#

The relation between Cartesian coordinates #(x,y)# and polar coordinates #(r,theta)# is given by

#x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#

Hence, #y=(x-y)^2+xy#

#hArrrsintheta=(rcostheta-rsintheta)^2+rcosthetaxxrsintheta#

or #sintheta/r=cos^2theta+sin^2theta-2sinthetacostheta+sinthetacostheta#

or #r=1-sinthetacostheta#

or #r=sintheta/(1-sinthetacostheta)#
graph{y=(x-y)^2+xy [-2.344, 2.656, -0.43, 2.07]}