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

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Answer 1 · 2016-05-03T08:52:00

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

If #(r,theta)# is in polar form and #(x,y)# in Cartesian form the relation between them is as follows:

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

Hence, #-y=3y^2-x^2-2x# can be written as

#-rsintheta=3r^2sin^2theta-r^2cos^2theta-2rcostheta# or

#-sintheta=3rsin^2theta-rcos^2theta-2costheta# or

#r(3sin^2theta-cos^2theta)=2costheta-sintheta# or

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