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

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Answer 1 · 2017-07-21T10:12:01

#rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta-sintheta=0#

Let;s expand this first :

#y=(x-2y)^2-2x^2y-2y^2=#

#x^2-4xy+4y^2-2x^2y-2y^2=x^2-4xy+2y^2-2x^2y=>#

#y=x^2-4xy+2y^2-2x^2y#

Now to swith to polar coordinates we do the following substitutions :

#y=rsintheta#
#x=rcostheta#

#rsintheta=r^2cos^2theta-4r^2sinthetacostheta+2r^2sin^2theta-2r^3cos^2thetasintheta#

#=>sintheta=rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta=>#

#rcos^2theta-2rsin2theta+2rsin^2theta-2r^2cos^2thetasintheta-sintheta=0#