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

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Answer 1 · 2018-02-11T08:23:32

#r^2(2.5-1.5cos2theta-2sin2theta)-r(sintheta+costheta)-2=0#

Given:
#2=(-x+2y)^2-y-x#
#x=rcostheta#
#y=rsintheta#

Now,
#2=(-rcostheta+2rsintheta)^2-rsintheta-rcostheta#

#2=r^2(-costheta+2sintheta)^2-r(sintheta+costheta)#

#2=r^2((costheta)^2-2costheta(2sintheta)+(2sintheta)^2)-rsintheta-rcostheta#

#2=r^2cos^2theta-4r^2costhetasintheta+4r^2sin^2theta-rsintheta-rcostheta#

#2=r^2(1+cos2theta)/2-4r^2(sin2theta)/2+4r^2(1-cos2theta)/2-r(sintheta+costheta)#

Simplifying further,
#2=r^2(0.5+0.5cos2theta-2sin2theta+2-2cos2theta)-r(sintheta+costheta)#

#2=r^2(2.5-1.5cos2theta-2sin2theta)-r(sintheta+costheta)#
Or

#r^2(2.5-1.5cos2theta-2sin2theta)-r(sintheta+costheta)-2=0#