How do you convert #9=(2x+y)^2-3y-x# into polar form?

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Answer 1 · 2018-04-27T07:39:25

#r=9/(2(cos^2theta+1)+2sin(2theta)-3sintheta-costheta)#

We will use:
#x=rcostheta#
#y=rsintheta#

#9=(2rcostheta+rsintheta)^2-3rsintheta-rcostheta#

#9=r((2costheta+sintheta)^2-3sintheta-costheta)#

#r=9/((2costheta+sintheta)^2-3sintheta-costheta)#

#r=9/(4cos^2theta+4costhetasintheta+2sin^2theta-3sintheta-costheta)#

#r=9/(2(2cos^2theta+sin^2theta)+2sin(2theta)-3sintheta-costheta)#

#r=9/(2(cos^2theta+1)+2sin(2theta)-3sintheta-costheta)#