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

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Answer 1 · 2018-03-30T10:03:37

#r^2(9costheta+sintheta)^2-rcostheta-2=0#

Given:

#2=(9x+y)^2-x#

#x=rcostheta#

#y=rsintheta#

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

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

#r^2(9costheta+sintheta)^2-rcostheta-2=0#