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

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Answer 1 · 2016-09-16T15:56:13

#r(cos^2theta-3sin^2theta+sin2theta)+sinthetacos^2theta=0#

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is given by

#x=rcostheta#, #y=rsintheta# i.e. #x=r/costheta#, #y=r/sintheta# and #r^2=x^2+y^2#,

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

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

Now multiplying it by #sin^2thetacos^2theta#, we get

#rsinthetacos^2theta=-r^2cos^2theta+3r^2sin^2theta-2r^2sinthetacostheta# or

#sinthetacos^2theta=-r(cos^2theta-3sin^2theta+2sinthetacostheta)# or

#r(cos^2theta-3sin^2theta+sin2theta)+sinthetacos^2theta=0#