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

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Answer 1 · 2018-04-13T10:30:32

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

For this we will need:
#x=rcostheta#
#y=rsintheta#

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

#rsintheta=2r^2sin^2theta+3r^2cos^2theta-2r^2costhetasintheta#

#sintheta=2rsin^2theta+3rcos^2theta-2rcosthetasintheta#

#sintheta=2rsin^2theta+3rcos^2theta-rsin(2theta)#

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

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