How do you convert #y=(x-y)^2-xy # into a polar equation? Trigonometry The Polar System Converting Between Systems 1 Answer Sihan Tawsik Feb 2, 2016 here is how, Explanation: here, #y=(x-y)^2-xy# #or,y=x^2-2xy+y^2-xy# #or,rsintheta=(x^2+y^2)-3xy# #or,rsintheta=r^2-3rcosthetarsintheta# #or,2rsintheta=2r^2-3*r^2*2costhetasintheta# #or,2sintheta=2r-3rsin2theta# Answer link Related questions How do you convert rectangular coordinates to polar coordinates? When is it easier to use the polar form of an equation or a rectangular form of an equation? How do you write #r = 4 \cos \theta # into rectangular form? What is the rectangular form of #r = 3 \csc \theta #? What is the polar form of # x^2 + y^2 = 2x#? How do you convert #r \sin^2 \theta =3 \cos \theta# into rectangular form? How do you convert from 300 degrees to radians? How do you convert the polar equation #10 sin(θ)# to the rectangular form? How do you convert the rectangular equation to polar form x=4? How do you find the cartesian graph of #r cos(θ) = 9#? See all questions in Converting Between Systems Impact of this question 1298 views around the world You can reuse this answer Creative Commons License