How do you convert #-xy=x+4y^2 # into a polar equation? Trigonometry The Polar System Converting Between Systems 1 Answer Bdub Apr 15, 2016 #4r^2sin^2theta+r^2costhetasintheta+rcostheta=0# Explanation: #-xy=x+4y^2# Use equations #x=rcos theta,y=rsin theta# #-rcos theta xx rsintheta=rcos theta+4r^2sin^2theta# #-r^2costhetasintheta=rcostheta+4r^2sin^2theta# #0=4r^2sin^2theta+r^2costhetasintheta+rcostheta# 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 1178 views around the world You can reuse this answer Creative Commons License