What the is the polar form of #y^2 = (x-3)^2/y-x^2 #? Trigonometry The Polar System Converting Between Systems 1 Answer Kalit Gautam Apr 22, 2018 #r^3sin^3theta-r^2cos^2theta-9+6rcostheta+r^3cos^2thetasintheta=0# Explanation: #y^2=((x-3)^2)/y - x^2# Put #x=rcostheta# and #y=rsintheta# ; we get :- #r^2 sin^2theta=((rcostheta-3)^2)/(rsintheta) -r^2cos^2theta# #rArrr^3sin^3theta-r^2cos^2theta-9+6rcostheta+r^3cos^2thetasintheta=0# 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 1296 views around the world You can reuse this answer Creative Commons License