What the is the polar form of #y = 1/y-xy+x^2/y #? Trigonometry The Polar System Converting Between Systems 1 Answer RedRobin9688 Jun 27, 2018 #rsin(theta)=1/(rsin(theta))-r^2sin(theta)cos(theta)+(rcos^2(theta))/(sin(theta))# Simplified: #r^2(cos(theta)-cos(3theta))-4=(2r)^2cos(2theta)# Explanation: #x=rcos(theta), y=rsin(theta)# Substitute and Simplify #rsin(theta)=1/(rsin(theta))-r^2sin(theta)cos(theta)+(rcos^2(theta))/sin(theta)# Alternate: #r^2sin(theta)=csc(theta)-r^3sin(theta)cos(theta)+r^2cot(theta)cos(theta)# 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 1383 views around the world You can reuse this answer Creative Commons License