How do you convert #2=(9x+y)^2-x# into polar form? Trigonometry The Polar System Converting Between Systems 1 Answer Shiva Prakash M V Mar 30, 2018 #r^2(9costheta+sintheta)^2-rcostheta-2=0# Explanation: Given: #2=(9x+y)^2-x# #x=rcostheta# #y=rsintheta# #2=(9rcostheta+rsintheta)^2-rcostheta# #2=r^2(9costheta+sintheta)^2-rcostheta# #r^2(9costheta+sintheta)^2-rcostheta-2=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 1707 views around the world You can reuse this answer Creative Commons License