How do you convert #r=6/(2-3sin(theta))# into cartesian form? Precalculus Polar Coordinates Converting Equations from Polar to Rectangular 1 Answer A. S. Adikesavan Aug 15, 2016 #4x^2-5y^2+9y-9=0# Explanation: Use #r(cos theta, sin theta)=(x, y)#. #r = sqrt(x^2+y^2) and sin theta = y/r# Now the given equation becomes #r = (6r)/(2-(3x)/r)#. Canceling r and rearranging, #r=3-3/2y#. Squaring, , #x^2+y^2=9/4y^2-9y+9#. Simplifying. #4x^2-5y^2+9y-9=0# Answer link Related questions What is the polar equation of a horizontal line? What is the polar equation for #x^2+y^2=9#? How do I graph a polar equation? How do I find the polar equation for #y = 5#? What is a polar equation? How do I find the polar equation for #x^2+y^2=7y#? How do I convert the polar equation #r=10# to its Cartesian equivalent? How do I convert the polar equation #r=10 sin theta# to its Cartesian equivalent? How do you convert polar equations to rectangular equations? How do you convert #r=6cosθ# into a cartesian equation? See all questions in Converting Equations from Polar to Rectangular Impact of this question 3856 views around the world You can reuse this answer Creative Commons License