How do you change the polar equation #r(2+costheta)=1# into rectangular form? Trigonometry The Polar System Converting Between Systems 1 Answer Cesareo R. Oct 31, 2016 #(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0# Explanation: Giving the pass equations #{(x=rcostheta),(y=rsintheta):}# #r(2+costheta)=1 = r(2+x/r)=2r+x=1# then #r = (1-x)/2->x^2+y^2=(1-2x+x^2)/4# and finally #3x^2+4y^2+2x-1=0# or #(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0# which is an ellipse. 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 1082 views around the world You can reuse this answer Creative Commons License