How do you convert #R=6/(2+cos(theta))# into cartesian form?

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Answer 1 · 2017-11-28T12:55:13

#3x^2+12x+4y^2-36=0#

#r=6/(2+cos(theta))#

#2r+rcos(theta)=6#

For polar coordinates; #r# is equal to #sqrt(x^2+y^2)# and #rcos(theta)# is equal to #x# in cartesian ones. Hence,

#2sqrt(x^2+y^2)+x=6#

#2sqrt(x^2+y^2)=6-x#

#4*(x^2+y^2)=(6-x)^2#

#4x^2+4y^2=x^2-12x+36#

#3x^2+12x+4y^2-36=0#