How do you convert #r(2 - cosx) = 2# into cartesian form?

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Answer 1 · 2016-07-30T17:51:57

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

The relation between Cartesian coordinates #(x.y)# and #(r,theta)# is given by #x=rcostheta# and #y=rsintheta# and #r^2=x^2+y^2#

Hence, #r(2-costheta)=2# can be written as

#2r-rcostheta=2# or

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

#2sqrt(x^2+y^2)=2+x# and squaring

#4(x^2+y^2)=(2+x)^2=4+4x+x^2# or

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

graph{3x^2+4y^2-4x-4=0 [-1.708, 3.292, -1.3, 1.5]}