How do you convert #r(2 + costheta) = 1# into cartesian form?

graph{(2sqrt(x^2+y^2)+x-1)(x^2+y^2-.0005)((x+1/3)^2+y^2-.0005)=0 [-1.25, 1.25, -0.625, 0.625]} The equation has the form

#(1/2)/r=1+1/2costheta#.

This represents the ellipse with focus S as pole, #theta=0# for S-side

major axis, eccentricity #e = 1/2# and semi major axis a = 2/3, from

#a(1-e^2)=l=1/2#.

The conversion formula is

#r(costheta, sin theta)=(x, y)#.

So, #2r+rcostheta=2sqrt(x^2+y^2)+x=1 #, giving

#x^2+y^2=1/4(1-x)^2=x^2/4-x/2+1/4#.

In the standard form, this is

#(x+1/3)^2/(3/4)+y^2/(1/3)=1#

In converting from polar #(r,theta)# to Cartesian #(x,y)#
we have the relations:
#color(white)("XXX")color(red)(r=sqrt(x^2+y^2))#
and
#color(white)("XXX")color(blue)(cos(theta)=x/sqrt(x^2+y^2))#
(and others)

Polar form: #color(red)(r)(2+color(blue)(cos(theta)))=1#
becomes (in Cartesian form)
#color(white)("XXX")color(red)(sqrt(x^2+y^2)) * (2 +color(blue)(x/sqrt(x^2+y^2)))=1#

#color(white)("XXX")rarr 2sqrt(x^2+y^2)+x=1#

#color(white)("XXX")rarr sqrt(x^2+y^2)=(1-x)/2#

#color(white)("XXX")rarr x^2+y^2=(1-2x+x^2)/4#

#color(white)("XXX")rarr 4x^2+4y^2=1-2x+x^2#

#color(white)("XXX")rarr 3x^2+4y^2+2x=1#