SOLUTION
The given equation is:
[tex]r(1-2\cos\theta)=1[/tex]Rewrite the equation
[tex]r-2r\cos\theta=1[/tex]Recall that
[tex]r=\sqrt{x^2+y^2}x=rcos\theta[/tex]Substituting these values gives
[tex]\begin{gathered} \sqrt{x^2+y^2}-2x=1 \\ (\sqrt{x^2+y^2})^2=(1+2x)^2 \\ x^2+y^2=1+4x^2+4x \\ 1+4x^2+4x-x^2-y^2=0 \\ 3x^2-y^2+4x+1=0 \end{gathered}[/tex]Therefore the required equation is
[tex]3x^{2}-y^{2}+4x+1=0[/tex]