## Calculus (3rd Edition)

$r = \frac{{2\cos \theta }}{{\cos \theta - \sin \theta }}$ as an equation in rectangular coordinates: ${x^2} + {y^2} = \frac{{4{x^2}}}{{{{\left( {x - y} \right)}^2}}}$
Since $r = \frac{{2\cos \theta }}{{\cos \theta - \sin \theta }}$ we have ${r^2} = \frac{{4{{\cos }^2}\theta }}{{{{\cos }^2}\theta - 2\cos \theta \sin \theta + {{\sin }^2}\theta }}$ Since $x = r\cos \theta$ and $y = r\sin \theta$, we get ${x^2} + {y^2} = {r^2}$. $\cos \theta = \frac{x}{{\sqrt {{x^2} + {y^2}} }}$, ${\ \ \ }$ $\sin \theta = \frac{y}{{\sqrt {{x^2} + {y^2}} }}$. So, ${x^2} + {y^2} = \frac{{4{x^2}/\left( {{x^2} + {y^2}} \right)}}{{{x^2}/\left( {{x^2} + {y^2}} \right) - 2xy/\left( {{x^2} + {y^2}} \right) + {y^2}/\left( {{x^2} + {y^2}} \right)}}$ ${x^2} + {y^2} = \frac{{4{x^2}}}{{{x^2} - 2xy + {y^2}}}$ ${x^2} + {y^2} = \frac{{4{x^2}}}{{{{\left( {x - y} \right)}^2}}}$