Chilling Polar Forms

Well we can easily convert this polar equation in Cartesian coordinate equation. $p = \dfrac{1}{1-\cos \theta + \sin \theta}$ $p-p\cos\theta + p\sin\theta = 1$ Now convert into Cartesian coordinates by using the formulas:- $p = \sqrt{x^{2}+y^{2}}$ $p\sin\theta = y$ $p\cos\theta = x$ Hence we get the equation of curve as:- $\sqrt{x^{2}+y^{2}} -x+y = 1$ $\sqrt{x^{2} + y^{2}} = 1+x-y$ Squaring both sides:- $x^{2} + y^{2} = 1+x^{2} + y^{2} + 2x-2y-2xy$ $2xy+2y-2x=1$ $(x+1)(y-1) = \dfrac{-1}{2}$ This is the equation of a rectangular hyperbola.