What is the eccentricity of the conic given by $r = \frac{2}{1 + \cos \theta - \sin \theta}$?

Question

A very simple question. Usually, Conic section in polar graph given as, $$r = \frac{ke}{1 - e \cos\theta}$$ or $\sin\theta$, or something along this way.

However, given $$r = \frac{2}{1 + \cos \theta - \sin \theta}$$ how could I find its eccentricity?

Answer 1

You have \begin{align} \cos\theta-\sin\theta&=-\sqrt2\,\left(-\frac1 {\sqrt2}\cos\theta+\frac1{\sqrt2}\sin\theta\right)\\ \ \\&=-\sqrt2\,\left(\cos\frac{3\pi}4\cos\theta+\sin\frac{3\pi}4\sin\theta\right)\\ \ \\ & =-\sqrt2\,\cos\left(\theta-\frac{3\pi}4\right) \end{align}