Chapter 10 - Section 10.3 - Polar Coordinates - 10.3 Exercises - Page 692: 24

$\tan\theta=-2r\cos\theta$

$y=-2x^{2}$ Putting $x=r\cos\theta$ and $y=r\sin\theta$, we get $r\sin\theta=-2r^{2}\cos^{2}\theta$ $\implies \sin\theta=-2r\cos^{2}\theta$ $\implies \frac{\sin\theta}{\cos\theta}=-2r\cos\theta$ Or $\tan\theta=-2r\cos\theta$