## Calculus (3rd Edition)

$$\rho=2 \csc\phi\sqrt{ \sec\phi} .$$
We have $$4=x^2-y^2=\rho^2\sin^2\phi\cos^2\theta-\rho^2\sin^2\phi\sin^2\theta\\ =\rho^2\sin^2\phi \cos 2\theta .$$ We solve for $\rho$: $$\rho=\frac{2}{\sin \phi\sqrt{\cos 2\theta}}.$$