Answer
$$\rho=2 \csc\phi\sqrt{ \sec\phi} .$$
Work Step by Step
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 $\rho$:
$$\rho=\frac{2}{\sin \phi\sqrt{\cos 2\theta}}.$$
