= Solution
Let a slowly varying wave have phase $S(\mathbf x,t)$, <wavevector> $\mathbf k=\nabla S$, and <angular frequency> $\omega=-\partial_tS$. The local <dispersion relation> is the Hamilton-Jacobi equation $\omega=\Omega(\mathbf k;\mathbf x,t)$. Equality of mixed derivatives, followed along a curve with velocity $\dot{\mathbf x}=\nabla_{\mathbf k}\Omega$, gives the <Hamiltonian ray-tracing equations>
$$
\boxed{\frac{dx_i}{dt}=\frac{\partial\Omega}{\partial k_i},
\qquad
\frac{dk_i}{dt}=-\frac{\partial\Omega}{\partial x_i},
\qquad
\frac{d\omega}{dt}=\frac{\partial\Omega}{\partial t}.}
$$
Here $d/dt=\partial_t+\dot{\mathbf x}\mathbin{\cdot}\nabla$ is the total derivative evaluated along a ray, and $\dot{\mathbf x}$ is the <group velocity>.
For $\Omega=kc(z)$, $\omega$, $k_x$, and $k_y$ are conserved. If $\psi$ is the angle between the ray and the $z$-axis, then $k\sin\psi=(k_x^2+k_y^2)^{1/2}$ and $k=\omega/c$. Hence <Snell's law> is
$$
\boxed{\frac{\sin\psi}{c}=\frac{(k_x^2+k_y^2)^{1/2}}\omega=\text{constant}.}
$$
For the ray launched with $\mathbf k=k(\cos\phi,0,\sin\phi)$ at $z=0$, this constant is $|\cos\phi|/c_0$. Therefore
$$
\sin\psi=|\cos\phi|(1+\beta^2z^2)\leq1,
$$
which proves
$$
\boxed{|z|\leq z_m=\frac1\beta
\left(\frac1{|\cos\phi|}-1\right)^{1/2}.}
$$
For $0<\phi<\pi/2$, write $a=\cos\phi$. Since the ray and wavevector are parallel,
$$
\left(\frac{dz}{dx}\right)^2
=\left[\frac1{a(1+\beta^2z^2)}\right]^2-1.
$$
Near the upper <turning point>, put $z=z_m-\zeta$. Expansion gives
$$
\left(\frac{dz}{dx}\right)^2\simeq4a\beta^2z_m\zeta.
$$
If $x=x_m$ at the turn, integration yields the local ray path
$$
\boxed{z\simeq z_m-a\beta^2z_m(x-x_m)^2.}
$$
The ray is tangent to the line $z=z_m$, turns smoothly, crosses $z=0$, and repeats symmetrically between $-z_m$ and $z_m$. Increasing $\phi$ decreases $z_m$, while $\phi\downarrow0$ sends the turning level outward.
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