= Solution
Let the thermal-wind basic velocity be
$$
\mathbf U=\Lambda z\,\widehat{\mathbf y},
\qquad
\Lambda=\frac{M^2}{f},
$$
so that $fU_z=b_x=M^2$. Assume constant $f,M,N$, hydrostatic perturbations, $y$ independence, and normal modes proportional to $e^{i(kx+mz-\omega t)}$. The linearized equations are
$$
u_t-fv=-p_x,
\qquad
v_t+fu+\Lambda w=0,
\qquad
p_z=b',
$$
$$
b'_t+M^2u+N^2w=0,
\qquad
u_x+w_z=0.
$$
Incompressibility gives $w=-(k/m)u$. Eliminating $v,p,b'$ yields
$$
\boxed{
\omega^2=f^2-2M^2\frac{k}{m}
+N^2\frac{k^2}{m^2}}.
$$
This quadratic in $k/m$ is positive for every orientation precisely when
$$
\boxed{M^4<N^2f^2}.
$$
Otherwise some disturbances grow monotonically through <symmetric instability>.
For the stable case, the minimum occurs at
$$
\frac{k}{m}=\frac{M^2}{N^2},
\qquad
\boxed{\omega_{min}^2=f^2-\frac{M^4}{N^2}}.
$$
Constant-phase lines have slope $dz/dx=-k/m=-M^2/N^2$, exactly the slope of the basic isopycnals $M^2x+N^2z=\text{constant}$. The minimum-frequency displacement follows an isopycnal, allowing buoyancy and Coriolis restoring forces to oppose one another. Its frequency is below the inertial frequency $|f|$ of an unstratified rotating fluid.
If $\theta$ is the counter-clockwise angle of the wavevector from the horizontal, $k/m=\cot\theta$, and
$$
\boxed{
\cot\theta=\frac{M^2\pm
\sqrt{M^4+N^2(\omega^2-f^2)}}{N^2}}.
$$
For $\omega^2>f^2$, define $D=[M^4+N^2(\omega^2-f^2)]^{1/2}>M^2$. Taking $k>0$, the incident wave with group velocity down and right has
$$
\cot\theta_i=\frac{M^2+D}{N^2}>0.
$$
A horizontal reflection preserves $k$ and $\omega$ but selects the other vertical wavenumber, so
$$
\boxed{
\cot\theta_r=\frac{M^2-D}{N^2}<0}.
$$
The incident and reflected phase lines have slopes $-(M^2+D)/N^2$ and $(D-M^2)/N^2$, respectively. Their group velocities are perpendicular to these phase lines: the incident ray points down-right and the reflected ray up-right. The isopycnal slope $-M^2/N^2$ lies between the two phase-line orientations.
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