= Solution
The angle
$$
\phi=(p+q)\lambda_2-p\lambda_1-q\Omega_2
$$
obeys the <D'Alembert characteristic>. Because reflection in the reference plane makes the disturbing function even in inclination, the nodal coefficient must be even, and therefore
$$
\boxed{q\ \hbox{is even}}.
$$
At <astronomical conjunction>, $\lambda_1=\lambda_2=\lambda_c$, so
$$
\boxed{\frac\phi q=\lambda_c-\Omega_2}.
$$
Thus $\phi/q$ is the angular distance from the <ascending node> of $M_2$ to the conjunction. When $M_2$ is at that node, $\lambda_2=\Omega_2$, and
$$
\boxed{\frac\phi p=\Omega_2-\lambda_1}.
$$
Up to the chosen sign convention, $\phi/p$ is the angular separation between the inner body and the node when the outer body crosses the reference plane. These two views explain geometrically why libration confines both conjunction and node-crossing phases.
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