10. (12 points) Given a positive number $p$ and the parabola $C: y^{2}=2 p x$ $(p>0)$, $A\left(\frac{p}{6}, 0\right)$ is a point on the axis of symmetry of the parabola $C$, $O$ is the vertex of the parabola $C$, and $M$ is any point on the parabola $C$. Find the maximum value of $\frac{|O M|}{|A M|}$.
10. Let point $M(x, y)$. Then $x \geqslant 0$, and $y^{2}=2 p x$.
$$
\begin{array}{l}
\text { Hence }\left(\frac{|O M|}{|A M|}\right)^{2}=\frac{x^{2}+y^{2}}{\left(x-\frac{p}{6}\right)^{2}+y^{2}} \\
=\frac{x^{2}+2 p x}{x^{2}-\frac{p}{3} x+\frac{p^{2}}{36}+2 p x} \\
=\frac{x^{2}+2 p x}{x^{2}+\frac{5 p}{3} x+\frac{p^{2}}{36}} .
\end{array}
$$

Let $t=\frac{x^{2}+2 p x}{x^{2}+\frac{5}{3} p x+\frac{p^{2}}{36}}$. Then
$$
(t-1) x^{2}+p\left(\frac{5}{3} t-2\right) x+\frac{p^{2}}{36} t=0 \text {. }
$$

When $t \neq 1$, by $x \in \mathbf{R}$, we know
$$
\begin{array}{l}
\Delta=p^{2}\left[\left(\frac{5}{3} t-2\right)^{2}-\frac{1}{9} t(t-1)\right] \\
=p^{2}\left(\frac{24}{9} t^{2}-\frac{59}{9} t+4\right) \geqslant 0 \\
\Leftrightarrow t \geqslant \frac{4}{3} \text { or } t \leqslant \frac{9}{8} .
\end{array}
$$

When $t=\frac{9}{8}$, we have $x=\frac{p}{2}$; and when $x \geqslant 0$,
$$
\frac{x^{2}+2 p x}{x^{2}+\frac{5}{3} p x+\frac{p^{2}}{36}} \leqslant \frac{9}{8} \Leftrightarrow\left(x-\frac{p}{2}\right)^{2} \geqslant 0,
$$

i.e., for all $x \geqslant 0$, we have $t \leqslant \frac{9}{8}$ (thus, $t \geqslant \frac{4}{3}$ is impossible).
Therefore, the maximum value of $t$ is $\frac{9}{8}$.
Hence, $\left(\frac{|O M|}{|A M|}\right)_{\max }=\frac{3 \sqrt{2}}{4}$.