Ni.(16 points) The vertex of the parabola is $O$ and the focus is $F$. When the moving point $P$ moves on the parabola, find the maximum value of the distance ratio $\left|\frac{P O}{P F}\right|$.
Let the equation of the parabola be $y^{2}=4 a x(a>0)$. The vertex is $O(0,0)$, and the focus is $F(a, 0)$.
If the coordinates of a moving point on the parabola are $P(x, y)$, then
$$
\begin{array}{l}
\left(\frac{P O}{P F}\right)^{2}=\frac{x^{2}+y^{2}}{(x-a)^{2}+y^{2}} \\
=\frac{x^{2}+4 a x}{(x-a)^{2}+4 a x}=\frac{x^{2}+4 a x}{x^{2}+2 a x+a^{2}} . \\
\text { Let } \frac{x^{2}+4 a x}{x^{2}+2 a x+a^{2}}=t . \text { Then, } \\
(t-1) x^{2}+2 a(t-2) x+t a^{2}=0 .
\end{array}
$$

If $t$ makes the equation have real solutions for $x$, then
$$
\begin{array}{l}
\Delta=4 a^{2}(t-2)^{2}-4 a^{2} t(t-1) \geqslant 0 \\
\Rightarrow 4-3 t \geqslant 0 \Rightarrow t \leqslant \frac{4}{3} .
\end{array}
$$

When the equality holds, by
$$
\begin{array}{l}
\frac{x^{2}+4 a x}{x^{2}+2 a x+a^{2}}=\frac{4}{3} \\
\Rightarrow x=2 a \Rightarrow y=2 \sqrt{2} a .
\end{array}
$$

At this point, $P(2 a, 2 \sqrt{2} a)$.
Therefore, $\left(\frac{P O}{P F}\right)^{2}=t \leqslant \frac{4}{3} \Rightarrow\left|\frac{P O}{P F}\right| \leqslant \frac{2 \sqrt{3}}{3}$.