(9) The vertex of the parabola is $O$, and the focus is $F$. When the moving point $P$ moves on the parabola, try to find the maximum value of the distance ratio $\left|\frac{P O}{P F}\right|$.
9 In the Cartesian coordinate system, let the equation of the parabola be $y^{2}=4 a x, a>0$, then the vertex is $O(0,0)$, the focus is $F(a, 0)$, if the coordinates of a moving point on the parabola are $P(x, y)$, then
$$
\begin{aligned}
\left(\frac{P O}{P F}\right)^{2} & =\frac{x^{2}+y^{2}}{(x-a)^{2}+y^{2}} \\
& =\frac{x^{2}+4 a x}{(x-a)^{2}+4 a x}=\frac{x^{2}+4 a x}{x^{2}+2 a x+a^{2}} .
\end{aligned}
$$

Let
$$
\frac{x^{2}+4 a x}{x^{2}+2 a x+a^{2}}=t,
$$

then
$$
(t-1) x^{2}+2 a(t-2) x+t a^{2}=0,
$$

If $t$ makes the equation have real solutions for $x$, then the discriminant
$$
\Delta=4 a^{2}(t-2)^{2}-4 a^{2} t(t-1) \geqslant 0,
$$

which means $4-3 t \geqslant 0$, so
$$
t \leqslant \frac{4}{3},
$$

When the equality holds, by
$$
\frac{x^{2}+4 a x}{x^{2}+2 a x+a^{2}}=\frac{4}{3},
$$

we solve to get $x=2 a$, then $y=2 \sqrt{2} a$, at this time $P(2 a, 2 \sqrt{2} a)$.
Therefore, $\left(\frac{P O}{P F}\right)^{2}=t \leqslant \frac{4}{3},\left|\frac{P O}{P F}\right| \leqslant \frac{2 \sqrt{3}}{3}$.