20 On the parabola $y^{2}=2 p x$, does there exist a point $M$ such that the ratio of the distance from $M$ to the vertex of the parabola and the distance from $M$ to the focus of the parabola is maximized? If the point $M$ exists, find its coordinates and the aforementioned maximum ratio; if the point $M$ does not exist, please explain the reason.
20 Let $M\left(x_{1}, y_{1}\right)$ be any point on the parabola $y^{2}=2 p x$, $F\left(\frac{p}{2}, 0\right)$ be the focus of the parabola, and $O$ be the origin. Then,
$$
\frac{|M O|^{2}}{|M F|^{2}}=\frac{x_{1}^{2}+y_{1}^{2}}{\left(x_{1}-\frac{p}{2}\right)^{2}+y_{1}^{2}}=\frac{x_{1}^{2}+2 p x_{1}}{x_{1}^{2}+p x_{1}+\frac{p^{2}}{4}} .
$$

Since $x_{1}^{2}+p^{2} \geqslant 2 p x_{1}$, adding $3 x_{1}^{2}+4 p x_{1}$ to both sides gives
$$
4 x_{1}^{2}+4 p x_{1}+p^{2} \geqslant 3 x_{1}^{2}+6 p x_{1},
$$

which means $\frac{x_{1}^{2}+2 p x_{1}}{x_{1}^{2}+p x_{1}+\frac{p^{2}}{4}} \leqslant \frac{4}{3}$. (Equality holds when $x_{1}=p$). Therefore, $\frac{|M O|^{2}}{|M F|^{2}} \leqslant \frac{4}{3}$. So, $\frac{|M O|}{|M F|} \leqslant \frac{2}{\sqrt{3}}$. (Equality holds when $x_{1}=p$).
Thus, on the parabola $y^{2}=2 p x$, there are two points $(p, \sqrt{2} p)$ and $(p,-\sqrt{2} p)$, for which the ratio of the distances to the vertex and the focus of the parabola has a maximum value of $\frac{2}{\sqrt{3}}$.