10. (20 points) Find a point $A$ on the parabola $y^{2}=2 p x$ such that the normal line at point $A$ intersects the parabola to form the shortest chord $A B$.
10. Let $A\left(2 p t^{2}, 2 p t\right), B\left(2 p s^{2}, 2 p s\right)$.

It is easy to know that the slopes of the normal line $A B$ and the tangent line at point $A$ are $\frac{1}{t+s}$ and $\frac{1}{2 t}$, respectively.
From $\frac{1}{t+s} \cdot \frac{1}{2 t}=-1 \Rightarrow t+s=-\frac{1}{2 t}$
$\Rightarrow s=-\frac{2 t^{2}+1}{2 t}$.
Then $A B^{2}=\left(2 p t^{2}-2 p s^{2}\right)^{2}+(2 p t-2 p s)^{2}$
$=4 p^{2}\left(\left(t^{2}-s^{2}\right)^{2}+(t-s)^{2}\right)$
$=4 p^{2}(t-s)^{2}\left((t+s)^{2}+1\right)$
$=4 p^{2}\left(t+\frac{2 t^{2}+1}{2 t}\right)^{2}\left(\left(-\frac{1}{2 t}\right)^{2}+1\right)$
$=4 p^{2} \cdot \frac{\left(4 t^{2}+1\right)^{2}}{4 t^{2}} \cdot \frac{4 t^{2}+1}{4 t^{2}}$
$=\frac{p^{2}}{4} \cdot \frac{\left(4 t^{2}+1\right)^{3}}{t^{4}}$.
By the AM-GM inequality,
$$
\begin{array}{l}
\frac{4 t^{2}+1}{t^{\frac{4}{3}}}=4 t^{\frac{2}{3}}+t^{-\frac{4}{3}} \\
\geqslant 3 \sqrt[3]{2 t^{\frac{2}{3}} \cdot 2 t^{\frac{2}{3}} \cdot t^{-\frac{4}{3}}}=3 \sqrt[3]{4},
\end{array}
$$

Equality holds if and only if $2 t^{\frac{2}{3}}=t^{-\frac{4}{3}}$, i.e., $t= \pm \frac{\sqrt{2}}{2}$.
Thus, $A B^{2} \geqslant \frac{p^{2}}{4}(3 \sqrt[3]{4})^{3}=27 p^{2}$
$\Rightarrow A B \geqslant 3 \sqrt{3} p$.
Therefore, the shortest normal chord $A B=3 \sqrt{3} p$.
When $t= \pm \frac{\sqrt{2}}{2}$,
point $A(p, \sqrt{2} p)$ or $A(p,-\sqrt{2} p)$.
Hence, the required point $A$ is one of the two intersection points of the line $x=p$ and the parabola.