10. (20 points) Given the parabola $y^{2}=2 p x(p>0)$ with a chord $A B$ of length $l(l>0)$. Find the minimum distance from the midpoint of the chord $A B$ to the $y$-axis.

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10. Let the endpoints of the chord be \( A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) \), and the midpoint be \( M(x, y) \). Then
\[
\begin{array}{l}
y_{1}^{2}=2 p x_{1}, y_{2}^{2}=2 p x_{2}, \\
2 x=x_{1}+x_{2}, 2 y=y_{1}+y_{2} . \\
\text { And } l^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2} \\
=\left(x_{2}+x_{1}\right)^{2}-4 x_{2} x_{1}+2\left(y_{2}^{2}+y_{1}^{2}\right)-\left(y_{2}+y_{1}\right)^{2} \\
=\left(x_{2}+x_{1}\right)^{2}-\left[\frac{\left(y_{2}+y_{1}\right)^{2}-\left(y_{2}^{2}+y_{1}^{2}\right)}{2 p}\right]^{2}+ \\
\quad 2\left(y_{2}^{2}+y_{1}^{2}\right)-\left(y_{2}+y_{1}\right)^{2} \\
=(2 x)^{2}-\left[\frac{(2 y)^{2}-4 p x}{2 p}\right]^{2}+8 p x-(2 y)^{2},
\end{array}
\]

Thus, \( x=\frac{y^{2}}{2 p}+\frac{p l^{2}}{8\left(y^{2}+p^{2}\right)}\left(y^{2}<2 p x\right) \).
\[
\text { (1) } l \geqslant 2 p \text {. }
\]
\[
\begin{array}{l}
\text { Then } x=\frac{y^{2}}{2 p}+\frac{p l^{2}}{8\left(y^{2}+p^{2}\right)} \\
=\frac{y^{2}+p^{2}}{2 p}+\frac{p l^{2}}{8\left(y^{2}+p^{2}\right)}-\frac{p}{2} \\
\geqslant 2 \sqrt{\frac{y^{2}+p^{2}}{2 p} \cdot \frac{p l^{2}}{8\left(y^{2}+p^{2}\right)}}-\frac{p}{2} \\
=\frac{l-p}{2} .
\end{array}
\]

The equality holds if and only if \( y= \pm \sqrt{p\left(\frac{l}{2}-p\right)} \), hence \( x_{\min }=\frac{l-p}{2} \).
(2) \( 0<l<2 p \).
Then \( x=\frac{t}{2 p}+\frac{p l^{2}}{8 t}-\frac{p}{2}\left(t=y^{2}+p^{2} \geqslant p^{2}\right) \).
It is easy to see that this function is increasing on \(\left[p^{2},+\infty\right)\).
When \( t=p^{2} \), i.e., \( y=0 \), \( x_{\min }=\frac{l^{2}}{8 p} \).
In summary, when \( l \geqslant 2 p \), \( x_{\min }=\frac{l-p}{2} \); when \( 0<l<2 p \), \( x_{\min }=\frac{l^{2}}{8 p} \).