10. Let the focus of the parabola $y^{2}=4 x$ be $F$, and draw any line through point $F$ intersecting the parabola at points $A$ and $B$. Does there exist a line $l$ such that for any point $M$ on line $l$, the slopes of lines $M A$, $M F$, and $M B$ always form an arithmetic sequence? If it exists, find the equation of line $l$; if not, explain the reason.
10. Solution: $F(1,0)$, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, the equation of the line passing through $F$ is $x=m y+1$ combined with $y^{2}=4 x$ gives: $y^{2}=4 m y+4, y_{1}+y_{2}=4 m, \quad y_{1} \cdot y_{2}=-4$,
Assume there exists a point $M(a, b)$ on the line $l$ such that $2 k_{M F}=k_{M A}+k_{M B}$ always holds, i.e.,
$\frac{2 b}{a-1}=\frac{b-y_{1}}{a-x_{1}}+\frac{b-y_{2}}{a-x_{2}}=\frac{2 a b-b\left(x_{1}+x_{2}\right)-a\left(y_{1}+y_{2}\right)+x_{1} y_{2}+x_{2} y_{1}}{a^{2}-a\left(x_{1}+x_{2}\right)+x_{1} x_{2}}$ always holds.
And $x_{1}+x_{2}=4 m^{2}+2, x_{1} x_{2}=1, x_{1} y_{2}+x_{2} y_{1}=-4 m, \frac{2 b}{a-1}=\frac{2 a b-b\left(4 m^{2}+2\right)-4 m(a+1)}{a^{2}-a\left(4 m^{2}+2\right)+1}$.
Expanding and comparing the coefficients of $m$ gives $a=-1, b \in \mathbf{R}$, hence there exists a line $l$, its equation is $x=-1$.