Example 11 Given the curve $y=x^{2}+(2 m-1) x+$ $m^{2}-2 m+1(m \in \mathbf{R})$. Does there exist a line $l$ with a defined slope that intersects the curve at only one point? If it exists, find this line; if not, explain the reason.
To solve the problem of converting the intersection issue into an equation issue, since $\Delta=0$, isolate the parameter $m$ and determine the values of $k$ and $b$.

Solution: Assume there exists a line $l$ with a defined slope that meets the requirements. Let the equation of line $l$ be $y=k x+b$.
$$
\text { From }\left\{\begin{array}{l}
y=k x+b, \\
y=x^{2}+(2 m-1) x+m^{2}-2 m+1
\end{array}\right.
$$

we get
$$
\begin{array}{l}
x^{2}+(2 m-1-k) x+m^{2}-2 m+1-b=0 \text {. } \\
\therefore \Delta=(2 m-1-k)^{2}-4\left(m^{2}-2 m+1\right. \\
-b)=0 \text {. } \\
\end{array}
$$

Simplifying, we get
$$
4(1-k) m+\left(k^{2}+2 k+4 b-3\right)=0
$$

which holds for all $m \in \mathbf{R}$.
Thus, we have $\left\{\begin{array}{l}4(1-k)=0, \\ k^{2}+2 k+4 b-3=0 .\end{array}\right.$
Solving these, we get $k=1, b=0$.
Therefore, there exists a line $l: y=x$ that meets the requirements.