13.6.3 * The equation $y^{2}-4 x-2 y+1=0$, represents what kind of curve? If the line $y=k x+2$ is tangent to the curve represented by the equation, what should $k$ be?
Analysis one: Transform the equation $y^{2}-4 x-2 y+1=0$ into $x=\frac{1}{4}(y-1)^{2}$. Therefore, the original equation represents $(2 k-4) x+1=0$. (3) For the line to be a tangent to the parabola, the system of equations must have the same real solution, meaning equation (3) must have two identical real roots (i.e., a double root). Thus, the discriminant $\Delta=(2 k-4)^{2}-4 k^{2}-16 k+16=0$. Solving for $k$ yields $k=1$. Therefore, when $k=1$, the line $y=x+2$ is tangent to the parabola represented by the original equation.

Analysis two: The equation $y^{2}-4 x-2 y+1=0$ represents a parabola. Let $P\left(x_{1}, y_{1}\right)$ be a point on the parabola, then the equation of the tangent line at point $P$ is:
$$
y_{1} y-4 \cdot \frac{x+x_{1}}{2}-2 \cdot \frac{y+y_{1}}{2}+1=0 .
$$

Rearranging gives $2 x-\left(y_{1}-1\right) y+2 x_{1}+y_{1}-1=0$, (1) and $k x-y+2=0$.
If (1) and (2) represent the same line, then $\frac{k}{2}=\frac{1}{y_{1}-1}=\frac{2}{2 x_{1}+y_{1}-1}$, (3) and since point $P$ is on the parabola, $y_{1}^{2}-4 x_{1}-2 y_{1}+1=0$. (4) Solving the system of equations (3) and (4), we get $k=1$.

Therefore, when the line $y-k x+2$ is tangent to the parabola represented by the equation $y^{2}-4 x-2 y+1=0$, the value of $k$ should be 1.