6. Let the parabola $C: y^{2}=2 x$ have its directrix intersect the $x$-axis at point $A$. Draw a line $l$ through point $B(-1,0)$ that is tangent to the parabola $C$ at point $K$. Draw a line parallel to $l$ through point $A$, intersecting the parabola $C$ at points $M, N$. Then the area of $\triangle K M N$ is $\qquad$.
Answer: $\frac{1}{2}$.
Solution: Let the slope of line $l$ and $M N$ be $k$, then $l: x=\frac{1}{k} y-1, M N: x=\frac{1}{k} y-\frac{1}{2}$. By combining $l$ with $C$, we get the equation $y^{2}-\frac{2}{k} y+2=0$. According to the condition, its discriminant is zero, hence $k= \pm \frac{\sqrt{2}}{2}$. By combining $M N$ with $C$, we get the equation $y^{2}-\frac{2}{k} y+1=0$, thus
$$
\left|y_{M}-y_{N}\right|=\sqrt{\left(y_{M}+y_{N}\right)^{2}-4 y_{M} y_{N}}=\sqrt{\frac{4}{k^{2}}-4}=2,
$$

Given that $l$ is parallel to $M N$, we know
$$
S_{\triangle K A N}=S_{\triangle B M N}=\left|S_{\triangle B A M}-S_{\triangle B A N}\right|=\frac{1}{2} \cdot|A B| \cdot\left|y_{M}-y_{N}\right|=\frac{1}{2} \cdot \frac{1}{2} \cdot 2=\frac{1}{2} \text {. }
$$