6. Given a parabola with focus at $(0,0)$, and passing through points $(-1,-1)$ and $(2,2)$. Then the slope of its directrix is $\qquad$

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6. $\frac{9 \pm 4 \sqrt{2}}{7}$.

By the definition of a parabola, the directrix is the 
external common tangent of 
$\odot C_{1}:(x+1)^{2}+(y+1)^{2}=2$
and $\odot C_{2}:(x-2)^{2}+(y-2)^{2}=8$.
Let the intersection point of the two external common tangents be $P$. Then $P$ lies on the line $C_{1} C_{2}$ (i.e., $y=x$), and $\frac{P C_{1}}{P C_{2}}=\frac{1}{2}$.
Solving this, we get $P(-4,-4)$.
Thus, the equation of the common tangent line is $y=k(x+4)-4$.
Since the distance from point $C_{1}$ to this line is $\sqrt{2}$
$$
\begin{array}{l}
\Rightarrow \frac{|3 k-3|}{\sqrt{1+k^{2}}}=\sqrt{2} \\
\Rightarrow k=\frac{9 \pm 4 \sqrt{2}}{7} .
\end{array}
$$