6. Let $A B$ be a chord of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passing through its left focus $F$, with point $A$ above the $x$-axis. If $F$ divides $A B$ in the ratio $\lambda(\lambda \neq 1)$, then the slope of $A B$ $k=$ $\qquad$
6. $\frac{\sqrt{4 \lambda a^{2}-(\lambda+1)^{2} b^{2}}}{(\lambda-1) a}$.

Let the left directrix of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ be $l$, and the projections of points $A$ and $B$ on $l$ be $A_{1}$ and $B_{1}$, respectively.
$$
\begin{array}{l}
\text { Then } \frac{\left|A A_{1}\right|}{\left|B B_{1}\right|}=\frac{|A F|}{|B F|}=\lambda . \\
\text { Let }\left|B B_{1}\right|=t(t>0) .
\end{array}
$$

Then $\left|A A_{1}\right|=\lambda t$.
It is easy to see that, $|A B|=e(\lambda+1) t$,
$$
\begin{array}{l}
x_{A}-x_{B}=(\lambda-1) t, \\
y_{A}-y_{B}=\sqrt{|A B|^{2}-\left(x_{A}-x_{B}\right)^{2}} \\
=t \sqrt{e^{2}(\lambda+1)^{2}-(\lambda-1)^{2}} .
\end{array}
$$

Therefore, $k=\frac{y_{A}-y_{B}}{x_{A}-x_{B}}=\frac{\sqrt{4 \lambda a^{2}-(\lambda+1)^{2} b^{2}}}{(\lambda-1) a}$.