6. Let the right focus of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F$, the eccentricity be $e$, and the line passing through point $F$ with a slope of 1 intersect the two asymptotes of the hyperbola at points $A$ and $B$. If the midpoint of $A B$ is $M,|F M|=c$, then $e=$ $\qquad$
6. $\sqrt[4]{2}$.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(x_{0}, y_{0}\right)$. Then
$$
\frac{\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)}{a^{2}}-\frac{\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)}{b^{2}}=0 \text {. }
$$

Substituting the coordinates of the midpoint $M$ and the slope 1, we get
$$
\frac{x_{0}}{a^{2}}-\frac{y_{0}}{b^{2}}=0 \text {. }
$$

Clearly, point $M$ is in the first quadrant.
Then $x_{0}=c+\frac{\sqrt{2}}{2} c, y_{0}=\frac{\sqrt{2}}{2} c$
$$
\begin{array}{l}
\Rightarrow \frac{b^{2}}{a^{2}}=\frac{y_{0}}{x_{0}}=\frac{\frac{\sqrt{2}}{2} c}{c+\frac{\sqrt{2}}{2} c}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1 \\
\Rightarrow e^{2}-1=\sqrt{2}-1 \\
\Rightarrow e=\sqrt[4]{2} .
\end{array}
$$