6. Given the ellipse $C: \frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ with left and right foci $F_{1}$ and $F_{2}$, and left and right vertices $A$ and $B$, the line $l: x=m y+1$ passing through the right focus $F_{2}$ intersects the ellipse $C$ at points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)\left(y_{1}>0, y_{2}<0\right)$. If $M A$ $\perp N F_{1}$, then the real number $m=$ $\qquad$.
6. $\frac{\sqrt{3}}{12}$.

Let $l_{M N}: x=m y+1$, substituting into the ellipse equation yields
$$
\left(8 m^{2}+9\right) y^{2}+16 m y-64=0 \text {. }
$$

Then $y_{1}+y_{2}=-\frac{16 m}{8 m^{2}+9}, y_{1} y_{2}=-\frac{64}{8 m^{2}+9}$
$\Rightarrow y_{1}=\frac{-8 m+24 \sqrt{m^{2}+1}}{8 m^{2}+9}$,
$$
y_{2}=\frac{-8 m-24 \sqrt{m^{2}+1}}{8 m^{2}+9} \text {. }
$$

Also, $k_{A M} k_{N F_{1}}=-1$
$$
\begin{array}{l}
\Leftrightarrow \frac{y_{1}}{x_{1}+3} \cdot \frac{y_{2}}{x_{2}+1}=-1 \\
\Leftrightarrow\left(m^{2}+1\right) y_{1} y_{2}+2 m y_{1}+4 m y_{2}+8=0
\end{array}
$$
$$
\begin{aligned}
\Rightarrow & \left(m^{2}+1\right)\left(-\frac{64}{8 m^{2}+9}\right)+2 m\left(-\frac{16 m}{8 m^{2}+9}\right)+ \\
& 2 m\left(\frac{-8 m-24 \sqrt{m^{2}+1}}{8 m^{2}+9}\right)+8=0 \\
\Rightarrow & 6 m^{2}+6 m \sqrt{m^{2}+1}-1=0 \\
\Rightarrow & 6 m=\frac{1}{\sqrt{m^{2}+1}+m}=\sqrt{m^{2}+1}-m \\
\Rightarrow & 7 m=\sqrt{m^{2}+1} \Rightarrow m=\frac{\sqrt{3}}{12} .
\end{aligned}
$$