176 Given the parabola $y^{2}=x, A B$ is a chord passing through the focus $F$, $A$ and $B$ are in the $\mathrm{I}$ and $\mathrm{N}$ quadrants respectively, if $A F-F B=1$, then the slope of $A B$ $=$ $\qquad$
$176 \frac{\sqrt{5}+1}{2}$. The coordinates of the focus $F$ are $\left(\frac{1}{4}, 0\right) . A F-F B=1$ which means $x_{A}-x_{B}=1$. Let the slope be $k$, then the equation of $A B$ is
$$
y=k\left(x-\frac{1}{4}\right) .
$$

Eliminating $y$ with $y^{2}=x$ gives
$$
k^{2}\left(x-\frac{1}{4}\right)^{2}=x
$$

which simplifies to
$$
\begin{array}{c}
k^{2} x^{2}-\left(\frac{k^{2}}{2}+1\right) x+\frac{k^{2}}{16}=0 \\
1=x_{A}-x_{B}=\frac{\sqrt{\Delta}}{k^{2}}=\frac{1}{k^{2}} \sqrt{\left(\frac{k^{2}}{2}+1\right)^{2}-4 k^{2} \cdot \frac{k^{2}}{16}}=\frac{\sqrt{1+k^{2}}}{k^{2}} .
\end{array}
$$

Therefore,
$$
k^{4}=k^{2}+1 . k=\frac{1+\sqrt{5}}{2} \text { (only the positive value is taken). }
$$