8. Given the line $\sqrt{3} x-y-\sqrt{3}=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$ (point $A$ is above the $x$-axis), and intersects the $x$-axis at point $F$. If $\overrightarrow{O F}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}$, then $\mu^{2}-\lambda^{2}=$ $\qquad$ .
8. $\frac{1}{2}$.

It is easy to know that $F(1,0)$ is the focus of the parabola $y^{2}=4 x$.
Substituting $y=\sqrt{3} x-\sqrt{3}$ into $y^{2}=4 x$, we get
$$
3 x^{2}-10 x+3=0 \text {. }
$$

Then $A(3,2 \sqrt{3})$ and $B\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$.
By the given condition,
$(1,0)=\lambda(3,2 \sqrt{3})+\mu\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$.
Solving, we get $\lambda=\frac{1}{4}, \mu=\frac{3}{4}$.
Thus, $\mu^{2}-\lambda^{2}=\frac{1}{2}$.