8 The line $\sqrt{3} x-y-\sqrt{3}=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$ ($A$ 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 $\mu^{2}-$ $\lambda^{2}=$ $\qquad$ .
$8 \frac{1}{2}$ Hint: The line $\sqrt{3} x-y-\sqrt{3}=0$ passes through the focus $(1, 0)$ of the parabola $y^{2}=4 x$. Substituting $y=\sqrt{3} x-\sqrt{3}$ into $y^{2}=4 x$, we get $3 x^{2}-10 x+3=0$. Therefore, $A(3, 2 \sqrt{3})$ and $B\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$. $\overrightarrow{O F}=(1,0), \overrightarrow{O A}=(3,2 \sqrt{3}), \overrightarrow{O B}=\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$. Since $\overrightarrow{O F}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B},(1,0)=\lambda(3,2 \sqrt{3})+$ $\mu\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$, solving gives $\lambda=\frac{1}{4}, \mu=\frac{3}{4}$, so $\mu^{2}-\lambda^{2}=\frac{1}{2}$.