8. Given the hyperbola $x^{2}-y^{2}=t(t>0)$ with its right focus at $F$, any line passing through $F$ intersects the right branch of the hyperbola at points $M$ and $N$. The perpendicular bisector of $M N$ intersects the $x$-axis at point $P$. When $t$ is a positive real number not equal to zero, $\frac{|F P|}{|M N|}=$ $\qquad$ .
8. $\frac{\sqrt{2}}{2}$.

For the hyperbola $x^{2}-y^{2}=t$ with the right focus $F(\sqrt{2 t}, 0)$, the line $M N: y=k(x-\sqrt{2 t})$,
given the asymptotes are $y= \pm x$, then $|k|>1$.
From $\left\{\begin{array}{l}x^{2}-y^{2}=t \\ y=k(x-\sqrt{2 t})\end{array}\right.$, we get $\left(k^{2}-1\right) x^{2}-2 \sqrt{2 t} k^{2} x+$ $\left(2 k^{2}+1\right) t=0$

Thus, $|M N|=\frac{2 \sqrt{t}\left(k^{2}+1\right)}{k^{2}-1}$, and the midpoint $R$ of $M N$ is $\left(\frac{\sqrt{2 t} k^{2}}{k^{2}-1}, \frac{\sqrt{2 t} k}{k^{2}-1}\right)$,

The x-coordinate of the intersection point $P$ of the perpendicular bisector with the x-axis is $x_{p}=$ $\frac{2 \sqrt{2 t} k^{2}}{k^{2}-1}$,
Thus, $|P \dot{F}|=\frac{\sqrt{2 t}\left(k^{2}+1\right)}{k^{2}-1}$, so $\frac{|F P|}{|M N|}=\frac{\sqrt{2}}{2}$.