Given the hyperbola $\frac{x^{2}}{a}-\frac{y^{2}}{a}=1$ with its right focus at $F$, any line passing through $F$ intersects the right branch of the hyperbola at $M, N$. The perpendicular bisector of $M N$ intersects the $x$-axis at $P$. When $a$ takes any real number, the value of $\frac{\vdots F P}{|M N|}$ is $\qquad$
12. $\frac{\sqrt{2}}{2}$.

It is known that the eccentricity $e=\sqrt{2}$. Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, and the inclination angle of $M N$ be $\theta$, with the midpoint of $M N$ being $Q$. Then, $\left|M F \vdots=e x_{1}-a,\right| N F \mid=\mathrm{e} x_{2}-a$, so $2 ; F Q|=| M F-N F|=\mathrm{e}| x_{1}-x_{2} \mid$.
$$
\begin{aligned}
|F P| & =\frac{|F Q|}{\cos \theta}=\frac{e\left|x_{1}-x_{2}\right|}{2 \cos \theta} \\
& \left.=\frac{e}{2 \cos \theta} \right\rvert\, M N^{\prime} \cos \theta,
\end{aligned}
$$

Therefore,
$$
\frac{F P}{M N} \cdots=\frac{1}{2} \mathrm{e}=\frac{\sqrt{2}}{2} .
$$