2. Given the hyperbola $\frac{x^{2}}{t}-\frac{y^{2}}{t}=1$ with the right focus $F$, any line passing through $F$ intersects the right branch of the hyperbola at $M, N$, and the perpendicular bisector of $M N$ intersects the $x$-axis at $P$. When $t$ takes any positive real number except 0, $\frac{|F P|}{|M N|}=$ $\qquad$ .
2. $\frac{\sqrt{2}}{2}$.

Obviously, the eccentricity $e=\sqrt{2}$. Let $M\left(x_{1}, y_{1}\right), N\left(x_{3}\right.$, $\left.y_{2}\right)$, the slope angle of $M N$ be $a$, the midpoint of $M N$ be $Q$, and the distance from $M$ to the right directrix be $d$. Then, by the definition of the hyperbola, we have $\frac{|M F|}{d}=e$, i.e., $\frac{|M F|}{x_{1}-\frac{a^{2}}{c}}=e,|M F|=$ $e x_{1}-a$.
Similarly, $|N F|=e x_{2}-$
$a$,
$$
2|F Q|=\mid M F-
$$
$N F \mid$
$$
\begin{array}{l}
=\left|e x_{1}-a-e x_{2}+a\right|=e\left|x_{1}-x_{2}\right| . \\
|F P|=\frac{|F Q|}{\cos \alpha}=\frac{e\left|x_{1}-x_{2}\right|}{\cos \alpha} . \\
\text { Also, }\left|x_{1}-x_{2}\right|=M N \cdot \cos \alpha,
\end{array}
$$

Thus, $\frac{|F P|}{|M N|}=\frac{1}{2} e=\frac{\sqrt{2}}{2}$.