11. Let $F$ be the left focus of the hyperbola $x^{2}-\frac{y^{2}}{3}=1$. Draw a line $AB$ through $F$ intersecting the hyperbola at points $A$ and $B$. Construct the tangents $l_{1}$ and $l_{2}$ to the hyperbola at points $A$ and $B$ respectively, and let $l_{1}$ and $l_{2}$ intersect at point $M$.
(1) Find the angle between the vector $MF$ and $AB$; (2) Find the minimum value of the area of $\triangle MAB$.
11. Solution: (1) It is easy to know that $F(-2,0)$, let the equation of the line be $x=m y-2, A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right), M\left(x_{M}, y_{M}\right)$,

Substituting into the hyperbola equation yields: $\left(m^{2}-\frac{1}{3}\right) y^{2}-4 m y+3=0$, hence $y_{A}+y_{B}=\frac{12 m}{3 m^{2}-1}, y_{A} y_{B}=\frac{9}{3 m^{2}-1}$; the equation of line $l_{1}$ is: $x x_{A}-\frac{y y_{A}}{3}=1$, the equation of line $l_{2}$ is: $x x_{B}-\frac{y y_{B}}{3}=1$;

Since $M F \bullet A B=0$, the angle between them is $90^{\circ}$;
(2) $S_{\triangle M A B}=\frac{1}{2} A B \bullet M F=\frac{3}{4}\left(m^{2}+1\right)\left|y_{A}-y_{B}\right|=\frac{9}{2}\left(m^{2}+1\right) \frac{\sqrt{m^{2}+1}}{\left|3 m^{2}-1\right|}$

Let $t=\frac{1}{\sqrt{m^{2}+1}}$, then $t \in(0,1], S_{\triangle M A B}=\frac{9}{2} \frac{1}{\left|3 t-4 t^{3}\right|}$, it is easy to know that $f(t)=3 t-4 t^{3}$ is monotonically increasing on $\left(0, \frac{1}{2}\right)$ and monotonically decreasing on $\left(\frac{1}{2}, 1\right)$, hence when $t \in(0,1]$, $f(t) \in[-1,1]$, i.e., $S_{\triangle M A B} \in\left[\frac{9}{2},+\infty\right)$, when $m= \pm \sqrt{3}, 0$, i.e., the equation of line $A B$ is $x= \pm \sqrt{3} y-2$ or $x=-2$, $S_{\triangle M A B}$ reaches its minimum value $\frac{9}{2}$.