Hyperbola and the directix

Suppose that $A,B$ have coordinates $(-a\cosh u,b\sinh u)$ and $(a\cosh v,b\sinh v)$ respectively, It is a standard calculation that $FA \; = \; a(e\cosh u + 1) \hspace{2cm} FB \; = \; a(e\cosh v - 1)$ where $e$ is the eccentricity of the hyperbola. Let $C$ be the point of intersection of the line segment $AB$ with the angle bisector of $\angle AFB$ . Then we now that $C$ divides the line $AB$ in the ratio $AC \,:\, CB \; = \; AF\,:\, FB$ Thus $\overrightarrow{OC} \; = \; \frac{FB}{FA+FB}\overrightarrow{OA} + \frac{FA}{FA+FB}\overrightarrow{OB}$ and so, in particular, the $x$ -coordinate of $C$ is $\frac{e\cosh v - 1}{e(\cosh u + \cosh v)}(-a\cosh u) + \frac{e\cosh u + 1}{e(\cosh u + \cosh v)} a \cosh v \; = \; ae^{-1}$ so that $C$ lies on the directrix of the hyperbola that is closer to $F$ .

Thus, in this case, we must have $\angle AFC = \angle CFB = 30^\circ$ , and hence $\angle AFB = \boxed{60^\circ}$ .