56 Given that the coordinates of point $P$ $(x, y)$ satisfy
$$
\begin{array}{c}
(x-a) \cos \theta_{1}+y \sin \theta_{1}=a, \\
(x-a) \cos \theta_{2}+y \sin \theta_{2}=a, \\
\tan \frac{\theta_{1}}{2}-\tan \frac{\theta_{2}}{2}=2 c(c>1),
\end{array}
$$

where $\theta_{1}, \theta_{2}$ are parameters, then the trajectory of point $P$ is
A. Circle
B. Ellipse
C. Hyperbola
D. Parabola
56 C. Let $\tan \frac{\theta_{1}}{2}=t_{1}, \tan \frac{\theta_{2}}{2}=t_{2}$. Since $c>1$, we have $t_{1} \neq t_{2}, a \neq 0$. By the universal formula, the equations can be transformed into $x t_{1}^{2}-2 y t_{1}-(x-2 a)=0, x t_{2}^{2}-2 y t_{_{2}}-(x-2 a)=0$. Consider the auxiliary equation $x t^{2}-2 y t-(x-2 a)=0$. Since $\theta_{1} \neq \theta_{2}$, $t_{1}, t_{2}$ are the two roots of this equation. Thus, $t_{1}+t_{2}=\frac{2 y}{x}, t_{1} t_{2}=-\frac{x-2 a}{x}$. Therefore, $4 c^{2}=\left(t_{1}-t_{2}\right)^{2}=\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}=\left(\frac{2 y}{x}\right)^{2} +\frac{4(x-2 a)}{x}$, which simplifies to $\left(x+\frac{a}{c^{2}-1}\right)^{2}-\frac{y^{2}}{c^{2}-1}=\frac{a^{2}}{\left(c^{2}-1\right)^{2}}$.