13.6.13 ** Let the two tangents to the parabola $y^{2}=4 a x$ intersect at an angle $\theta$ (where $\theta$ is a constant), find the equation of the locus of the vertex of the angle.
Let the tangents be $l_{1}, l_{2}$, with points of tangency $P_{1}(x_{1}, y_{1})$ and $P_{2}(x_{2}, y_{2})$, respectively. Then the equation of $l_{1}$ is $y_{1} y = 2a(x + x_{1})$, (1) with the slope $k_{1} = \frac{2a}{y_{1}} = \frac{2ay_{1}}{y_{1}^{2}} = \frac{y_{1}}{2x_{1}}$, (1) which can be rewritten as $y = k_{1} x + \frac{a}{k_{1}}$. Thus, $k_{1}$ is a root of the equation $x \cdot u^{2} - y u + a = 0$ (2), where $(x, y)$ is any point on $l_{1}$. Therefore, $k_{1}$ and $k_{2}$ are the two roots of (2), where $(x, y)$ is the intersection point of $l_{1}$ and $l_{2}$. By Vieta's formulas, $k_{1} + k_{2} = \frac{y}{x}$ and $k_{1} k_{2} = \frac{a}{x}$. Hence,
$$
\tan^{2} \theta = \left(\frac{k_{2} - k_{1}}{1 + k_{1} k_{2}}\right)^{2} = \frac{\left(\frac{y}{x}\right)^{2} - 4 \cdot \frac{a}{x}}{\left(1 + \frac{a}{x}\right)^{2}} = \frac{y^{2} - 4ax}{(x + a)^{2}},
$$

Simplifying, we get
$$
\tan^{2} \theta \cdot x^{2} - y^{2} + 2a(2 + \tan^{2} \theta)x + a^{2} \tan^{2} \theta = 0.
$$

From this, we can see that in general, this equation represents a hyperbola; when $\theta = \frac{\pi}{2}$, $1 + k_{1} k_{2} = 0$, the trajectory is a vertical line parallel to the $y$-axis, $x = -a$.