Example 1. $\mathrm{P}, \mathrm{Q}$ are two moving points on the parabola $\mathrm{y}^{2}=2 \mathrm{mx}$, $\mathrm{O}$ is the vertex of the parabola, $\angle \mathrm{POQ}=\alpha$ (a constant). The tangents at points $P, Q$ intersect at point $T$. Find the equation of the locus of point $T$.
Solve for points $P$ and $Q$ with their y-coordinates $y_1$ and $y_2$ as parameters.
Since points $P$ and $Q$ are on the parabola, the coordinates of points $P$ and $Q$ are:
$$
\left(\frac{y_1^2}{2 m}, y_1\right) \left(\frac{y_2^2}{2 m}, y_2\right).
$$

The equations of the tangents $\mathrm{PT}$ and $\mathrm{QT}$ are:
$$
\begin{array}{l}
y_1 y = m\left(x + \frac{y_1^2}{2 m}\right) \\
y_2 y = m\left(x + \frac{y_2^2}{2 m}\right).
\end{array}
$$
The coordinates of point $\mathrm{T}$ are: $\mathrm{x} = \frac{y_1 y_2}{2 m}$,
$$
y = \frac{y_1 + y_2}{2}.
$$

If $\alpha \neq \frac{\pi}{2}$, the slopes of $\mathrm{PO}$ and $\mathrm{QO}$ are:
$$
\begin{array}{c}
\frac{2 m}{y_1}, \frac{2 m}{y_2}, \\
\therefore \operatorname{tg} \alpha = \left|\frac{\frac{2 m}{y_1} - \frac{2 m}{y_2}}{1 + \frac{2 m}{y_1} \cdot \frac{2 m}{y_2}}\right| \\
= \left|\frac{2 m (y_2 - y_1)}{y_1 y_2 + 4 m^2}\right|.
\end{array}
$$

From (1): $y_1 y_2 = 2 m x, y_1 + y_2 = 2 y$,
$$
\begin{aligned}
\therefore |y_2 - y_1| & = \sqrt{(y_1 + y_2)^2 - 4 y_1 y_2} \\
& = \sqrt{4 y^2 - 8 m x}.
\end{aligned}
$$

Substituting into (2) gives:
$\operatorname{tg} \alpha = \left|\frac{2 m \sqrt{4 y^2 - 8 m x}}{2 m x + 4 m^2}\right|$, squaring and rearranging
yields:
$$
\begin{array}{l}
\operatorname{tg}^2 \alpha \cdot x^2 - 4 y^2 + 4 m \operatorname{tg}^2 \alpha \cdot x + 8 m x \\
+ 4 m^2 \operatorname{tg}^2 \alpha = 0.
\end{array}
$$

This is the equation of the trajectory of point $\mathrm{T}$.
$$
\begin{array}{l}
\alpha = \frac{\pi}{2} \text{ when, } \\
\frac{2 m}{y_1} \cdot \frac{2 m}{y_2} = -1, \\
y_1 y_2 + 4 m^2 = 0.
\end{array}
$$

Substituting $y_1 y_2 = 2 m x$ gives:
$2 m x + 4 m^2 = 0$, which simplifies to $x + 2 m = 0$ as the required equation.