Question: As shown in Figure 1, from a point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ outside the parabola $\mathrm{y}^{2}=2 \mathrm{p} \times$, two tangent lines $\mathrm{PA}, \mathrm{PB}$ are drawn to the curve, with $A$ and $B$ being the points of tangency. Try to find: the equation of the chord $A B$ of the points of tangency.
As shown in Figure 1, let the points of tangency be $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$,
$$
\begin{array}{c}
B\left(x_{2}, y_{2}\right) \text {, then } \\
\text { the equation of PA is } y_{1} y=p\left(x_{1}+x\right), \\
\text { the equation of PB is } y_{2} y=p\left(x_{2}+x\right) \text {. } \\
\because P\left(x_{0}, y_{0}\right) \text { lies on PA and PB, } \\
\therefore y_{0} y_{1}=p\left(x_{1}+\right. \\
\left.x_{0}\right), \\
y_{0} y_{2}=p\left(x_{2}+x_{0}\right) .
\end{array}
$$

From (1) and (2), we know that $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ lie on the line $\mathrm{y}_{0} \mathrm{y}=\mathrm{p}\left(\mathrm{x}_{0}+\mathrm{x}\right)$, so the equation of the chord of contact $A B$ is:
$$
y_{0} y=p\left(x+x_{0}\right) \text {. }
$$