Example 1. Find the equation of the chord of contact $\mathrm{AB}$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ from an external point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$.
Let the points of tangency be $A \left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and the equation of the tangent line PA be:
$$
\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1
$$
Since $P \in PA$, we have $\frac{x_{1} x_{0}}{a^{2}}+\frac{y_{1} y_{0}}{b^{2}}=1$.
Similarly, $\frac{x_{2} x_{0}}{a^{2}}+\frac{y_{2} y_{0}}{b^{2}}=1$.
Therefore, points $A$ and $B$ lie on the line $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$,
Thus, the equation of the chord of contact $AB$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
Using the same method, we can prove:
(1) For a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, the equation of the chord of contact from an external point $P\left(x_{0}, y_{0}\right)$ is $\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1$.
(2) For a parabola $y^{2}=2px$, the equation of the chord of contact from an external point $P\left(x_{0}, y_{0}\right)$ is:
$$
y_{0} y=p\left(x+x_{0}\right).
$$

In general, if $P\left(x_{0}, y_{0}\right)$ is an external point of a conic section, using the substitution rule for finding tangents (see the table below), we can obtain the equation of the chord of contact from $P$ to the conic section.
\begin{tabular}{|c|c|c|c|c|c|}
\hline Conic Section & $x^{2}$ & $y^{2}$ & $xy$ & $x$ & $y$ \\
\hline Substitution: & $x_{0} x$ & $y_{0} y$ & \begin{tabular}{l}
$x_{0} y+y_{0} x$
\end{tabular} & \begin{tabular}{l}
$\frac{x_{0}+x}{2}$
\end{tabular} & \begin{tabular}{l}
$\frac{y_{0}+y}{2}$
\end{tabular} \\
\hline
\end{tabular}