Example 3. Draw a chord $AB$ of the curve $\left\{\begin{array}{l}x=4 \cos \theta, \\ y=2 \sin \theta\end{array}\right.$ (where $\theta$ is a parameter) through the point $M(2,1)$, and $M$ is the point that trisects $AB$. Find the equation of $AB$.
Here, the curve should use the ordinary equation: $x^{2}+4 y$ $=16, AB$ should use the standard parametric form (as it involves chord length): $\left\{\begin{array}{l}x=2+t \cos a, \\ y=1+t \sin \alpha .\end{array}\right.$
( $t$ is the parameter) Substituting it into the curve equation and simplifying:
$$
\begin{array}{l}
\left(\cos ^{2} \alpha+4 \sin ^{2} \alpha\right) t^{2}+4(\cos \alpha \\
+2 \sin \alpha) t-8=0 .
\end{array}
$$

According to the problem, $M$ trisects $AB$, so $t_{1}=-2 t_{2}$, hence $t_{1}+t_{2}=-t_{2}, t_{1} t_{2}=-2 t_{2}^{2}=-2\left(t_{1}+t_{2}\right)^{2}$,

which means
$$
\begin{aligned}
& -\frac{8}{\cos ^{2} \alpha+4 \sin ^{2} \alpha} \\
= & -2\left[\frac{4(\cos \alpha+2 \sin \alpha)}{\cos ^{2} \alpha+4 \sin ^{2} \alpha}\right]^{2} .
\end{aligned}
$$

Simplifying: $\square$
$$
\begin{array}{l}
12 \sin ^{2} \alpha+16 \sin \alpha \cos \alpha+3 \cos ^{2} \alpha=0, \\
12 \operatorname{tg}^{2} \alpha+16 \operatorname{tg} \alpha+3=0, \\
\operatorname{tg} \alpha=-\frac{4 \pm \sqrt{7}}{6} .
\end{array}
$$

Therefore, the equation of line $AB$ is
$$
y-1=-\frac{4 \pm \sqrt{7}}{6}(x-2) .
$$
(Author's affiliation: Tianjin Nanlou High School)