Example 1 Try to determine the range of values for $m$ such that for the line $y=4x+m$, there are two distinct points $A, B$ on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ that are symmetric with respect to this line.

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Let $P\left(x_{0}, y_{0}\right)$ be the midpoint of the chord $AB$. By Property 10, the curve $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ is symmetric about point $P$ to the curve $\frac{\left(2 x_{0}-x\right)^{2}}{4}+\frac{\left(2 y_{0}-y\right)^{2}}{3}=1$. Subtracting and rearranging the two equations yields the equation of the common chord:
$$
2 x_{0} x+4 y_{0} y-3 x_{0}^{2}-4 y_{0}^{2}=0.
$$

The slope of the common chord should be $-\frac{1}{4}$, so we have $k=-\frac{3 x_{0}}{4 y_{0}}=-\frac{1}{4}$, which implies $y_{0}=3 x_{0}$. Since $P\left(x_{0}, y_{0}\right)$ lies on the line $y=4 x+m$, we have $y_{0}=4 x_{0}+m$. Solving these two equations gives $x_{0}=-m, y_{0}=-3 m$.
Since $P\left(x_{0}, y_{0}\right)$ is inside the ellipse, we have $\frac{x_{0}^{2}}{4}+\frac{y_{0}^{2}}{3}<1$, which means $\frac{(-m)^{2}}{4}+\frac{(-3 m)^{2}}{3}<1$. Therefore, $-\frac{2 \sqrt{13}}{13}<m<\frac{2 \sqrt{13}}{13}$ is the solution.