Example 5 (Fujian Province Competition Question) Find the line that is intersected by each curve of the family $4 x^{2}+5 y^{2}-8 m x-20 m y+24 m^{2}-20=0$ (where $m$ is a real parameter) with a segment length of $\frac{5}{3} \sqrt{5}$.
The equation of the curve can be rewritten as $\frac{(x-m)^{2}}{5}+\frac{(y-2 m)^{2}}{4}=1$.
From this, we can see that the family of curves is a family of ellipses, with their centers $(m, 2 m)$ lying on the line $y=2 x$, and the major axis $2 a=2 \sqrt{5}$, minor axis $2 b=4$ are fixed. Therefore, to make the segments intercepted by the ellipses on the line all equal to $\frac{5}{3} \sqrt{5}$, the line must be parallel to the line $y=2 x$. Hence, we can assume the equation of the required line to be $y=2 x+b$, and substitute it into $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, yielding
$$
24 x^{2}+20 b x+5 b^{2}-20=0 .
$$
Let the intersection points of $l$ with the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$ be $M_{1}\left(x_{1}, y_{1}\right), M_{2}\left(x_{2}, y_{2}\right)$, then $x_{1}, x_{2}$ are the two roots of (1). By Vieta's formulas, we get $x_{1}+x_{2}=-\frac{5}{6} b, x_{1} x_{2}=\frac{5 b^{2}-20}{24}$.
Thus, by $y_{1}-y_{2}=2\left(x_{1}-x_{2}\right)$ and the given condition, we have
$$
\begin{aligned}
\left(\frac{5}{3} \sqrt{5}\right)^{2} & =\left|M_{1} M_{2}\right|^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=5\left(x_{1}-x_{2}\right)^{2} \\
& =5\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]=5\left[\left(-\frac{5}{6} b\right)^{2}-4 \times \frac{5 b^{2}-20}{24}\right],
\end{aligned}
$$

Solving for $b$ gives $b= \pm 2$. Therefore, the equations of the required lines are $y=2 x \pm 2$.