13.4.15 ** 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 $\frac{5}{3} \sqrt{5}$.
Given a family of curves
$$
4 x^{2}+5 y^{2}-8 m x-20 m y+24 m^{2}-20=0 .
$$

Let the equation of the required line be $y=k x+b$.
(2) Substituting (2) into (1) and simplifying, we get
$$
\left(4+5 k^{2}\right) x^{2}+(10 k b-8 m-20 m k) x+\left(5 b^{2}-20 m b+24 m^{2}-20\right)=0 \text {. }
$$

Let $x_{1} 、 x_{2}$ be the roots of equation (3). According to the problem, we have
$$
\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}=\left(\frac{5}{3} \sqrt{5}\right)^{2} \text {. }
$$

From (2), we get $\left(x_{2}-x_{1}\right)^{2}+\left(k x_{2}-k x_{1}\right)^{2}=\left(1+k^{2}\right)\left(x_{2}-x_{1}\right)^{2}=\frac{125}{9}$, which means
$$
\left(1+k^{2}\right)\left[\left(\frac{10 k b-8 m-20 m k}{4+5 k^{2}}\right)^{2}-4\left(\frac{5 b^{2}-20 m b+24 m^{2}-20}{4+5 k^{2}}\right)\right]=\frac{125}{9} \text {, }
$$

Simplifying, we get
$$
\begin{array}{l}
{\left[4(2+5 k)^{2}-24\left(4+5 k^{2}\right)\right] m^{2}-\left[20 k b(2+5 k)-20 b\left(4+5 k^{2}\right)\right] m} \\
+\left[25 k^{2} b^{2}-\left(4+5 k^{2}\right)\left(5 b^{2}-20\right)-\frac{125\left(4+5 k^{2}\right)^{2}}{36\left(1+k^{2}\right)}\right]=0 .
\end{array}
$$

Since for all $m$ the roots of equation (3) satisfy condition (4), (5) is an identity in $m$, so we have $4(2+$ $5 k)^{2}-24\left(4+5 k^{2}\right)=0$, and $20 k b(2+5 k)-20 b\left(4+5 k^{2}\right)=0$, and
$$
25 k^{2} b^{2}-\left(4+5 k^{2}\right)\left(5 b^{2}-20\right)-\frac{125\left(4+5 k^{2}\right)^{2}}{36\left(1+k^{2}\right)}=0 .
$$

Solving, we get $k=2, b= \pm 2$. Therefore, the required line equations are $y=2 x \pm 2$.
Analysis two The family of curves can be rewritten as
$$
\frac{(x-m)^{2}}{5}+\frac{(y-2 m)^{2}}{4}=1 \text {. }
$$

From this, we see that the family of curves is a family of ellipses, with centers $(m, 2 m)$ on the line $y=2 x$, and the axes have fixed lengths and are parallel to the coordinate axes. Therefore, any ellipse in the family can be obtained by translating the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, and the line that intercepts equal segments from each ellipse must be parallel to the line $y=2 x$. Let the equation of the required line be $y=2 x+b$, substituting into $\frac{x^{2}}{5}$ $+\frac{y^{2}}{4}=1$ we get
$$
24 x^{2}+20 b x+5 b^{2}-20=0 .
$$

Let $x_{1} 、 x_{2}$ be the roots of equation (2). According to the problem, we have
$$
\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}=\left(\frac{5}{3} \sqrt{5}\right)^{2},
$$

which means $5\left(x_{2}-x_{1}\right)^{2}=\frac{125}{9}$, or $\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\frac{25}{9}$. From (2) $\left(\frac{20 b}{24}\right)^{2}-4\left(\frac{5 b^{2}-20}{24}\right)=$ $\frac{25}{9}$, solving gives $b= \pm 2$. Therefore, $y=2 x \pm 2$ are the required line equations.