Example 4 Given the family of curves $2(2 \sin \theta-\cos \theta+3)$. $x^{2}-(8 \sin \theta+\cos \theta+1) y=0, \theta$ is a parameter. Find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
(1995, National High School Mathematics Competition)
Solution: Clearly, the family of curves always passes through the origin, and the line $y$ $=2 x$ also passes through the origin, so the length of the chord intercepted by the family of curves on $y=2 x$ depends only on the coordinates of the other intersection point of the family of curves with $y=2 x$.
Substituting $y=2 x$ into the equation of the family of curves, we get
$$
\begin{array}{l}
(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta-\cos \theta \\
+1) x=0 .
\end{array}
$$

Also, $2 \sin \theta-\cos \theta+3$
$$
=\sqrt{5} \sin \left(\theta-\arctan \frac{1}{2}\right)+3 \neq 0 \text {. }
$$

When $x \neq 0$, we have $x=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3}$.
Let $\sin \theta=\frac{2 u}{1+u^{2}}, \cos \theta=\frac{1-u^{2}}{1+u^{2}}$.
Then $x=\frac{8 u+1}{2 u^{2}+2 u+1}$.
$$
\therefore 2 x u^{2}+2(x-4) u+(x-1)=0 \text {. }
$$

Since $u \in \mathbf{R}$, when $x \neq 0$,
$$
\begin{aligned}
\Delta & =[2(x-4)]^{2}-8 x(x-1) \\
& =4\left(-x^{2}-6 x+16\right) \geqslant 0 .
\end{aligned}
$$

Solving this, we get $-8 \leqslant x \leqslant 2$, and $x \neq 0$.
$$
\therefore|x|_{\max }=8 \text {. }
$$

Since $y=2 x$, we get $|y|_{\max }=16$.
Therefore, the maximum length of the chord is $\sqrt{8^{2}+16^{2}}=8 \sqrt{5}$.