Example 12 (1995 National High School League Question) 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 family of curves on the line $y=2 x$.


The curve 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 on the coordinates of the other intersection point of the curve with $y=2 x$.
Substituting $y=2 x$ into the curve equation, we get $(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) x=0$.

Since $2 \sin \theta-\cos \theta=\sqrt{5} \sin \left(\theta-\arctan \frac{1}{2}\right)+3 \neq 0$, when $x \neq 0$,
$$
x=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3} .
$$

Let $\sin \theta=\frac{2 u}{1+u^{2}}, \cos \theta=\frac{1-u^{2}}{1+u^{2}}$, then $x=\frac{8 u+1}{2 u^{2}+2 u+1}$,
$$
2 x u^{2}+2(x-4) u+(x-1)=0 .
$$

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

Solving this, we get $-8 \leqslant x \leqslant 2$ and $x \neq 0$, so $|x|_{\max }=8$. From $y=2 x$, we get $|y|_{\max }=16$. Therefore, the maximum length of the chord is $\sqrt{8^{2}+16^{2}}=8 \sqrt{5}$.