$$
2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0,
$$
where $\theta$ is a parameter.
Find the maximum value of the chord length of this family of curves on the line $y=2 x$.
Obviously, 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
$(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta-\cos \theta+1) x=0$.

Also, $2 \sin \theta-\cos \theta+3=\sqrt{5} \sin \left(\theta-\operatorname{arctg} \frac{1}{2}\right)+3 \neq 0$,
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 $x=\frac{8 u+1}{2 u^{2}+2 u+1}$.
We have $2 x u^{2}+2(x-4) u+(x-1)=0$.
Since $u \in R$, when $x \neq 0$,
$$
\begin{aligned}
\Delta & =[2(x-4)]^{2}-8 x(x-1) \\
& =4\left(-x^{2}-6 x+16\right) \geqslant 0,
\end{aligned}
$$

That is
$$
\begin{array}{l}
x^{2}+6 x-16 \leqslant 0 \text { and } x \neq 0, \\
(x+8)(x-2) \leqslant 0 \text { and } x \neq 0, \\
-8 \leqslant x<2 \text { and } x \neq 0 . \\
\therefore|x|_{\operatorname{sos}}=8 .
\end{array}
$$

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