One, (25 points) Given the family of curves $2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0, \theta$ being the parameter, find the maximum length of the chord intercepted by the line $y=2 x$ on this curve.

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$8 \sqrt{5}$
I. 【Analysis and Solution】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 $y=2 x$ only depends 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
$$
(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta-\cos \theta+1) x=0 .
$$

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

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

i.e., $x^{2}+6 x-16 \leqslant 0$ and $x \neq 0$,
$$
\begin{array}{l}
(x+8)(x-2) \leqslant 0 \text { and } x \neq 0, \\
-8 \leqslant x \leqslant 2 \text { and } x \neq 0, \therefore|x|_{\text {max }}=8 .
\end{array}
$$

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