$6 \cdot 109$ Given the family of curves
$$2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0$$

where $\theta$ is a parameter. Find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
[Solution] Obviously, the family of curves always passes through the origin, and the line $y=2 x$ also passes through the origin. Therefore, 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 family of curves with $y=2 x$.

Substituting $y=2 x$ into the equation of the family of curves, we get
$$(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) x=0$$

Noting that
$$2 \sin \theta-\cos \theta+3=\sqrt{5} \sin \left(\theta-\operatorname{arctg} \frac{1}{2}\right)+3 \neq 0,$$

so 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 $\square$
$$\begin{array}{l}
x=\frac{8 u+1}{2 u^{2}+2 u+1} \\
2 x u^{2}+2(x-4) u+(x-1)=0
\end{array}$$

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

i.e., $\square$
$$\begin{array}{l}
x^{2}+6 x-16 \leqslant 0 \quad \text { and } x \neq 0, \\
-8 \leqslant x \leqslant 2 \quad \text { and } x \neq 0 .
\end{array}$$

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