1. 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 value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then
\[
\left\{\begin{array}{l}
2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0 \\
y=2 x
\end{array}\right.
\]
\[
\Rightarrow (2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) x=0 \Rightarrow |A B|=\left|\frac{\sqrt{5}(8 \sin \theta+\cos \theta+1)}{2 \sin \theta-\cos \theta+3}\right|
\]
Let $t=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3} \Rightarrow 8 \sin \theta+\cos \theta+1=t(2 \sin \theta-\cos \theta+3)$
\[
\Rightarrow (8-2 t) \sin \theta+(1+t) \cos \theta=\sqrt{(8-2 t)^{2}+(1+t)^{2}} \sin (\theta+\varphi)=3 t-1
\]
\[
\Rightarrow (8-2 t)^{2}+(1+t)^{2} \geqslant (3 t-1)^{2} \Rightarrow t^{2}+6 t-16=(t-2)(t+8) \leqslant 0 \Rightarrow -8 \leqslant t \leqslant 2.
\]
When $t=-8$, $24 \sin \theta-7 \cos \theta=-25 \Rightarrow \sin \theta=-\frac{24}{25}, \cos \theta=\frac{7}{25}$.
Therefore, the maximum length of the chord intercepted by the family of curves on the line $y=2 x$ is $8 \sqrt{5}$.