Example 3 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. Try to find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
$(1995$, National High School Mathematics Competition)
Analysis: Substituting $y=2 x$ into the given family of curve equations yields $2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) 2 x=0$.
Solving for $x_{1}=0, x_{2}=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3}$.
To maximize the length of the chord intercepted, the absolute value of $x_{2}$ must be maximized. To utilize the boundedness of sine and cosine functions, the above equation is transformed into
$$
\left(2 x_{2}-8\right) \sin \theta-\left(x_{2}+1\right) \cos \theta=1-3 x_{2} \text {. }
$$

We get
$$
\begin{array}{l}
\sqrt{\left(2 x_{2}-8\right)^{2}+\left(x_{2}+1\right)^{2}} \sin (\theta+\varphi) \\
=1-3 x_{2} .
\end{array}
$$

Since $|\sin (\theta+\varphi)| \leqslant 1$, we have
$$
\begin{array}{l}
\sqrt{5 x_{2}^{2}-30 x_{2}+65} \geqslant 11-3 x_{2} \mid . \\
\text { Hence }-8 \leqslant x_{2} \leqslant 2 .
\end{array}
$$

Therefore, the maximum length of the chord intercepted by the family of curves on the line $y=2 x$ is
$$
l=\sqrt{\left(2^{2}+1\right)\left(x_{1}-x_{2}\right)^{2}}=8 \sqrt{5} .
$$