Example 2 Given the family of curves $2(2 \sin \theta-\cos \theta) x^{2}-(8 \sin \theta+\cos \theta+1) y=0, \theta$ 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.
(1995 National High School League Question)
Solution: It is evident that the family of curves and the line $y=2x$ both pass through the origin. Substituting $y=2x$ into the family of curves yields $(2 \sin \theta - \cos \theta + 3)x^2 - (8 \sin \theta - \cos \theta + 1)x = 0$. Noting that $2 \sin \theta - \cos \theta + 3 = \sqrt{5} \sin \left(\theta - \arctan \frac{1}{2}\right) + 3 \neq 0$, the roots of the equation are $x_1 = 0, x_2 = \frac{8 \sin \theta - \cos \theta + 1}{2 \sin \theta - \cos \theta + 3}$.
Let the length of the intercepted chord be $l$, then $l = \sqrt{1^2 + 2^2} |x_2 - x_1| = \sqrt{5} |x_2|$.
Let $x_2 = l$, then $(8 - 2t) \sin \theta + (1 + t) \cos \theta = 3t + 1$, which simplifies to $\sin (\theta + \varphi) = \frac{3t - 1}{\sqrt{(8 - 2t)^2 + (1 + t)^2}}$. Therefore, by $\left|\frac{3t - 1}{\sqrt{(8 - 2t)^2 + (1 + t)^2}}\right| \leq 1$, we have $4t^2 + 24t - 64 \leq 0$, which simplifies to $-8 \leq t \leq 2 (t \neq 0)$. Hence, $|t| \leq 8$, with equality holding if and only if $t = -8$, i.e., when $\cos \varphi = \frac{24}{25}, \sin \varphi = \frac{7}{25}$, and $\sin (\theta + \varphi) = -1$. Therefore, when $\theta = \frac{3\pi}{2} + \arcsin \frac{7}{25}$, $|t|_{\min} = 8\sqrt{5}$, which means the maximum chord length is $8\sqrt{5}$.