15. 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.
15. It is easy to see that the curve intersects the line $y=2 x$ at the origin. The x-coordinate of the other intersection point is
$$
x=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3} \text {. }
$$

Thus, the chord length $l=\sqrt{5}|x|=\sqrt{5} \cdot\left|\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3}\right|$.
Let $y=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3},(2 y-8) \sin \theta-(y+1) \cos \theta=1-3 y, \sqrt{(2 y-8)^{2}+(y+1)^{2}} \sin (\theta-\varphi)=$ $1-3 y, \sqrt{(2 y-8)^{2}+(y+1)^{2}} \leqslant|1-3 y|,-(y+8)(y-2) \geqslant 0, y \in[-8,2]$, hence $|y|_{\max }=8$, $l_{\max }=8 \sqrt{5}$.