10. Let the line $y=\sqrt{3} x+b$ intersect the parabola $y^{2}=2 p x(p>0)$ at points $A, B$. The circle passing through $A, B$ intersects the parabola $y^{2}=2 p x(p>0)$ at two other distinct points $C, D$. Find the angle of intersection between the lines $A B$ and $C D$.
10. Solution: Let the equation of the circle passing through points $A, B, C, D$ be: $x^{2}+y^{2}+d x+e y+f=0$. Then the coordinates of $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$ are the solutions of the system of equations $\left\{\begin{array}{l}x^{2}+y^{2}+d x+e y+f=0 \\ y^{2}=2 p x\end{array}\right.$, eliminating $x$ gives $y^{4}+\left(4 p^{2}+2 p d\right) y^{2}+4 p^{2} e y+4 p^{2} f=0$.
Since $y_{1}, y_{2}, y_{3}, y_{4}$ are the four roots of the above equation, by Vieta's formulas we have: $y_{1}+y_{2}+y_{3}+y_{4}=0$.
And $k_{A B}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{y_{1}-y_{2}}{\frac{y_{1}^{2}}{2 p}-\frac{y_{2}^{2}}{2 p}}=\frac{2 p}{y_{1}+y_{2}}$, similarly $k_{C D}=\frac{2 p}{y_{3}+y_{4}}$. Thus, $k_{A B}+k_{C D}=\frac{2 p}{y_{1}+y_{2}}+\frac{2 p}{y_{3}+y_{4}}=\frac{2 p\left(y_{1}+y_{2}+y_{3}+y_{4}\right)}{\left(y_{1}+y_{2}\right)\left(y_{3}+y_{4}\right)}=0$.
Therefore, the slopes of lines $A B$ and $C D$ are complementary, and since the slope of the line $y=\sqrt{3} x+b$ is $60^{\circ}$, the slope of the other line is $120^{\circ}$.
Hence, the angle between lines $A B$ and $C D$ is $60^{\circ}$.