25~26高三下·安徽江南十校3月联考·第8题

A

如图，作出抛物线$C$和圆$E$在点$M$处的公共切线$l$，同时过$M$作射线$MN \parallel x$轴，

则有$EM \perp l$，由抛物线的光学性质，可得$\angle FME = \angle EMN = \angle FEM = \dfrac{\pi}{6}$，

$\therefore FE = FM = x_M + \dfrac{p}{2} = 3 + \dfrac{p}{2}$，$F\left(\dfrac{p}{2},0\right)$，$\therefore E(3+p,0)$，

且$k_{EM} = \tan\angle EMN = \dfrac{\sqrt{3}}{3} = \dfrac{0 – y_M}{(3+p) – 3}$，

又$y_M^2 = 2p \times 3 = 6p$，代入得$\dfrac{1}{3}p^2 = 6p$，解得$p = 18$.