设 $A\left(2pm,2pm^2\right)$，$B\left(2pn,2pn^2\right)$，则由抛物线的定义，可得\[\left(2pm^2+\dfrac p2\right)+\left(2pn^2+\dfrac p2\right)=4\cdot \dfrac p2,\]即\[m^2+n^2=\dfrac 12.\]双曲线的弦 $AB$ 的中点为 $\left(pm+pn,pm^2+pn^2\right)$，由双曲线的垂径定理，可得\[\dfrac{2pm^2-2pn^2}{2pm-2pn}\cdot \dfrac{pm^2+pn^2}{pm+pn}=\dfrac{b^2}{a^2},\]即\[m^2+n^2=\dfrac{b^2}{a^2},\]因此 $\dfrac{b^2}{a^2}=\dfrac 12$，双曲线的渐近线方程为 $y=\pm \dfrac{\sqrt 2}2x$．
其他方法：联立双曲线与抛物线的方程消去 $x$ 得$$\dfrac 1{b^2}y^2-\dfrac {2p}{a^2}y+1=0,$$设 $A(x_1,y_1),B(x_2,y_2)$，则有$$y_1+y_2=\dfrac {2pb^2}{a^2}.$$由抛物线的定义知$$|AF|+|BF|=y_1+\dfrac p2+y_2+\dfrac p2=\dfrac {2pb^2}{a^2}+p=4\cdot\dfrac p2,$$所以 $a^2=2b^2$，从而双曲线的渐近线方程为 $y=\pm\dfrac {\sqrt 2}2x$．