
Place the parabola as $y^2=4px$ with vertex $V=(0,0)$ and focus $F=(p,0)$. At $x=p$ we get $y=\pm2p$, so $A=(p,2p)$ and $B=(p,-2p)$.
- In the right triangle $VFA$: $VF=p$, $FA=2p$, so $VA=\sqrt{p^2+4p^2}=p\sqrt5$. Let $\theta=\angle AVF$; then $\cos\theta=\frac{p}{p\sqrt5}=\frac1{\sqrt5}$ and $\sin\theta=\frac2{\sqrt5}$.
- The figure is symmetric about the axis, so $\angle AVB=2\theta$.
- $\cos2\theta=\cos^2\theta-\sin^2\theta=\frac15-\frac45=\boxed{-\frac35}$.
- Answer is (D).