2011 AMC 12B Problem 14

Let $p=FV$ and let the directrix be $\ell.$ Projecting $F$ and $B$ onto $\ell,$ the focus-directrix property gives $FB=2p$ (the distance from $B$ to $\ell$), and by the Pythagorean Theorem $$\begin{aligned} VB&=\sqrt{FV^2+FB^2} \\ &=\sqrt{p^2+4p^2}=\sqrt5\,p. \end{aligned}$$ Then $\cos(\angle FVB)=\dfrac{FV}{VB}$ $=\dfrac{p}{\sqrt5\,p}$ $=\dfrac{1}{\sqrt5}.$ Since $\angle AVB=2\angle FVB,$ $$\begin{aligned} \cos(\angle AVB) &=2\cos^2(\angle FVB) \\ &\quad {}-1 \\ &=2\cdot\dfrac15-1 \\ &=-\dfrac35. \end{aligned}$$ Thus, the correct answer is D.