smc

Let $\ell$ be the directrix of $\mathcal{P}$; recall that $\mathcal{P}$ is the set of points $T$ such that the distance from $T$ to $\ell$ is equal to $TF$. Let $P$ and $Q$ be the orthogonal projections of $F$ and $A$ onto $\ell$, and further let $X$ and $Y$ be the orthogonal projections of $F$ and $V$ onto $AQ$. Because $AF<AV$, there are two possible configurations which may arise, and they are shown below. Set $d=FV$, which by the definition of a parabola also equals $VP$. Then as $AQ=AF=20$, we have $AY=20-d$ and $AX=|20-2d|$. Since $FXYV$ is a rectangle, $FX=VY$, so by Pythagorean Theorem on triangles $AFX$ and $AVY$, $$\begin{array}{rl}21^2-(20-d{)}^2& =A{V}^2-A{Y}^2=V{Y}^2\\ & =F{X}^2=A{F}^2-A{X}^2=20^2-(20-2d{)}^2\end{array}$$ This equation simplifies to $3d^2-40d+41=0$, which has solutions $d=\frac{20\pm \sqrt{277}}{3}$. Both values of $d$ work - the smaller solution with the bottom configuration and the larger solution with the top configuration - and so the requested answer is $\boxed{\frac{40}{3}}$.