jmc

Since all parabolas are similar, we may assume that $\mathcal{P}$ is the curve $y=x^2,$ so $V_1=(0,0).$ Then, if $A=(a,a^2)$ and $B=(b,b^2)$, the slope of line $A{V}_1$ is $a,$ and the slope of line $B{V}_1$ is $b.$ Since $\angle A{V}_{1}B=90^{\circ },$ $ab=-1$. Then, the midpoint of $\overline{AB}$ is $$\left(\frac{a+b}{2},\frac{a^2+b^2}{2}\right)=\left(\frac{a+b}{2},\frac{(a+b{)}^2-2ab}{2}\right)=\left(\frac{a+b}{2},\frac{(a+b{)}^2}{2}+1\right).$$(Note that $a+b$ can range over all real numbers under the constraint $ab=-1$.) It follows that the locus of the midpoint of $\overline{AB}$ is the curve $y=2x^2+1$.

Recall that the focus of $y=a{x}^2$ is $\left(0,\frac{1}{4a}\right)$. We find that $V_1=(0,0)$, $V_2=(0,1)$, $F_1=\left(0,\frac{1}{4}\right)$, $F_2=\left(0,1+\frac{1}{8}\right)$. Therefore, $\frac{F_1{F}_2}{V_1{V}_2}=\boxed{\frac{7}{8}}$.