Let $\mathcal P$ be a parabola, and let $V_1$ and $F_1$ be its vertex and focus, respectively. Let $A$ and $B$ be points on $\mathcal P$ so that $\angle AV_1 B = 90^\circ$. Let $\mathcal Q$ be the locus of the midpoint of $\overline{AB}$. It turns out that $\mathcal Q$ is also a parabola, and let $V_2$ and $F_2$ denote its vertex and focus, respectively. Determine the ratio $\frac{F_1F_2}{V_1V_2}$.
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 $AV_1$ is $a,$ and the slope of line $BV_1$ is $b.$  Since $\angle AV_1 B = 90^\circ,$ $ab = -1$. 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 $y = ax^2$ is $\left(0, \frac{1}{4a} \right)$. We find that $V_1 = (0,0)$, $V_2 = (0,1)$, $F_1 = \left( 0, \frac 14 \right)$, $F_2 = \left( 0, 1 + \frac18 \right)$. Therefore, $\frac{F_1F_2}{V_1V_2} = \boxed{\frac78}$.