I have been searching for any published work on this definition of a parabolic arc. The latus rectum of a parabola is a defining chord drawn parallel to the directrix and passes thru the focus.
In the example below I have labeled the latus rectum as segment $\overline{S_1S_2}$ .
Take segment $\overline{S_1S_2}$ and subdivide it into 2 line segments using point $B$, which then creates the 2 segments $\overline{S_1B}$ and $\overline{BS_2}$. Point $B$ lies on $\overline{S_1S_2}$ and is able to traverse the entire latus rectum.
Let segment $\overline{S_1B}$ = $m$, segment $\overline{BS_2}$ = $n$, and the product of $m \cdot n$ = $s$
Next, construct segment $\overline{BD}$ perpendicular to $\overline{S_1S_2}$ with a magnitude of $s$. The locus of point $D$ defines a parabolic arc from the but not including the endpoints of the latus rectum, thru the vertex. This excludes the endpoints. i.e. when $S_1$ = $B$, and $S_2$ = $B$, $|S_1B|$ or $|BS_2|$= $0$
This definition can also be shown in Cartesian coordinates. Let $S_1$ = $(s_1,0)$, $B=(b,0)$ and $S_2$ = $(s_2,0)$. The function $f(x)=(x-s_1)(x-s_2)$ gives a one-to-one correspondence for values $s_1<b<s_2$
A GeoGebra link:
Defining the length of $BD$ as an area $s=mn$ is not scale-invariant. As a result the locus of $D$ is always a parabola, but $S_1S_2$ is its latus rectum only if $S_1S_2=1$. To see why, compute $BD$ when $m=n$: $BD=m^2$. But for $m=n$ point $D$ is the vertex of the parabola and $BD={1\over4}\cdot\ \text{latus rectum}$. If $S_1S_2$ is the latus rectum then we have $m^2={1\over2}m$, which is true only if $m={1\over2}$.
You can correct this by setting $$ BD={mn\over m+n}. $$ With this definition the locus of $D$ is a parabola with $S_1S_2$ as latus rectum.