= Solution
Let $\eta$ be a smooth <variation> with $\eta(a)=\eta(b)=0$. Assuming the usual continuous second derivatives of $F$, stationarity of the <functional> under $y+t\eta$ gives its <first variation>
$$
0=\left.\frac d{dt}J[y+t\eta]\right|_{t=0}=\int_a^b(F_y\eta+F_{y'}\eta')\,dx.
$$
<Integration by parts> removes the derivative of $\eta$; its boundary term vanishes because the endpoint values are fixed. Thus
$$
\int_a^b\left(F_y-\frac d{dx}F_{y'}\right)\eta\,dx=0
$$
for every such <variation>. The <fundamental lemma of the calculus of variations> implies the <Euler-Lagrange equation> $F_y=(d/dx)F_{y'}$.
For <arc length>, $F=\sqrt{1+(y')^2}$, so
$$
\frac d{dx}\frac{y'}{\sqrt{1+(y')^2}}=0.
$$
The function $s\mapsto s/\sqrt{1+s^2}$ is strictly increasing; hence $y'$ is constant. With endpoints $(a,A)$ and $(b,B)$, $a<b$, the extremal is
$$
\boxed{y(x)=A+\frac{B-A}{b-a}(x-a)}.
$$
It is also the global shortest curve: integrating the unit displacement vector dotted with the curve velocity gives the endpoint distance, which is at most the integrated speed by <Cauchy-Schwarz inequality>; the straight segment attains equality. A vertical segment is described parametrically or after rotating axes, since it cannot be represented by $y(x)$ on a nontrivial horizontal interval.
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