= Solution
Introduce a constant <Lagrange multiplier> $\lambda$ for the fixed <arc length>. Up to an irrelevant constant term, the augmented <functional> is
$$
J[y]=\int_{-a}^a(y+\lambda)\sqrt{1+y'^2}\,dx.
$$
It has no explicit $x$ dependence, so the <Beltrami identity> gives
$$
(y+\lambda)\sqrt{1+y'^2}-\frac{(y+\lambda)y'^2}{\sqrt{1+y'^2}}=\frac{y+\lambda}{\sqrt{1+y'^2}}=k.
$$
For the hanging minimum choose $k>0$. Writing $y_0=-\lambda$, this becomes $1+y'^2=(y-y_0)^2/k^2$. Its nonconstant smooth solution is
$$
\boxed{y(x)-y_0=k\cosh\left(\frac{x-x_0}{k}\right).}
$$
One may verify it without dividing by $y'$ at the lowest point: the full <Euler-Lagrange equation> gives $(y+\lambda)y''=1+y'^2$, which the expression satisfies everywhere. The branch with negative $k$ is the inverted stationary chain rather than the hanging minimum. The <catenary minimizes the integral of height along a fixed-length curve>, so the positive branch is also the global minimum, not merely a stationary candidate.
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