= Solution
At a regular constrained stationary curve, choose a constant multiplier $\lambda$ and make the augmented functional $I-\lambda J$ stationary under endpoint-fixed variations. Its integrand is $H=F-\lambda G$ and the equation is $dH_{y'}/dx-H_y=0$. This normal multiplier formulation requires the <first variation> of the constraint not to vanish identically; the usual abnormal case would require two multipliers. When $H_x=0$,
$$
\frac d{dx}(y'H_{y'}-H)=y'\left(\frac d{dx}H_{y'}-H_y\right)-H_x=0,
$$
so $\boxed{y'H_{y'}-H=c_0}$ is constant, proving the requested first integral.
For the chain, $ds=\sqrt{1+(y')^2}dx$, so the fixed physical length is exactly $J[y]$. Its gravitational potential, apart from the constant factor $\rho g$, is $I[y]=\int_{-a}^ay\sqrt{1+(y')^2}dx$. Thus $H=(y-\lambda)\sqrt{1+(y')^2}$, and the Euler–Lagrange equation is
$$
\boxed{\frac d{dx}\left(\frac{(y-\lambda)y'}{\sqrt{1+(y')^2}}\right)-\sqrt{1+(y')^2}=0.}
$$
The first integral becomes $(y-\lambda)/\sqrt{1+(y')^2}=c$, after renaming the constant with the opposite sign. The differential equation also gives $(y-\lambda)y''=1+(y')^2$, so the nondegenerate solution integrates to $y=\lambda+c\cosh((x-x_0)/c)$. Equal endpoint heights force $x_0=0$: the two cosh values agree only when their arguments are equal or opposite, and $a>0$ excludes the former. Hence
$$
\boxed{y(x)=\lambda+c\cosh(x/c).}
$$
For the hanging energy <local minimum>[minimum] take $c>0$. Since $\sqrt{1+(y')^2}=\cosh(x/c)$,
$$
\boxed{L=2c\sinh(a/c),\qquad\lambda=b-c\cosh(a/c).}
$$
For $L>2a$ the length equation has exactly one positive solution: its right side decreases from infinity to $2a$ as $c$ increases from zero to infinity. For $L=2a$ the only feasible shape is the straight segment, obtained as a limiting case; for $L<2a$ no chain can span the endpoints. A negative $c$ gives the inverted stationary <catenary>, rather than the physically hanging <local minimum>[minimum].
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