= Solution
{c}
Set $A(y)=y^{-1/2}+\lambda$ and $F(y,y')=A(y)\sqrt{1+y'^2}$. Since there is no explicit $x$ dependence, the <Beltrami identity> says $F-y'F_{y'}$ is constant along a stationary curve. One can verify it directly:
$$
\frac d{dx}(F-y'F_{y'})=y'\left(F_y-\frac d{dx}F_{y'}\right)=0
$$
by the <Euler-Lagrange equation>. Here $F_{y'}=A(y)y'/\sqrt{1+y'^2}$, so
$$
\boxed{H=F-y'F_{y'}=\frac{y^{-1/2}+\lambda}{\sqrt{1+y'^2}}\quad\text{is constant}.}
$$
This identity applies on the open arch where $y>0$. The constant is finite by evaluating at any interior point. Near an endpoint $A(y)>0$, hence $H>0$ everywhere on such an arch. Thus also $A(y)>0$ throughout it.
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