 Solution

Set $A(y)=y^{-1/2}+\lambda$ and $F(y,y^{\prime })=A(y)\sqrt{1+y^{\prime 2}}$. Since there is no explicit $x$ dependence, the Beltrami identity says $F-y^{\prime }{F}_{y^{\prime }}$ is constant along a stationary curve. One can verify it directly:by the Euler-Lagrange equation. Here $F_{y^{\prime }}=A(y)y^{\prime }/\sqrt{1+y^{\prime 2}}$, soThis 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.