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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.

For light speed $c(y)=y>0$, the travel-time integrand is $\sqrt{1+y^{\prime 2}}/y$. The Beltrami identity gives $y\sqrt{1+y^{\prime 2}}=R$, so nonvertical rays are semicircles centred on $y=0$. Boundary endpoints at $y=0$ have infinite travel time: along any rectifiable path entering positive height, $ds/y\ge |dy|/y$ and the integral diverges. These semicircles are therefore ideal-endpoint hyperbolic geodesics, not finite-time optical minimizers joining zero-speed endpoints.

For positive $\lambda$ and light speed $c(y)=c_0{e}^{\lambda y}$, the Fermat principle uses the time integrand $c_0^{-1}{e}^{-\lambda y}\sqrt{1+y^{\prime 2}}$. The Beltrami identity sets $e^{-\lambda y}/\sqrt{1+y^{\prime 2}}$ constant. Solving this first integral gives the displayed symmetric ray between $(\pm a,0)$ when $0<\lambda a<\pi /2$. Its travel time is $2\sin (\lambda a)/(c_0\lambda )$. The ray's central height diverges as $\lambda a\uparrow \pi /2$, while its travel time tends to $2/(c_0\lambda )$.

For positive twice differentiable functions $y$, the Euler-Lagrange equation for this functional reduces to $y{y}^{\prime \prime }+1+y^{\prime 2}=0$. Thus $(y{y}^{\prime }{)}^{\prime }=-1$ and integration gives $(x-h{)}^2+y^2=R^2$. The upper circular arc is a graph wherever its radicand is positive. A fixed horizontal endpoint with free height imposes $F_{y^{\prime }}=y^{\prime }/(y\sqrt{1+y^{\prime 2}})=0$, so that endpoint must lie above the centre. Distinct horizontal endpoints cannot both satisfy this natural boundary conditions for a free endpoint. The Beltrami identity alone also admits constant functions, which are excluded by the full Euler-Lagrange equation.

For refractive index $n(y)=c\sqrt{y}$ with $c>0$, the optical-length integrand is $c\sqrt{y}\sqrt{1+y^{\prime 2}}$. The Beltrami identity gives $\sqrt{y}/\sqrt{1+y^{\prime 2}}=\sqrt{h}$, while the full Euler-Lagrange equation gives $2y{y}^{\prime \prime }=1+y^{\prime 2}$. Together these yield the displayed parabolas and exclude a spurious constant solution of the squared first integral. Equal-height endpoints $(\pm L,Y)$ force $x_0=0$ and $h^2-Yh+L^2/4=0$. There are two positive stationary rays for $Y>L$, one coalesced ray for $Y=L$, and none of this smooth family for $Y<L$. For $L=1,Y=5/4$, the two paths are $y=1+x^2/4$ and $y=1/4+x^2$. Fermat principle requires stationary optical length, not a global minimum over every competing path.