 Two optical rays in a square-root refractive-index profile

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.