= Two optical rays in a square-root refractive-index profile
{title2=$y=h+(x-x_0)^2/(4h),\quad h>0$}
For <refractive index> $n(y)=c\sqrt y$ with $c>0$, the optical-length integrand is $c\sqrt{y}\sqrt{1+y'^2}$. The <Beltrami identity> gives $\sqrt y/\sqrt{1+y'^2}=\sqrt h$, while the full <Euler-Lagrange equation> gives $2yy''=1+y'^2$. 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.
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