 Solution

The original PDF places both endpoints at height $5/4$; the vertically stacked fraction can be misread in extracted text. The coefficient in the index is $4$, multiplying $\sqrt{y}$.

For the optical-length integrand $F=4\sqrt{y}\sqrt{1+y^{\prime 2}}$, the Beltrami identity applies because there is no explicit $x$ dependence. It giveswhere $h>0$ is constant. Therefore $y^{\prime 2}=y/h-1$. The full Euler-Lagrange equation is $2y{y}^{\prime \prime }=1+y^{\prime 2}$, and combining these gives $y^{\prime \prime }=1/(2h)$. This also excludes a constant curve, which would satisfy the squared first integral but not the full variational equation. Integration yieldsEqual endpoint heights at $x=-1,1$ require $x_0=0$. The height condition then becomes $h+1/(4h)=5/4$, or $4h^2-5h+1=0$. Its two positive roots are $h=1$ and $h=1/4$. Hence the two possible stationary rays areBoth stay in $y>0$, have the required endpoints and satisfy the variational equation. These are the two optical rays in a square-root refractive-index profile. Fermat stationarity supplies two rays; it does not assert that both globally minimize travel time.