= Solution
The <Euler-Lagrange equation> is $f_y-d(f_{y'})/dx=0$. Since $f$ has no explicit $x$ dependence,
$$
\frac d{dx}(f-y'f_{y'})=y'\left(f_y-\frac d{dx}f_{y'}\right)=0.
$$
Thus a stationary curve satisfies the <Beltrami identity>, $f-y'f_{y'}=C$.
By <Fermat principle>, the <light ray> makes the travel time $\int\sqrt{1+y'^2}/(y+c_0)\,dx$ stationary. Its <Beltrami identity> becomes
$$
\frac1{(y+c_0)\sqrt{1+y'^2}}=C>0.
$$
Writing $R=1/C$ and integrating the resulting first-order equation gives $(x-x_0)^2+(y+c_0)^2=R^2$. Both endpoints have height zero, so subtracting their equations gives $x_0=0$, and then $R^2=a^2+c_0^2$. The arc lying in $y>0$ is
$$
\boxed{y(x)=\sqrt{a^2+c_0^2-x^2}-c_0,\qquad -a\leq x\leq a.}
$$
Its maximum height is $\sqrt{a^2+c_0^2}-c_0$. The ray rises into the faster region before returning to the boundary; its <circle> centre is $(0,-c_0)$. The prescribed boundary endpoints are understood by continuity from the open half-plane. This is the <circular light ray in a linear speed profile>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-2-light-ray.png]
{title=Circular light ray arching into a region where propagation speed increases with height}
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