= Solution
\b[First integral.] If $F$ has no explicit $x$ dependence, the <chain rule> gives
$$
\frac d{dx}\left(F-y'F_{y'}\right)=F_yy'+F_{y'}y''-y''F_{y'}-y'\frac d{dx}F_{y'}=y'\left(F_y-\frac d{dx}F_{y'}\right).
$$
The <Euler-Lagrange equation> makes this zero without requiring division by $y'$. Hence the <Beltrami identity> is
$$
\boxed{F-y'\frac{\partial F}{\partial y'}=k\quad\text{constant}.}
$$
\b[Optical paths.] By <Fermat principle>, a ray makes the <optical path length> stationary. Within the strip $0<y\leq1$, its <functional> for a graph $y(x)$ is
$$
I=\int\mu(y)\,ds=\int\frac{\sqrt{1+(y')^2}}y\,dx.
$$
The <Beltrami identity> gives
$$
F-y'F_{y'}=\frac1{y\sqrt{1+(y')^2}}=k.
$$
For a nonvertical segment $k>0$, write $R=1/k$. Then $(y')^2=(R^2-y^2)/y^2$, so
$$
\frac{dx}{dy}=\pm\frac y{\sqrt{R^2-y^2}},\qquad x-c=\mp\sqrt{R^2-y^2}.
$$
Thus the <optical rays in a reciprocal-height refractive index> satisfy
$$
\boxed{(x-c)^2+y^2=R^2,\qquad y>0,}
$$
restricted to the part in the strip. They are arcs of <circles> centered on $y=0$. Vertical lines are the additional solutions, obtained as a limiting case or directly from the constant horizontal optical momentum. The same curves are <geodesics> of the <Poincaré half-plane model>, whose length element is $ds/y$.
\b[Horizontal displacement.] Let the ray enter at $(x_0,1)$, travelling to the right and downward at angle $0<\theta<\pi/2$ to the vertical. The <refractive index> is continuous at $y=1$, so there is no instantaneous angular jump at entry. The tangent has $(dx/ds,dy/ds)=(\sin\theta,-\cos\theta)$; hence $y'(x_0)=-\cot\theta$. The first integral gives $k=\sin\theta$, or $R=\csc\theta$. The circle derivative $y'=-(x-c)/y$ then fixes $c=x_0-\cot\theta$.
As $y\downarrow0$, the rightward ray approaches $x=c+R$. The total limiting horizontal distance is therefore
$$
\boxed{D=R-\cot\theta=\csc\theta-\cot\theta=\tan\frac\theta2.}
$$
For vertical incidence, $D=0$ by continuity. For a leftward ray the displacement has the opposite sign, with the same magnitude. Since $\tan(\theta/2)$ increases on $[0,\pi/2]$,
$$
\boxed{D\leq1,\qquad D_{\max}=1\text{ in the grazing-incidence limit }\theta\to\pi/2.}
$$
If entry is required to have a strictly downward component, this is a supremum rather than an attained maximum. A horizontal tangent at entry gives the limiting unit-radius arc, which immediately bends downward in the medium.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-1-reciprocal-height-rays.png]
{title=Light rays entering a medium with refractive index equal to reciprocal height}
{description=Original circular-ray construction for entry at x equal to zero and y equal to one. Rays bend toward the singular lower boundary; their limiting horizontal displacements are tan of half the entry angle, bounded by one.}
The lower boundary is a limit, rather than a point reached in finite optical travel time. On a nonvertical circular ray, $ds/dy=R/\sqrt{R^2-y^2}\to1$ as $y\downarrow0$. Thus the <optical path length> contains a divergent integral proportional to $\int dy/y$. The horizontal distance stays finite even though propagation toward $y=0$ takes infinite time.
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