= Solution
For endpoint-preserving variations of a <functional>, the <Euler-Lagrange equation> is
$$
\boxed{F_y-\frac{d}{dx}F_{y'}=0.}
$$
Along an extremal, the <chain rule> gives $d(F-y'F_{y'})/dx=F_x+y'[F_y-dF_{y'}/dx]=F_x$. Thus when $F$ has no explicit $x$ dependence, the <Beltrami identity> is
$$
\boxed{F-y'F_{y'}=A.}
$$
For a positive speed profile, the elapsed-time integrand in <Fermat principle> is $F=\sqrt{1+(y')^2}/c(y)$. Its first integral gives
$$
\frac1{c(y)\sqrt{1+(y')^2}}=A>0,\qquad\frac{dx}{dy}=\pm\frac{Ac(y)}{\sqrt{1-A^2c(y)^2}}.
$$
Integrating each monotone-height branch yields
$$
\boxed{\int^y\frac{Ac(\widehat y)}{\sqrt{1-A^2c(\widehat y)^2}}d\widehat y=\pm(x+B).}
$$
Real branches require $Ac(y)\leq1$. Turning points are joined in the original graph $y(x)$, even though the inverse derivative is singular there.
For the inverse-height speed profile, this becomes $dx/dy=\pm a/\sqrt{y^2-a^2}$, with $a=A>0$. Integration gives $x-b=\pm a\operatorname{arcosh}(y/a)$, so the <light rays> are <catenaries>
$$
y(x)=a\cosh\frac{x-b}{a}.
$$
Equal endpoint heights at opposite $x$ values force $b=0$, since the real hyperbolic cosine is even and strictly increases with the absolute value of its argument. The remaining endpoint condition is $a\cosh(1/a)=\cosh1$. One exact solution is $a=1$, giving
$$
\boxed{y(x)=\cosh x,\qquad y_{\min}=1,\qquad T=\int_{-1}^1\cosh^2x\,dx=1+\frac12\sinh2\simeq2.813430204.}
$$
Here $ds/dx=\cosh x$ and $1/c(y)=y=\cosh x$ account for the squared hyperbolic cosine in the time integral.
\b[The stated endpoints also allow a second stationary ray.] Set $t=1/a$. The function $\cosh t/t$ decreases and then increases: its derivative has the sign of $t\sinh t-\cosh t$, whose derivative is $t\cosh t>0$. Its unique minimum occurs where $t\tanh t=1$, at $t\simeq1.199679$. Since the known root $t=1$ lies below this minimum's location, there is exactly one further root, $t\simeq1.4242930655$. Thus the <two stationary catenary rays in an inverse-height speed profile> also include
$$
\boxed{y(x)=a\cosh(x/a),\qquad a\simeq0.7021026952,\qquad y_{\min}=a,\qquad T=a+\frac{a^2}{2}\sinh\frac2a\simeq2.822449145.}
$$
No extra condition in the question selects uniquely between these rays. The minimum height is a minimum along the chosen ray, not a global travel-time assertion. Indeed, competitors descending vertically to height $\eta>0$, crossing horizontally, and rising again have time $\cosh^2(1)-\eta^2+2\eta$. As $\eta\downarrow0$ this tends to $\cosh^2(1)\simeq2.381098$, below both stationary times. Their corners can be smoothed, so <Fermat principle> here must be understood as stationarity rather than unrestricted global minimization.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1-q9-two-catenary-rays.png]
{title=Two stationary light rays through the same endpoints in an inverse-height speed profile}
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