= Solution
For $c>1$, the <catenary> has a positive minimum height $x_0=(c-1)/\beta$ above the road. Its derivatives are
$$
x'(y)=\sinh\left[\frac\beta c(y-y_0)\right],\qquad x''(y)=\frac\beta c\cosh\left[\frac\beta c(y-y_0)\right]>0.
$$
Thus the <light ray> is symmetric about $y=y_0$, descends towards a horizontal tangent at its minimum, and then rises. It never intersects the road. This smooth upward bending in the <refractive index> gradient produces an inferior <mirage>, without reflection at the road.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-q9-mirage-ray.png]
{title=A turning light ray above a hot road in a linearly increasing refractive index}
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