Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 9 / ii / b

For the displayed form take $c>0$ and $1+\beta x\ge c$. Substitute $s=(1+\beta x)/c$ in the first integral:Inverting the hyperbolic cosine gives the catenary light ray in a linear refractive indexThe minimum occurs at $y=y_0$ and $x=x_0=(c-1)/\beta$, so the integration constant has the specified meaning. The integral above describes the $y\ge y_0$ graph branch; the other sign of the square root continues it smoothly to the other side of the turning point in $x(y)$.

The signed-constant qualification matters: for $c<0$ in part (a), integration reverses horizontal direction, and the positive prefactor in the full catenary is $|c|/\beta$, with minimum height $(|c|-1)/\beta$. The printed expression presupposes $c>0$. The separate case $c=0$ gives $y$ constant, a vertical ray. At the turning point $dy/dx$ diverges, so $y(x)$ is not a single regular graph across it, although $x(y)$ is smooth.