= Solution
For the displayed form take $c>0$ and $1+\beta x\geq c$. Substitute $s=(1+\beta x)/c$ in the <first integral>:
$$
y-y_0=\frac c\beta\int_1^{(1+\beta x)/c}\frac{ds}{\sqrt{s^2-1}}=\frac c\beta\operatorname{arcosh}\frac{1+\beta x}{c}.
$$
Inverting the <hyperbolic cosine> gives the <catenary light ray in a linear refractive index>
$$
\boxed{x(y)=-\frac1\beta+\frac c\beta\cosh\left[\frac\beta c(y-y_0)\right].}
$$
The 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 $y\geq y_0$ 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.
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