Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 9 / b / ii / Solution

Equal endpoint heights implyso $x_0=0$. The height condition then gives $y_0=b-k\cosh (a/k)$ and the hanging catenary isSince $\sqrt{1+y^{\prime 2}}=\cosh (x/k)$, its length fixes $k$ throughFor the requested graphical argument, put $u=a/k>0$ and $q=L/a>1$. Intersections of $Y=\sinh u$ with $Y=qu$ determine the solution. Near zero the line has larger slope, while the hyperbolic sine eventually grows faster. Its slope $\cosh u$ increases strictly, so after turning upwards relative to the line it can cross it only once. The intersection at $u=0$ is excluded: it corresponds to infinite $k$ and the limiting length $L=a$.

More explicitly, $h(u)=\sinh u/u$ has derivative $(u\cosh u-\sinh u)/u^2>0$, because the numerator vanishes at zero and has derivative $u\sinh u>0$. Since $h(0+)=1$ and $h(u)\to \infty$, every $q>1$ yields exactly one positive $u$, hence exactly one positive $k$.