= Solution
Equal endpoint heights imply
$$
\cosh\frac{a-x_0}{k}=\cosh\frac{a+x_0}{k},
$$
so $x_0=0$. The height condition then gives $y_0=b-k\cosh(a/k)$ and the hanging <catenary> is
$$
y(x)=b+k\left[\cosh(x/k)-\cosh(a/k)\right].
$$
Since $\sqrt{1+y'^2}=\cosh(x/k)$, its length fixes $k$ through
$$
2L=\int_{-a}^a\cosh(x/k)\,dx=2k\sinh(a/k),\qquad\boxed{L=k\sinh(a/k).}
$$
For 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 $L=a$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-q9-catenary-uniqueness.png]
{title=Unique positive catenary parameter from a hyperbolic-sine intersection, with an example hanging chain}
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 \b[exactly one positive $u$, hence exactly one positive $k$].
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