$y = a \cosh\!\left(\dfrac{x}{a}\right) = \dfrac{a}{2}\!\left(e^{x/a} + e^{-x/a}\right)$
The catenary is the curve formed by a uniform chain hanging under gravity.
- At the vertex $(0,a)$, the curvature is $\kappa = 1/a$.
- The matching parabola $y = \dfrac{x^2}{2a} + a$ shares this curvature at the vertex.
- Chain length from $-L$ to $L$: $\;\; S = \displaystyle\int_{-L}^{L}\!\sqrt{1 + (y')^2}\,dx = 2\,|a|\,\sinh\!\left(\dfrac{L}{|a|}\right)$