Computes the catenary passing by the two (symmetrical) points (−X, Y) and (+X, Y) and the origin. The length of this catenary is obtained. A parabola with the same length passing by the two points and the origin is then computed, in order to observe the different (although approximate) shapes of the catenary and parabola.

The formula for a catenary passing by the two points and the origin is y_C = Y + a [cosh(x⁄a) − cosh(X⁄a)]. For a parabola passing by the two points and the origin, the formula is y_P = Y (x⁄X)². To solve y_C = 0 (the origin) in order to determine a, there is no analytical solution, so a Newton-Raphson (NR)† (numerical) method is used, hence the need for a "convenient" initial guess for a (as many others, this numerical method can fail).

The lengths of the catenary between −X and +X, L_C, and of the parabola, L_P, are:

        L_C = 2 a sinh(X ⁄ a)     L_P = √(X² + 4 Y²) + X²⁄(2 Y) asinh(2 Y ⁄ X).

Thus, finally, from L_P = L_C, and solving (easier) for L_C to get a (again by NR), a new catenary passing by the two points is computed, to show the difference between the two curves, with the parabola being sharper‡ than the catenary.

A plot is shown for the catenary and the parabola.

Other suggested data: Y = 3, 5, 5.5 (no convergence from proposed a).