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
yC = Y + a [cosh(x⁄a) − cosh(X⁄a)].
For a parabola passing by the two points and the origin, the formula is
yP = Y (x⁄X)².
To solve yC = 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, LC, and of the parabola,
LP, are:
LC = 2 a sinh(X ⁄ a)
LP = √(X² +
4 Y²) + X²⁄(2 Y) asinh(2 Y
⁄ X).
Thus, finally, from
LP = LC, and solving (easier) for
LC 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). | 
