Find the evolute of the catenary y=acosh(x)/(a)

Find the evolute of the catenary $y=a\cosh\dfrac{x}{a}$.

Step 1: Re-derive the parametric evolute independently

With $y=a\cosh(x/a)$, \texttt{sympy.diff} gives $y'=\sinh(x/a)$ and $y''=\cosh(x/a)/a$. Substituting into $X=x-y'(1+y'^2)/y''$ and $Y=y+(1+y'^2)/y''$ and calling \texttt{sympy.simplify} gives, exactly, \[X=x-\frac{a}{2}\sinh\frac{2x}{a},\qquad Y=2a\cosh\frac{x}{a},\] matching the source's parametric result term-for-term. This part of the source is correct.

Step 2: Numerically test the source's boxed implicit formula — it fails

Using the verified parametric formulas with $a=1$, \texttt{sympy} evaluates $(X,Y)$ at several $x$ and checks $4a^2X^2\overset{?}{=}(Y^2-4a^2)^2$:

x=0.5: X=-0.0876, Y=2.2553  LHS=0.0307   RHS=1.1797
x=1.0: X=-0.8134, Y=3.0862  LHS=2.6467   RHS=30.5189
x=2.0: X=-11.645, Y=7.5244  LHS=542.42   RHS=2768.49

The two sides disagree by orders of magnitude at every sample point — the source's boxed formula $4a^2X^2=(Y^2-4a^2)^2$ is numerically false. The source itself only asserted this "after some algebraic manipulation" without ever showing the elimination; that elimination does not in fact exist as claimed.

Step 3: Complete the elimination the source abandoned, and verify it

The source's own Step 4 (before it gave up and jumped to an unjustified guess) correctly gets as far as \[\cosh\frac{x}{a}=\frac{Y}{2a},\qquad x=a\,\operatorname{arcosh}\frac{Y}{2a},\qquad \sinh\frac{x}{a}=\frac{\sqrt{Y^2-4a^2}}{2a}\ \ (x\ge 0),\] and substituting into $X=x-a\sinh(x/a)\cosh(x/a)$ gives the genuine semi-explicit implicit relation \[X=a\,\operatorname{arcosh}\frac{Y}{2a}-\frac{Y\sqrt{Y^2-4a^2}}{4a}\qquad(x\ge0).\] \texttt{sympy} confirms this reproduces the parametric $X(x)$ exactly (residual $<10^{-15}$) at every tested $x\ge0$. For $x\le 0$ (where $\sinh(x/a)\le0$), the correct branch flips sign: \[X=-a\,\operatorname{arcosh}\frac{Y}{2a}+\frac{Y\sqrt{Y^2-4a^2}}{4a}\qquad(x\le0),\] also confirmed exactly by \texttt{sympy} at negative sample points. Because $x$ itself appears through $\operatorname{arcosh}$, there is no purely algebraic (polynomial) closed form eliminating the parameter — a genuinely transcendental relation, not the source's incorrect polynomial guess.