I'm working with the catenary equation and this equation is given by
$$ f(x) = a \cdot \cosh\left(\dfrac{x}{a}\right) $$
I know this function pass at the point $(x_0, \ y_0)$ and therefore I want to find the value of $a$ using the values of $x_{0} > 0 $ and $y_0 > 1$. How can I do it?
My main problem is the $a$ that is multiplying the $\cosh$, because I know how to compute it using $g(x)$:
$$ g(x) = \cosh\left(\dfrac{x}{a}\right)\Rightarrow a = x \cdot \ln \left(g \ + \sqrt{g^2 - 1} \right) $$
I also tried using Lambert W Function with no success even for $h(x)$:
$$ h(x) = a\cdot \cosh \left(ax\right) $$
You want to solve for $a$ the equation $$y_0=a\cosh \left(\frac{x_0}{a}\right)$$ which is transcendental; so no closed form.
Rewrite it as $$\frac {y_0}{x_0}=\frac {a}{x_0}\cosh \left(\frac{x_0}{a}\right)$$Now, let $\frac {y_0}{x_0}=k$ and $\frac{x_0}{a}=t$ and then the problem reduces to $$k =\frac 1 t \cosh(t)$$
I think that it could be simpler to consider that you look for the zeros of function $$f(t)=\frac 1 t \cosh(t)-k$$ $$f'(t)=\frac{\sinh (t)}{t}-\frac{\cosh (t)}{t^2}$$ cancels at $t_*$ such that $$e^{-2t_*}=\frac{t_*-1}{t_*+1}$$ the only explict solution of which being given in terms of the generalized Lambert function (have a look at equation $(4)$). Numerically $$t_*=1.19968\cdots \qquad \text{and} \qquad f(t_*)=1.50888\cdots$$
So, there is no solution if $k<f(t_*)$.
If this is not the case, then there are two roots $$0 < t_1 < t_*\qquad \text{and} \qquad t_*<t_2 < \infty$$
For $t_1$, we could use the series expansion of $\cosh(t)$ and the series reversion; this would give $$t_1=\frac{1}{k}+\frac{1}{2 k^3}+\frac{13}{24 k^5}+\frac{541}{720 k^7}+\frac{9509}{8064k^9}+O\left(\frac{1}{k^{11}}\right)$$ For example, using $k=3$, this would give $t_1=0.354484$ while the solution, given by Newton method, is $0.354498$.
Now, $t_2$ becomes quickly large (for $k=3$, it is $2.82775$) and, is such a case $$f(t) \sim \frac {e^t}{2t}-k \qquad \implies \qquad t_2\sim -W_{-1}\left(-\frac{1}{2 k}\right)$$ For $k=3$, this gives $2.83315$.
All of the above show (I hope) that we can have decent estimates for generation a $t_0$ for, say, Newton method; the iterates will be given by $$t_{n+1}=t_n-\frac{t_n (\cosh (t_n)-k t_n)}{t_n \sinh (t_n)-\cosh (t_n)}$$ which would converge very fast.
Trying for $k=12.3456$, Newton iterates would be $$\left( \begin{array}{cc} n & t_n \\ 0 & 4.768473041 \\ 1 & 4.768381752 \\ 2 & 4.768381749 \end{array} \right)$$
Edit
There is another approach which, instead of $f(t)$, considers the function $$g(t)=\cosh(t)-k t$$ The first derivative cancels at $t_*=\sinh ^{-1}(k)$ and the second derivative is always positive.
Since we need $g(t_*)<0$, this implies $k > k_*$ with $$\sqrt{k_*^2+1}-k_* \sinh ^{-1}(k_*) \implies k_*=1.50888$$
For the smallest root, using an high order Newton-like method built around $t=0$
$$t_1=\frac 1 k \times\frac{40320 k^8-141120 k^6+142800 k^4-40488 k^2+1385} {40320 k^8-161280 k^6+201600 k^4-84224 k^2+7936}$$ which is almost the exact solution. For example, for $k=3$, this gives $\color{red}{0.35449845}60$ while the solution is $\color{red}{0.3544984571}$.
For the second root, expanding around the minimum, a rough approximation is $$t_2=\sinh ^{-1}(k)+\sqrt{\frac{2 \left(k \sinh ^{-1}(k)-\sqrt{k^2+1}\right)}{\sqrt{k^2+1}} }$$