I need to find minimum point of hanging rope with two known points $p_1, p_2$ (start and end point of the rope) and known rope length.
I want to model all rope shapes with different length and start and end point.
Do I have to use numerical methods?
Does it have any closed form solution?
Known $\to L,(x_1,y_1), (x_2,y_2)$ then $a=$?
I use a general catenary equation like below:
$$ f(x)= a\cosh\left(\frac {x-b}a\right)+c\\ L= a\sinh\left(\frac{x_2}{a}\right)-a\sinh\left(\frac{x_1}{a}\right) $$
Starting from @Cesareo answer, considering $$L^2-v^2 = 4a^2\sinh^2\left(\frac{h}{2a}\right)$$ let $x=\frac{h}{2a}$ to make the equation $$\frac{L^2-v^2}{h^2}=\frac{\sinh^2(x) }{x^2}\implies k=\sqrt{\frac{L^2-v^2}{h^2}}=\frac{\sinh(x) }{x}$$
Consider that you look for the zero of $$f(x)=\frac{\sinh(x) }{x}-k$$ It varies very quickly. Then, it would be better to look for the zero of $$g(x)=\log \left(\frac{\sinh (x)}{x}\right)-\log(k)$$ which is much better conditioned.
Edit
To get a starting value $x_0$ for Newton method, using Padé approximant built at $x=0$ $$\frac{\sinh (x)}x\simeq \frac{7 x^2+60}{60-3 x^2} \implies x_0=\frac{2 \sqrt{15} \sqrt{k-1}}{\sqrt{3 k+7}}$$ which would be good for $1 \leq k \leq 3$.
For larger values of $k$ $$\frac{\sinh (x)}x\simeq \frac{e^x}{2 x}\implies x_0=-W_{-1}\left(-\frac{1}{2 k}\right)$$ where appears Lambert function.
The table below shows some results $$\left( \begin{array}{ccc} k & x_0 & x \\ 1.25 & 1.18125 & 1.18273 \\ 1.50 & 1.61515 & 1.62213 \\ 1.75 & 1.91663 & 1.93300 \\ 2.00 & 2.14834 & 2.17732 \\ 2.25 & 2.33550 & 2.37963 \\ 2.50 & 2.49136 & 2.55265 \\ 2.75 & 2.62398 & 2.70395 \\ 3.00 & 2.73861 & 2.83845 \\ & & \\ 3.00 & 2.83315 & 2.83845 \\ 3.25 & 2.95545 & 2.95952 \\ 3.50 & 3.06642 & 3.06962 \\ 3.75 & 3.16801 & 3.17058 \\ 4.00 & 3.26169 & 3.26380 \\ 4.25 & 3.34861 & 3.35037 \\ 4.50 & 3.42970 & 3.43117 \\ 4.75 & 3.50567 & 3.50693 \\ 5.00 & 3.57715 & 3.57823 \end{array} \right)$$
Edit
Looking at this question (which I did not remember - problem of age, I guess), I noticed that I was able to generate a quite good estimates building at $x=0$ the $[3,4]$ Padé approximant of $\sinh(x)-k x$.
From this, I considered building the $[3,2n]$ Padé approximants which write $$\sinh(x)-k x=x \frac{(1-k)+a^{(n)}_1 x^2 }{1+\sum_{m=1}^n b_m x^{2m} }$$ leading to an approximate solution $$x=\sqrt{\frac {k-1}{a^{(n)}_1 }}$$ For sure, this was done using a CAS. The longest result able to fit on a single line corresponds to $n=6$ and the result is $$x=\frac{\sqrt{6} \sqrt{(k-1)(105 k^5+60705 k^4+1365738 k^3+5507466 k^2+5665509 k+1414477 )}}{\sqrt{3 k^6+6120 k^5+307017 k^4+2586544 k^3+5952621 k^2+4301640 k+860055}}$$ which seems to be very good even for large values of $k$ (checked up to $k=500$).
Calling
$$ x_2-x_1 = h\\ y_2-y_1 = v $$
we have
$$ v = a\cosh\left(\frac{x_2}{a}\right)-a\cosh\left(\frac{x_1}{a}\right)\\ L = a\sinh\left(\frac{x_2}{a}\right)-a\sinh\left(\frac{x_1}{a}\right) $$
so after some trigonometric transformations we arrive at
$$ L^2-v^2 = 4a^2\sinh^2\left(\frac{x_2-x_1}{2a}\right) $$
so we have finally
$$ L^2-v^2 = 4a^2\sinh^2\left(\frac{h}{2a}\right) $$
with this last equation we can determine the $a = a_0$ value using an iterative procedure.
NOTE
Calling now
$$ y(x) = y_0 + a_0\cosh\left(\frac{x-x_0}{a_0}\right) $$
such that
$$ y(x_1) = y_1\\ y(x_2) = y_2 $$
The minimum point is obtained by solving for $x$
$$ y'(x) = 0 $$
so if $x_1 \le x_0 \le x_2$ follows
$$ \min y(x) = y_0 + a_0 $$
otherwise
$$ \min y(x) = \min(y_1, y_2) $$
Now, regarding the iterative process we have
$$ \frac{L^2-v^2}{4a^2} = \sinh^2\left(\frac{h}{2a}\right) $$
or
$$ C_0^2\lambda^2=\sinh^2\lambda,\ \ \lambda = \frac{h}{2a},\ \ C_0 = \frac{\sqrt{L^2-v^2}}{h} $$
or
$$ C_0 \lambda = \sinh(\lambda) $$
which should be solved numerically.
