catenary equation with additional weigth Is there a trivial solution for the catenary equation with an additional weight attached to a given position on the rope? Like you would hang something from a rope at some given point and the cable weight can not be neglected. For example:

*

*Pole distance w: 10m

*Cable length l: 12m

*Distributed load Fl = 1N/mm

*Additional weight Fg = 300N

*Weights position x: 8.5m

x is measured in the horizontal between the poles. My goal is not to find the exact position of the rope at every position, but to find the exact position of the added weight.
As you can see in the image. The blue line from D to E is the rope, hanging without any attached weight with the length $l$, the gray lines $i$ and $j$ represent a stage, where a weight is added at the point B assuming, that the rope is fully stretched, and the cable weight can be neglected.

*

*$l = i + j $
Now in the real world, the rope will not be fully stretched, if the weight is not big enough. So I would like to now how much this will reposition B with some given values.
The blue line is given by the catenary equation

*

*$f(x) = \frac{e^{ax} + e^{-ax}}{2a}$

 A: There is a solution, but it is not trivial at all. Well, there are non-linear equations that give you the solution, and an analytical solution maybe for special cases.
Catenary Equations
Basically, you have two catenary curves on both sides of the cable. Where they meet there is a slope discontinuity such as to support the added weight.

The above is the free body diagram of the weight $W$. On each side of the cable has common constant horizontal tension $H$, and different vertical tension components $V_1$ and $V_2$. If the attached weight was zero then $V_1+V_2=0$. For the general case $V_1+V_2 = W$ is the vertical force balance equation. Each vertical force component of a catenary is related to the horizontal force component via the slope since tension has to act in a tangent fashion to the curve. In equation form this is $$ V(x) = H \frac{\rm d}{{\rm d}x} y(x)$$ and it sets the constraint (2) to describe the force balance below.
The equations for the two curves are
$$ \begin{aligned}y_{1}(x) & =a\left(\cosh\left(\frac{x-x_{1}}{a}\right)-\cosh\left(\frac{x_{1}}{a}\right)\right)\\
y_{2}(x) & =a\left(\cosh\left(\frac{x-x_{2}}{a}\right)-\cosh\left(\frac{S-x_{2}}{a}\right)\right)
\end{aligned} $$
where both share the same catenary constant $a=H/w$, with $w$ is the weight per length unit. Also, the two curves have different curve minimum positions $x_1$ and $x_2$. Finaly, $S$ is the horizontal span of the cable.
The above are subject to the constraints $y_1(0)=0$ and $y_2(S)=0$ already. What is missing is $x_1$ and $x_2$, the lowest points on the two curves. Once those are known, the curves are fully defined.
There is a continuity constraint at the weight location $x=x_M$
$$ y_1 (x_M) = y_2 (x_M)  $$
or specifically
$$ \cosh\left(\frac{x_M-x_{1}}{a}\right)-\cosh\left(\frac{x_{1}}{a}\right) = \cosh\left(\frac{x_M-x_{2}}{a}\right)-\cosh\left(\frac{S-x_{2}}{a}\right) \tag{1} $$
The additional weight $W$ adds the constraint of slope between the two curves as
$$ W = H \left. \left(  \frac{{\rm d}y_2}{{\rm d}x}  - \frac{{\rm d}y_1}{{\rm d}x}  \right) \right|_{x=x_M}  $$
or
$$ W = H \left( \sinh \left( \frac{x_M-x_2}{a} \right) - \sinh \left( \frac{x_M-x_1}{a} \right) \right) \tag{2} $$
These two constraint equations are to be solved for the location of the lowest points of the two curves $x_1$ and $x_2$. The equations are highly non-linear and difficult to solve analytically.
Finally, note that the total length of the cable is determined by the integral $L=\int_{0}^{x_{M}}\sqrt{1+\left(\tfrac{{\rm d}}{{\rm d}x}y_{1}\right)^{2}}\,{\rm d}x+\int _{x_{M}}^{S}\sqrt{1+\left(\tfrac{{\rm d}}{{\rm d}x}y_{2}\right)^{2}}\,{\rm d}x$ which evaluates to
$$ \small L=a\left(\sinh\left(\frac{x_{M}-x_{1}}{a}\right)-\sinh\left(\frac{0-x_{1}}{a}\right)+\sinh\left(\frac{S-x_{2}}{a}\right)-\sinh\left(\frac{x_{M}-x_{2}}{a}\right)\right) \tag{3}$$
Parabolic Approximation
You can find a first-order approximation of the above problem by noting that $\sinh(x) \approx x$ and $\cosh(x) \approx 1 + \tfrac{x^2}{2}$
This will allow us to find the tensions from the length using
$$ H = \sqrt{ \frac{ x_M (S-x_M) (S w+W) W}{2 S^2 (L/S-1) } + \frac{w^2 S^2}{24 (L/S-1)} } $$
in your example $H=4761$
The lowest points on the curves are
$$\begin{aligned}
x_1 &= \frac{S}{2} + \frac{W (S-x_M)}{w S} \\
x_2 &= \frac{S}{2} - \frac{W x_M}{w S} 
\end{aligned}$$
which are $x_1 = 5.05$ and $x_2 = 4.75$ for the example above.
Here is what the approximate solution looks like:

As you can see there is a small discontinuity at $x_M = 8.5$
