Given a point $A$, describe those points to which a catenary cannot be drawn from $A$.

Question

Background

An elementary problem in the calculus of variations shows that among all curves joining two points $A$, and $B$ in the first quadrant, the one which generates the surface of minimum area when rotated about the $x$-axis is the catenary: \begin{equation} y=\cosh \frac{x+C_1}{C},\tag1 \end{equation}

where $C,C_1$ are to be determined by the endpoints $A$ and $B$. There are three cases:

1. If there is only one possible catenary through $A$, and $B$, it will be the solution we seek.
2. If there are two extremals through the points (from the Euler-Lagrange equation), then one of them is the minimum solution and the other is not.
3. If there is no catenary that can be drawn between the two points, then the solution curve will not be smooth (See Goldschmidt's discontinuous solution).

I am interested in the third case.

Question

Suppose we fix the start point of the catenary given above. I seek an equation describing the shape of the boundary that separates those end points through which the catenary can be drawn and those through which it cannot.

My attempt

$(1)$ can alternately be presented as $$C\cosh^{-1}\frac{y}{C}=x+C_1. \tag2$$ Suppose $A=(x_0,y_0)$ is fixed and $B=(x,y)$ is allowed to vary (sorry for reusing notation). Then using these conditions we obtain: \begin{align} C\cosh^{-1}\frac{y_0}{C}&=x_0+C_1 \\ C\cosh^{-1}\frac{y}{C}&=x+C_1 \end{align} Subtracting to eliminate the $C_1$ variable yields:$$\cosh^{-1}\frac{y}{C}-\cosh^{-1}\frac{y_0}{C}=\frac{x-x_0}{C}. \tag3$$

$\cosh^{-1}\frac{y_0}{C}$ ceases to be real when $C>y_0$ so I am tempted to set $C=y_0$ yielding the equation: $$y_0\cosh^{-1}\frac{y}{y_0}+x_0=x. \tag4$$

As an example I let $A=(0,1)$ and I used mathematica to plot $(3)$ for various values of $C$ and $(4)$ on the same screen:

Here the thick line at the end is my proposed boundary and the lines before that are plots for various values of C from $0$ onward. It seems that my solution is a bit too much to the right. Now, rather than subtracting, if I add the $\cosh$ terms in $(3)$, then I get, in my opinion, a better plot but then my solution is incorrect. For reference here is the plot with the terms added:

Here again the thick line is $(4)$ The reason why I think this is the answer is because it "looks better". A possible reason for this may be that $(1)$ satisfies the differential equation: $$y^2-C^2=C^2y'^{2}$$ and when the normal solution method is carried out using variation techniques, only the positive root is considered.

Edit I think that it should be the envelope of the family of catenaries $(1)$, but I do not know how to directly calculate it (other than the implicit $(4)$ with a minus sign instead of a plus).