Quote - However, the paper gives a formula which the writer seems to state will allow for the solution of a catenary when the endpoints are not level on Y. 

Quote - Although our figure shows the end-points at the same height, this is not necessary.
Indeed we could place the end-points anywhere on a given catenary curve,
snip the excess, and the resulting portion of the curve would hang unchanged. Put
differently, every part of a catenary curve is also a catenary curve.

Can you make heads or tails of this?  Might this help solve the problem?

Well the author does say that , but later says :-

Quote - Unfortunately, we cannot solve the equation h = (cosh(L/2)a − 1) /a for a, so
we are unable to write a as an explicit expression in terms of L and h alone. However,
we can put a value for L into the equation for h and then approximate a arbitrarily
closely using numerical methods

meaning that they didn't actually manage to find an analytical solution even for the easy case and I don't think they got close to one for other cases because there's no sign of them setting out a formulation of the non-equal height case.

The other reference supports my view, saying > Quote - ... when the supports of a catenary are at different evelations, the mathematical complexity preculdes a theoretically correct solution ...

When I get a chance I'll check my working and code to see if I can suggest a better method for the inital guess.