Smooth parametrization of curve $\cosh x+\cosh y = \operatorname{constant}$ I have a curve (image below)
$$
\cosh x+\cosh y = C,\qquad C>2.
$$
I would like to get its smooth parametrization of form
$$
x = f(t),\qquad y=g(t),\qquad t\in[a,b],
$$
so for every point on the curve there is a corresponding parameter $t$. (In the same manner, as for curve $x^2+y^2=1$, there is a smooth parametrization $x=\cos t$, $y = \sin t$).
I would appreciate any help. Thanks!


Here's something more symmetric. I'll write the steps if anyone wants to modify the idea to get something else.
First, we can remove the pesky hyperbolic functions simply by letting $x = \ln t,\ y = \ln s$ so $$\cosh x + \cosh y = c \iff t + \frac 1t + s + \frac 1s = 2c.$$
Now, we'd like to "invert" $z\mapsto z + 1/z$. More precisely, we want $t + 1/t = 2u^2+2$ and $s + 1/s = 2v^2 + 2$, so the whole thing becomes $ u^2 + v^2 = c - 2$ which we know how to parametrize.
Solving the appropriate quadratics, we get $$t = 1 + u^2\pm \sqrt{2u^2+u^4},\ s = 1+v^2 \pm \sqrt{2v^2 + v^4}.$$ Noting that we want $ u^2 + v^2 = c - 2$, we further choose $u = \sqrt {c-2} \cos\varphi$ and $v = \sqrt {c-2}\sin\varphi$.
All in all, $$x = \ln \left( 1 + (c-2)\cos^2\varphi \pm \sqrt{2(c-2)\cos^2\varphi + (c-2)^2\cos^4\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi \pm \sqrt{2(c-2)\sin^2\varphi + (c-2)^2\sin^4\varphi} \right).
$$
Now you must be wondering what to do with these $\pm$ signs. Well, if you choose one of the four possibilities, you will get the part of the curve in the corresponding quadrant. So, to deal with that:
$$x = \ln \left( 1 + (c-2)\cos^2\varphi + \cos\varphi \sqrt{2(c-2) + (c-2)^2\cos^2\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi + \sin\varphi \sqrt{2(c-2) + (c-2)^2\sin^2\varphi} \right),
$$ and we are finally done.
