The hyperbola is parametrised by $x=a\cosh(t),y=b\sinh(t)$. The intrinsic equations find $s(\psi)$, where $\psi$ is the angle of elevation from some reference axis. The arc length of a hyperbola is actually a very tricky integral, so I thought that maybe finding some function $\psi(s)$ and then taking its inverse might give an intrinsic equation of sorts. We know that $\frac{dx}{ds}=\sin(\psi)$, and that is:
$$\frac{dx}{ds}=\frac{dt}{ds}\frac{dx}{dt}=\frac{1}{\frac{ds}{dt}}\cdot\frac{dx}{dt}=a\sinh(t)\cdot\left(\sqrt{a\sinh^2(t)+b\cosh^2(t)}\right)^{-1}=\frac{a}{\sqrt{a+b\coth^2(t)}}=\sin(\psi)$$
So, very messily, could it be that $s=\int\psi(x)\,dx$? Where $\psi(x)$ is:
$$\psi(x)=\arcsin\left(\frac{a}{\sqrt{a+b\frac{\sqrt{x^2-1}}{x}}}\right)$$
I understand that this is possibly a worse integral to deal with, but I'm just curious about the intrinsic equation of the hyperbola, as an exercise. Is anything I've written correct? I also found out that $\tan(\psi)=\frac{b}{a}\coth(t)$.