Explicit form of hyperbolic parametrised curve

Question

Consider the parametrised curve given by \begin{align} \left(\frac{\sinh^{-1}(t)}{t},\frac{\sqrt{1+t^2}-1}{t}\right) \end{align} for $t\geq 0$. This answer suggests such parameterization has the following explicit equation $$x = \frac{\tanh^{-1}y}{y}\cdot(1-y^2)$$ This seems to be right, but I do not see how the author got here. Any ideas?

Answer 1

We have,

$ y = \dfrac{\sqrt{1+t^2}-1}{t}$.

Let's first try to simplify by squaring both sides.

$y^2 = \dfrac{1+t^2 + 1 - 2\sqrt{1+t^2}}{t^2} = 1 - \dfrac2t\left(\dfrac{\sqrt{1+t^2}-1}{t}\right) = 1-\dfrac {2y}{t}$

$\Rightarrow \boxed{t=\dfrac{2y}{1-y^2}}$

Now let's plug this in $x$.

$x=\dfrac{\sinh^{-1}\left(\dfrac{2y}{1-y^2}\right)}{\dfrac{2y}{1-y^2}}$

Let's try to make the hyperbolic term look nicer.

Let $y=\tanh p \Rightarrow\dfrac{2y}{1-y^2} = \dfrac{2\tanh p}{\text{sech}^2p} = 2\sinh p\cosh p = \sinh(2p) \\\Rightarrow\boxed{\sinh^{-1}\left(\dfrac{2y}{1-y^2}\right) =2p = 2\tanh^{-1} y}$

Finally,

$\boxed{x=\dfrac{2\tanh^{-1} y}{\dfrac{2y}{1-y^2}} = \dfrac{\tanh^{-1} y}{y}(1-y^2)}$