Approach 1 gave the cartesian equation \[x=a\ln\left(\frac{a+\sqrt{a^2-y^2}}{y}\right)-\sqrt{a^2-y^2}.\]
Approach 2 gave the parametric form \[\left(t-a\tanh\Bigl(\frac{t}{a}\Bigr), a\sech\Bigl(\frac{t}{a}\Bigr)\right).\]
Are these equivalent?
In the parametric form, we have \[(x,y)=\left(t-a\tanh\Bigl(\frac{t}{a}\Bigr),
a\sech\Bigl(\frac{t}{a}\Bigr)\right).\]
Therefore \(\sech(t/a)=y/a\), so \(a/y=\cosh(t/a)\) and \(t=a\arcosh(a/y)\).
Using the identity \(1-\tanh^2(t/a)=\sech^2(t/a)\), we also get \[a\tanh\Bigl(\frac{t}{a}\Bigr)=
a\sqrt{1-\Bigl(\frac{y}{a}\Bigr)^2}=\sqrt{a^2-y^2}.\]
This is definitely looking hopeful!
We just need to find a way of expressing \(\arcosh u\) in terms of \(u\) (where we’re writing \(u=a/y\) to make things look a little simpler).
If \(\arcosh u=w\), then \(u=\cosh w\), so \[u=\frac{e^w+e^{-w}}{2}.\] Multiplying by \(e^w\) and rearranging gives \[e^{2w}-2ue^w+1=0.\] Since \(u=y/a>0\), we take the root of this quadratic in \(e^w\) which is greater than \(1\), getting \[e^w=\frac{2u+\sqrt{(2u)^2-4}}{2}=u+\sqrt{u^2-1},\] so \[w=\arcosh u=\ln\bigl(u+\sqrt{u^2-1}\bigr).\] Now substituting \(u=a/y\) and taking logarithms gives
\[\begin{align*}
\arcosh\Bigl(\frac{a}{y}\Bigr)
&=\ln\left(\frac{a}{y}+\sqrt{\Bigl(\frac{a}{y}\Bigr)^2-1}\right)\\
&=\ln\left(\frac{a+\sqrt{a^2-y^2}}{y}\right)
\end{align*}\]
Putting everything together, using \(t=a\arcosh(a/y)\), then gives
\[\begin{align*}
x&=t-a\tanh\Bigl(\frac{t}{a}\Bigr)\\
&=a\ln\left(\frac{a+\sqrt{a^2-y^2}}{y}\right) - \sqrt{a^2-y^2}
\end{align*}\]
