\begin{cases} & \text{ } x(t)=t - a \frac{\sinh (\frac{t}{a})}{\cosh (\frac{t}{a})} \\ & \text{ } y(t)= \frac{a}{\cosh (\frac{t}{a})} \end{cases}
According to the exercise, I need to see the parametric view of traxis, i.e. bring to normal view of traxis using some substitution, but no idea which one
You have already parameterized the tractrix. in $ [(x(t),y(t) )]= (z(\theta),r(\theta))$ form in 3D suitable for a surface of revolution where geometrically angle $\theta $ is reckoned from cuspidal equator in polar coordinates $(r,\theta,z)$.
$$ z(\theta)/a= \theta-\tanh \theta \,;r(\theta)/a= sech \,\theta \tag1 $$
If slope $\phi$ to z-axis is a parameter. Tangent length is constant $=a$.
$$ z(\theta) = a( \log\, \tan \frac{\phi}{2}-\cos \phi ) \,;r(\theta)=a \sin \phi. \tag2 $$
Eliminating $\phi$
$$ z = a\, sech^{-1}(r/a) -\sqrt{a^2-r^2} \tag3 $$
Equn 1) can be rewritten by means of arc length $s= a \theta$ reckoned from cuspidal equator:
$$ z(s)= s-\tanh(s/a)\,;r(s) = sech \,(s/a) \tag4 $$