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The cartesian equation of the tractrix is:$$y=\pm\left(a\cdot \operatorname{arcsech}^{-1}\left(\frac{x}{a}\right)-\sqrt{a^2-x^2}\right)$$ where $a>0$ is a real parameter and $x$ varies from $0$ to $a$.

One parametric form is:$$\begin{cases}x(t)=a\cdot \operatorname{sech}(t)\\y(t)=a\cdot(t-\tanh(t)) \end{cases}$$

Why do the parametric form represent the same curve?; I mean why does the plus/minus sign disappear?

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Because $y$ is no longer a function of $x$ in the parametric equations, we can have two points on the same vertical line (that is, different values of $y$ can occur for the same value of $x$). If $y$ is a function of $x$, we can only have one point on each vertical line, necessitating a $\pm$ sign. See vertical line test.

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  • $\begingroup$ ok...how could I show the parametric form depicts the same curve (that I wrote above)? $\endgroup$ – user182587 Oct 11 '14 at 15:33
  • $\begingroup$ You could always solve for $t$ in one parametric equation and substitute that for $t$ in the other. Again, you'll get two answers. $\endgroup$ – Robin Goodfellow Oct 11 '14 at 15:37
  • $\begingroup$ once I've obtained $t=arcsech(\frac{x}{a})$ I substitute it in second equation...so why will I get two answers?$ $\endgroup$ – user182587 Oct 11 '14 at 15:40
  • $\begingroup$ Hint: $\operatorname{sech}(t)=\operatorname{sech}(-t)$. $\endgroup$ – Robin Goodfellow Oct 11 '14 at 15:46
  • $\begingroup$ that's right Robin $\endgroup$ – user182587 Oct 11 '14 at 15:50

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