The Missing Intercept: -1,0 in Parametric Equations

[soroban]

Consider the parametric equations: $$\:\begin{Bmatrix}x &=& \dfrac{t^2-1}{t^2+1} \\ y &=& \dfrac{2t}{t^2+1} \end{Bmatrix}$$

Then: $$\;x^2 + y^2 \:=\:\left(\frac{t^2-1}{t^2+1}\right)^2 + \left(\frac{2t}{t^2+1}\right)^2 \;=\;\frac{(t^2-1)^2 + (2t)^2}{(t^2+1)^2}$$

. . . . $$=\;\frac{t^4-2t^2+1 +4t^2}{(t^2+1)^2} \;=\;\frac{t^4+2t^2+1}{(t^2+1)^2} \;=\;\frac{(t^2+1)^2}{(t^2+1)^2} \;=\;1$$

We have: $$\;x^2+y^2\,=\,1$$, a unit circle centered at the origin.Find the $$y$$-intercepts.
Set $$x=0\!:\;\frac{t^2-1}{t^2+1} \,=\,0 \quad\Rightarrow\quad t\,=\,\pm1$$
. . Hence: $$\;y \:=\:\frac{2(\pm1)}{(\pm1)^2+1} \:=\:\pm1$$

The $$y$$-intercepts are: $$\;(0,1),\;(0,-1)$$Find the $$x$$-intercepts.
Set $$y = 0\!:\;\frac{2t}{t^2+1}\,=\,0 \quad\Rightarrow\quad t \,=\,0$$
. . Hence: $$\;x \:=\:\frac{0^2-1}{0^2+1} \:=\:-1$$

The $$x$$-intercept is: $$\;(-1,0)$$What happened to the other $$x$$-intercept?

 

[Fallen Angel]

A way of thinking on it.

Spoiler

You are giving to any point of the unit circle the parametrization given by the equation of a line whose gradient is rational and passes through $(1,0)$, and the parameter is when that line intersect the circle at a point diferent from $(1,0)$, then the point $(1,0)$ is going to infinity since it's line is tangent to the circle through $(1,0)$.

 

[jacobi1]

Another, purely analytic, way of looking at it:

Spoiler

The second "solution" for $y=0$ is $t= \infty$ because
$$\lim_{t \to \infty} \frac{2t}{t^2+1} = 0.$$
So taking the limit as $t \to \infty$ of $x$ gives
$$\lim_{t \to \infty} \frac{t^2-1}{t^2+1} = 1,$$
and $(x,y) = (1,0).$