Given a system of equations: $$x = {t^2 + 1\over 2t}\\ y = {t^2 - 1\over 2t}$$ Prove that it defines two continuous functions $x(y)$ on $\Bbb R$ and four continuous functions $y(x)$ on $\{x: |x|\ge 1\}$
I can not really understand what the author of the problem mean by two functions on $R$ and four functions on $|x| \ge 1$. Obviously for $t = 0$ neither $x(t)$ nor $y(t)$ is defined. So we only consider $t\in\Bbb R\setminus\{0\}$. For this domain the ranges are: $$ x(t) \in(-\infty, -1)\cup(1, +\infty)\\ y(t) \in \Bbb R $$
Now let's express $t$ in terms of $x$ and $y$: $$ x = {t^2 + 1\over 2} \iff t^2 - 2xt + 1 = 0 \\ t_{1,2} = x\pm\sqrt{x^2 -1} $$ Similarly: $$ y = {t^2 - 1\over 2} \iff t^2 - 2yt - 1 = 0 \\ t_{3,4} = y\pm\sqrt{y^2 + 1} $$
Since those are the only solutions for $t$ we need to consider 4 cases: $$ t_1 = t_3 \\ t_1 = t_4 \\ t_2 = t_3 \\ t_2 = t_4 $$ Start with $t_1 = t_3$, then it would mean: $$ x + \sqrt{x^2 - 1} = y + \sqrt {y^2 - 1} \\ x - y = \sqrt{y^2 + 1} - \sqrt{x^2 - 1} \\ (x - y)^2 = \left(\sqrt{y^2 + 1} - \sqrt{x^2 - 1}\right)^2 \\ xy = \sqrt{(y^2 + 1)(x^2 - 1)} \\ (xy)^2 = (y^2 + 1)(x^2 - 1) \\ x^2 - y^2 = 1 $$
Other cases yield the same result. So basically we've arrived at the equation: $$ x^2 - y^2 = 1 \tag 1 $$
Now no matter what function we define, either $y(x)$ or $x(y)$ each defines a single function, which are essentially the same: $$ |x| = \sqrt{y^2 + 1}\\ |y| = \sqrt{x^2 - 1} $$
So how could a parametric function define several functions? Is that because we need to split the domain somehow and consider various cases for $x<0, y<0$; $x<0, y>0$ and etc.?