Parametrization of Hyperbola

Question

I "know" that a parametrization of an Hyperbola ($x^2-y^2=1$) is given by: $$\gamma(t)=(\sec(t),\tan(t)),t\in\mathbb{R}$$ I know that $x=\sec(t)$ and $y=\tan(t)$ is a solution of the equation.

How can I show that for all $(x,y)\in\mathbb{R}^2$ such that $x^2-y^2=1$ there exists $t\in\mathbb{R}$ such that $x=\sec(t)$ and $y=\tan(t)$?

Answer 1

I will put an answer I thought about.

Let $(x,y)\in\mathbb{R}$ such that $x^2-y^2=1$.

As $u(v)=\tan v$ is surjective there exists $t\in\mathbb{R}$ such that $y=\tan t$. And as $x^2-y^2=1$ we get $$x^2=y^2+1=1+\tan^2 t=\sec^2 t$$ so $x=\pm\sec t$. If $x=\sec t$ we are done.

If not, $x=-\sec t$, and let $t'=\pi+t$. Then $\sec(t')=\sec(t+\pi)=-\sec(t)=x$ and $\tan t'=\tan (t+\pi)=\tan t=y$ (As the period of $\tan u$ is $\pi$).

So in any case, there exists a real number $q$ such that $x=\sec q$ and $y=\tan q$.

Thanks for the help, it was truly useful. And it is more useful to see the different ways we can use to get the answer.

Answer 2

For $x^2-y^2=1$, the domain is $|x|\ge 1$ and the range is $y\in R$.

We observe that for $0\le t < \pi/2$, the mapping $x=\sec t$ is one-to-one, onto for $x\ge 1$, while for $\pi/2<x\le \pi$, the mapping $x=\sec t$ is one-to-one, onto for $x\le -1$.

For these mappings, it is simply an identity to write $y=\tan t$ since $1+\tan^2 t=\sec^2t$. And the values of $\tan t$ for $0\le t<\pi/2$ and $\pi/2 <t\le \pi$ spans the reals.

Answer 3

If $x^2-y^2=1$ then $x>1$ or $x<-1$, so $0<1/x<1$, or $0>x>-1$ so there exists $u \in (0,\pi)$ with $\cos u = \cos (-u) =1/x$. Choose $t \in \{u,-u\}$ with $\text{sgn }(y)=\text { sgn }(\tan t).$ (Where $\text{ sgn }(z)=1$ if $z>0$,or $-1$ if $z<0$,or $0$ if $z=0$.) So $x=\sec t$ and $y=\text{ sgn }(y) \sqrt {x^2-1}$=$\text{ sgn }(\tan t) \sqrt {\sec^2 t -1}=\tan t.$