Finding Parametric Equations for the right side of Hyperbola

Question

I have this equation $x^2-y^2=1$.

I can understand that it is a Hyperbola. But my question is, how can I create/find a parametric equation for the right side of this hyperbola? It seems it doesn't work, using $\sin$ or $\cos$. Thanks

Answer 1

Let's find functions $x(t), y(t)$ that satisfy $x^2 - y^2 = 1$. Perhaps, if we rewrite as $x^2 = 1 + y^2$, that'll help.

This reminds me of Pythagorean identities. In particular, it reminds me of $1 + \tan^2(t) = \sec^2(t)$.

Trying $(x(t), y(t)) = (\sec(t), \tan(t))$ has the unfortunate side effect that we may get negative values for $x$, depending on what we allow for $t$ (but it can't be fixed by choosing a small range for $t$, without losing some of the right half). Try using $|\sec(t)|$ instead, to ensure that we only get positive $x$-values.

Answer 2

Whit $r(t)=(x(t), y(t)) = (\sec(t), \tan(t))$ get the right branch, with $t \in\,]-\pi/2,\; \pi/2[$ (you can try for example $]-1.4,1.4[$). For the left branch you can try $r(t+\pi)$, $t \in\,]-\pi/2,\; \pi/2[$.

Best is $r(t)=(x(t), y(t)) = (\pm\cosh(t), \sinh(t))$, one sign for each branch. And you can use any interval $[-s,s]$. For example $r(t) = (\cosh(t), \sinh(t)), \; t \in\, [-1,1]$ and $r(0)$ is the vertex of the branch