Why is $\cosh$ used in parametrization of $x^2-y^2=1$ when $\cosh$ can't be negative?

Question

I came across a question to parametrize $x^2-y^2=1$. From its similarity to the identity $\cosh^2(t)-\sinh^2(t)=1$, I parametrized as follows. $$x = \cosh(t), y = \sinh(t)$$ It all seemed fine and the book said it is the answer but I soon realized that $\cosh(t) \geq 1$ for all real number $t$. This made me question the parametrization I came up with, as there are points where $x$ is negative for the graph of $x^2-y^2=1$.

Answer 1

Your parameterization parameterizes one branch of the hyperbola $$\{x^2 - y^2 = 1\} ,$$ but like you say, it does not also trace out the other. The same must be true of any continuous parameterization with connected domain, since the image of such a domain under such a map is connected but the other does not.

In general, whether a parameterization $f : X \to Y$ whose image is not all of the target space $Y$ is sufficient for one's needs depends on the context. For example, a point mass in a central (inverse-square) gravitational field on an escape trajectory traces out one branch of a hyperbola: The trajectory function that gives its position as a function of time parameterizes that branch, but not the other (which in this case is nonphysical).

As N.S. suggests in the comments, it's still possible to parameterize the hyperbola if you're willing to tolerate a nonconnected domain, e.g., with the map $$\alpha: \{\pm 1\} \times \Bbb R , \qquad \alpha(\mu, t) := (\mu \cosh t, \sinh t) .$$