I have the parabola $x^2-y^2=1$. I am asked to first show that $\cosh(t)=x,\sinh(t)=y$ parameterizes the curve.
This, I have done; substituting to the implicit function expression gives $1$.
But I guess I am missing something from the concept as I cannot understand the following statement.
This parametrization covers only the right-hand side of $x^2-y^2=1$ because $\cosh(t)$ is always positive.
I don't see how $\cosh(t)$ always positive $=$ "covers only right-hand side of the function(graph)".
Can someone explain to me why? I simply don't understand why not cover the entire parabola with this particular parametrization.