Why does $x=e^t+2e^{-t},y=e^t-2e^{-t}$ plot to a straight line?

Question

This parametrization satisfies $x^2-y^2=8$, so I was expecting a hyperbola. But what I got was a straight line. Why though?

https://www.wolframalpha.com/input/?i=parametric+plot+%28e%5Et%2B2e%5E%28-t%29%2Ce%5Et-2e%5E%28-t%29%29

EDIT- I tried a different range for $t$ and the point (8,0) isn't even on the line that gets plotted

Answer 1

Try to make the plot for $t$ between for example -3 and 3, the you should see it. Also, the point $(\sqrt 8,0)$ should be on there, not $(8,0)$.

Regarding the comments, it does not give a straight line, since the constant term depends on $t$, which varies. However, the reason why it looks like a straight line is that for 'large' $t > 0$ we have that $x \approx e^t$ and $y \approx e^t$, so that $y \approx x$. A similar thing happens for large negative $t$.

Answer 2

If $x,y$ are expressible as a linear function of one parameter, then only it is a line in 2D. If they are non-linear function of the parameter it is 2D curve. Here $x=e^t+2e^{-t}, y=e^t-2r^{-t}$ cannot represent a line. Eliminate $t$ as $$(x+y)=2e^{t},(x-y)=4 e^{-t} \implies (x^2-y^2)=8,$$ which is a hyperbola.