Eliminating Parameter for Calculating Graphs

[kasse]

Homework Statement

Eliminate the parameter given
x=2e^t, y=2e^-t

The Attempt at a Solution

lnx = ln2 + t, so t = lnx - ln 2

This gives:

y=2e^(ln2-lnx)
y=2(e^ln2 * e^-lnx)
y= -4x

This does not, however, match the graph of the parametric function on my calculator. Have I made a mistake?

 

[neutrino]

$$e^{-\ln{x}} \neq -x$$

 

[kasse]

No, it's 1/x, so y=4/x must be the correct solution (for x bigger than 0)

The graph doesn't seem to start at x=0 though...

 

[HallsofIvy]

You didn't need to go through the "ln" business. If x= 2e^t then xe^-t= 2 so e^-t= 2/x and y= 2e^-t= 2(2/x)= 4/x.

Obviously, the can't "start at x= 0"- why does that bother you? x= 2e^t and e^t is never 0.