# Elimination of parameter

1. Jan 19, 2007

### kasse

1. The problem statement, all variables and given/known data

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

3. 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?

2. Jan 19, 2007

### neutrino

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

Last edited: Jan 19, 2007
3. Jan 19, 2007

### 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...

Last edited: Jan 19, 2007
4. Jan 19, 2007

### HallsofIvy

Staff Emeritus
You didn't need to go through the "ln" business. If x= 2et 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= 2et and et is never 0.