Eliminating the parameter??

Question

I have the problem

$\ x=2^t+2^{-t} ,y=2^t-2^{-t}$

Do I need to use a logarithm to solve this equation? If anyone can help me solve for the parameter I would greatly appreciate it. If this helps, it graphs as a horizontal parabola. Thank you!

Answer 1

$$x^2-y^2=(4^t+4^{-t}+2)-(4^t+4^{-t}-2)=4$$ Hence equation of curve : $x^2-y^2=4$

Answer 2

You can just define $z=2^t$ and rewrite the equations using that for whatever purpose you have. Then when you are done, $t=\log_2 z$, but you don't have to carry the powers along. You have two equations in three unknowns, so will have an infinite set of solutions. If (as I guess) the last term is supposed to be $2^{-t}$ you can just add and subtract the equations and things are simpler.

Answer 3

You have

$$x=2^t+\frac{1}{2^t}$$

and

$$ 2^t=y+\frac{1}{4}$$

thus

$$x=(y+\frac{1}{4})+(\frac{1}{y+\frac{1}{4}})$$

$$=\frac{16y^2+8y+17}{16y+4}$$