1. ## Eliminate 't'

The term "hyperbolic" coems from the fact that if we define the Cartesian coordinates of a point to be $x=\cosh t$ and $y=\sinh t$, where $t$ is so-called parametric variable, then elimination of $t$ leads to the equation of $x^2-y^2=1$, the equation of an hyperbola.
I know of the identity $\cosh^2 t - \sinh^2 t = 1$, which directly leads to hyperbola equation, but this isn't what is expected here.

So, how do you "eliminate $t$"?

2. Originally Posted by courteous
I know of the identity $\cosh^2 t - \sinh^2 t = 1$, which directly leads to hyperbola equation, but this isn't what is expected here.

So, how do you "eliminate $t$"?
$x^2 = \cosh^2 (t)$ and $y^2 = \sinh^2 (t)$. Now use the identity you say you know.

3. This surely isn't meant by "eliminate $t$"? Or is it?

4. Originally Posted by courteous
This surely isn't meant by "eliminate $t$"? Or is it?
Well, yes, it is .... What do you think happens when you take the difference and use the identity? Do you see any t's anymore ....?

5. I believe that you are supposed to use the defining equations for cosh(t) and sinh(t), the ones in terms of the expontials e^t and e^(-t).

Start with (cosh t)^2-(sinh t)^2, and substitute the exponential definitions for these expression. Multiply out, simplify, and then you will get 1. This says that these functions satisfy the equation for the hyperbola.

6. Originally Posted by woof
I believe that you are supposed to use the defining equations for cosh(t) and sinh(t), the ones in terms of the expontials e^t and e^(-t).

Start with (cosh t)^2-(sinh t)^2, and substitute the exponential definitions for these expression. Multiply out, simplify, and then you will get 1. This says that these functions satisfy the equation for the hyperbola.
I agree that your method works, however in my opinion it has two disadvantages:

1. The approach is more difficult than it needs to be, and

2. You need to know the answer in advance. Extremely unlikely in an exam situation.

Posts #2 and #4 say exactly how to do it - quick and simple.