# Eliminate 't'

• Nov 7th 2010, 12:15 PM
courteous
Eliminate 't'
Quote:

The term "hyperbolic" coems from the fact that if we define the Cartesian coordinates of a point to be \$\displaystyle x=\cosh t\$ and \$\displaystyle y=\sinh t\$, where \$\displaystyle t\$ is so-called parametric variable, then elimination of \$\displaystyle t\$ leads to the equation of \$\displaystyle x^2-y^2=1\$, the equation of an hyperbola.
I know of the identity \$\displaystyle \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 \$\displaystyle t\$"?
• Nov 7th 2010, 02:50 PM
mr fantastic
Quote:

Originally Posted by courteous
I know of the identity \$\displaystyle \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 \$\displaystyle t\$"?

\$\displaystyle x^2 = \cosh^2 (t)\$ and \$\displaystyle y^2 = \sinh^2 (t)\$. Now use the identity you say you know.
• Nov 7th 2010, 10:00 PM
courteous
This surely isn't meant by "eliminate \$\displaystyle t\$"? Or is it?
• Nov 7th 2010, 10:36 PM
mr fantastic
Quote:

Originally Posted by courteous
This surely isn't meant by "eliminate \$\displaystyle 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 ....?
• Nov 8th 2010, 03:03 AM
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.
• Nov 8th 2010, 10:25 AM
mr fantastic
Quote:

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.