Hyperbolic line, solving for a,b

Question

h(t,x) is an isometry of the hyperbolic line t^2-x^2=1 , t > 0 , of the form h(t,x)= a*t+b*x, -a*x-b*t for some constants a > 0 , b such that a^2-b^2=1 . Find a,b so that h(1.221, -0.7) = (1.092, -0.439)

Answer 1

You're told that h(t, x) yields two solutions: at + bx and -ax - bt.

 

Next, you're given a specific example: h(1.221, -0.7) = (1.092, -0.439).

 

This means that t = 1.221 and x = -0.7, so at + bx = 1.221a - 0.7b = 1.092 and -ax - bt = 0.7a - 1.221b = -0.439. (It's also possible to switch the solutions, so 1.221a - 0.7b = -0.439 and 0.7a - 1.221t = 1.092.)

 

 

 

You have a system of equations:

 

1.221a - 0.7x = 1.092

0.7a - 1.221b = -0.439

 

Solve the system for a and b.

 

 

 

Do the same for the other possible system of equations:

 

1.221a - 0.7x = -0.439
0.7a - 1.221b = 1.092

 

 

 

You should now have two possible pairs of values for a and b. They happen to be a = 1.639, b = 1.299 or a = -1.299, b = -1.639. However, you're told that a² - b² = 1. Plug the pairs of values into that equation to find that only the first pair works.