You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola \Delta t = \sqrt{(3^2 + \Delta x^2)} then for ANY velocity \Delta x/ \Delta t in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola. This is for the "horizontal hyperbola". For the vertical hyperbola, \Delta x = \sqrt{(3^2 + \Delta t^2)} the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended
). Start with the Minkowski relationship (c\tau)^2 = (ct^2) - x^2 and multiply by both sides by minus one to obtain -(c\tau)^2 = -(ct^2) + x^2 then we get ic\tau = \sqrt{(x^2-(ct)^2)}. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, dx_o = ic\tau. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated
Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.
