Talk:Unit hyperbola

Parametrization
I recommend changing the exponential function, from $$( e^t, \ e^{-t})$$ to $$( e^{-t}, \ e^t)$$, and the linear mapping, from

$$A = \tfrac {1}{2}\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}\ $$

to

$$A = \tfrac {1}{2}\begin{pmatrix}1 & 1 \\ -1 & 1 \end{pmatrix} = \tfrac {1}{\sqrt{2}}\begin{pmatrix}\cos (\pi /4) & \sin (\pi /4) \\ -\sin (\pi /4) & \cos (\pi /4) \end{pmatrix}\ $$, so that A will be a rotation matrix (and scaling) in standard form, unless there is some historical reason for the non-standard form. The linear mapping can thus be seen to rotate the hyperbola clockwise through an angle of π/4 and scale it by $$1/\sqrt{2}$$. See rotation matrix. Thoughts? — Anita5192 (talk) 03:40, 2 March 2014 (UTC)
 * Yes, Anita, the matrix reverses orientation so it is not a rotation. Yes again, it has something to do with the historical development of the subject. Look at hyperbolic sector where there is a standard position and positive hyperbolic angles evolve in the direction opposite to the direction of evolution of circular angles. Thus the reversal of orientation. Circular angles evolve consistent with the Sun rising over the Earth from the East. The hyperbolic sector evolves as used by Gregoire de Saint-Vincent in the discovery of logarithm. The unit hyperbola of this article not only normalizes the x y = 1 of Saint-Vincent, but also orients its evolution to agree with the familiar circular evolution. Your suggestion would do the normalization but require a revised format from the traditional basis of hyperbolic angle.Rgdboer (talk) 21:57, 3 March 2014 (UTC)
 * Thank you! — Anita5192 (talk) 03:33, 4 March 2014 (UTC)