Parametric equation - of a hyperbola I know that the parametric equation for points on a hyperbola($\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$) is:
$$x = a\sec \theta$$ $$y = b\tan \theta$$
However, what does the parameter $\theta$ actually represent? In a circle, it is quite obvious, what the parameter $\theta$ represents. However, how should I visualize it for a hyperbola?
 A: Here is an animated gif of $\theta$ changing.
Excuse the red frames (90 and 270 degree). That's my software hitting the discontinuity.

A: Let's consider the special case $a = b = 1$. Instead of the trigonometric parametrization
$$
x(\theta) = \sec\theta,\quad
y(\theta) = \tan\theta,\qquad -\pi/2 < \theta < \pi/2,
\tag{1}
$$
consider the hyperbolic parametrization
$$
\begin{aligned}
x(t) &= \cosh t \\
  &= \tfrac{1}{2}(e^{t} + e^{-t}),
\end{aligned}\quad
\begin{aligned}
y(t) &= \sinh t \\
  &= \tfrac{1}{2}(e^{t} - e^{-t}),
\end{aligned}\qquad -\infty < t < \infty.
\tag{2}
$$
Just as the point
$$
x(t) = \cos t,\quad
y(t) = \sin t,\qquad -\pi < t < \pi
$$
on the unit circle subtends a sector of signed area $t/2$, the hyperbolic parametrization (2) subtends a "hyperbolic sector" of signed area $t/2$:


The slope of the segment from $(0, 0)$ to $(\cosh t, \sinh t)$ satisfies
$$
-1 < \frac{\sinh t}{\cosh t} = \tanh t < 1,
$$
so there exists a unique $\theta$ with $|\theta| < \pi/2$ such that $\sin\theta = \tanh t$. With this choice of $\theta$,
$$
(\cosh t, \sinh t) = (\sec\theta, \tan\theta).
$$
That is, (1) is the reparametrization of (2) arising from setting $\sin\theta = \tanh t$. The parameter $\theta$ doesn't have any particularly simple geometric interpretation; $\sin\theta$ is the slope of the segment from the origin to $(\cosh t, \sinh t)$.
A: $\theta$ is a parametric angle between the x-axis & the normal, passing through the origin & the point of tangency corresponding to the foot of perpendicular drawn from arbitrary point $(x, y)$   
(Here is a link)
Draw two auxiliary concentric circles, having radii $a$ & $b$, centered at the origin.
Now, select any arbitrary point say $P(x, y)$ on the right branch of hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as it's symmetrical about the y-axis  
then drop the perpendicular from the point $P(x, y)$ on the transverse axis i.e. x-axis say at the point $S$ (i.e. foot of perpendicular).  Now, draw a tangent (having negative slope if point $P(x, y)$ is above the x-axis)  to the circle of radius $a$ then the angle between the x-axis & the normal passing through the origin & the point of tangency is $\theta$ 
