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?
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$\begingroup$Here is an animated gif of $\theta$ changing.
Excuse the red frames (90 and 270 degree). That's my software hitting the discontinuity.
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 $(\cosh t, \sinh t)$.
$\endgroup$ $\begingroup$$\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 $\theta$
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