In case of an ellipse and its auxiliary circle (the circle with major axis as its diameter), the meaning of "corresponding points" is straightforward. Let us consider the following diagram which shows the upper half of an ellipse with its major axis horizontal:
Image Source : Florida Center for Instructional Technology
$QM$ is perpendicular to the horizontal. Here $Q$ and $P$ are called as "corresponding points" on the auxiliary circle and the ellipse respectively. And the angle $QOM$ is called the "eccentric angle" of the point $P$.
In case of an hyperbola, I understood that the auxiliary circle is the one which has its centre at the centre of hyperbola (usually origin) and diameter equal to the length of transverse axis. But I don't understand how "corresponding points" are defined in this case.
Or in other words, if we're given a hyperbola and its auxiliary circle, how to find the corresponding point on the hyperbola for a point on the circle? Further, how many corresponding points on the hyperbola exists for each point on the circle (I'm having this doubt because hyperbola has two branches and suspect a second "corresponding point" on the other branch)?
In my book, a diagram similar to the following one was given in the section "Auxiliary circle of a Hyperbola":
Image Source : Florida Center for Instructional Technology
Best Answer
Expanding a comment ...
As shown in OP's second figure: From $P$ on the hyperbola, drop a perpendicular to $M$ on the transverse axis, $Q$ be one of the points for which $\overline{MQ}$ is tangent to the circle. (We'll discuss which one of the points below.) Then $P$ and $Q$ are "corresponding points". (So, we've traded "transfer $M$ perpendicularly to the circle" in the ellipse case to "transfer $M$ tangentially to the circle" in the hyperbola case, which makes some sense in a "pole and polar" context.)
The construction can be reversed: From $Q$ on the circle, let $M$ be such that $\overline{QM}$ is tangent to the circle, then let $P$ be one of the points on the hyperbola such that $\overline{MP}$ is perpendicular to the hyperbola's transverse axis. (Again, there's ambiguity in the choice of $P$.)
Ambiguities aside, we find that every finite point $P$ on either branch of the hyperbola corresponds to some point on the unit circle except its top- and bottom-most points. The two "points at infinity" on the hyperbola correspond to those last two points on the circle.
As for those ambiguities ... This animation shows the "natural" way of resolving them. As $Q$ travels normally around the circle through Quadrants 1, 2, 3, 4, the corresponding $P$ travels along the hyperbola in Quadrants 1, 3, 2, 4; Quadrants 2 and 3 are "flipped".
This is because as $Q$ passes from Q1 to Q2 through the top-most point of the circle, $P$ passes from Q1 to Q3 "via the blue asymptote". Likewise, as $Q$ passes from Q3 to Q4, $P$ passes from Q2 to Q4 "via the red asymptote".
This quadrant-flipping notion happens to arise naturally from the equations, too. Let the hyperbola have equation $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{1}$$ so that the auxiliary circle's equation is $$x^2+y^2=a^2 \tag{2}$$ For a point $Q = (x_Q,y_Q)$ on the circle, one can show that $M = (a^2/x_Q,0)$. Of course, $P$ shares its $x$-coordinate with $M$; the $y$-coordinate, solved-for in $(1)$ has a sign ambiguity: $$\begin{align}P &= \left(\frac{a^2}{x_Q}, \pm b \sqrt{\frac{(a^2/x_Q)^2}{a^2}-1}\right) = \left(\frac{a^2}{x_Q},\pm b\sqrt{\frac{a^2-x_Q^2}{x_Q^2}}\right) = \left(\frac{a^2}{x_Q},\pm b\sqrt{\frac{y_Q^2}{x_Q^2}}\right) \\[4pt] &= \left(\frac{a^2}{x_Q},\pm b\left| \frac{y_Q}{x_Q}\right|\right)\tag{3}\end{align}$$
So, we strip $y_Q/x_Q$ of its sign, only to immediately apply an ambiguous sign. That seems somewhat silly. "Quadrant-flipping" arises by letting $y_Q/x_Q$ determine its own fate, so that we have $$P = \left(\frac{a^2}{x_Q},b\frac{y_Q}{x_Q} \right) \tag{4} $$ Thus, $P$'s $y$-coordinate is positive when $Q$'s coordinates have the same sign; that is, $P$ is in Quadrants 1 and 2 when $Q$ is in Quadrants 1 and 3; similarly, $P$ is in Quadrants 3 and 4 when $Q$ is in Quadrants 2 and 4. Again, Quadrants 2 and 3 are "flipped" for $P$ and $Q$.