Hyperbola with Vertical Transverse Axis

Transverse axis along the y-axis, length 2a. Vertices at (0, ±a). Foci at (0, ±c) where c² = a² + b². Asymptotes: y = ±(a/b)x.

Place the foci at $S(0, -c)$ and $S'(0, c)$. The condition $|PS - PS'| = 2a$ with $c > a$ gives, by the same derivation with $x$ and $y$ interchanged:

$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, \quad b^2 = c^2 - a^2$$

Elements:

Distinguishing the two forms: In $x^2/A^2 - y^2/B^2 = 1$, the positive term determines the transverse axis. If $x^2$ is positive, transverse axis is along x; if $y^2$ is positive, it is along y.

Note: unlike the ellipse, there is no constraint that $A > B$ or $A < B$ — a hyperbola can have $a > b$, $a = b$, or $a < b$.