Hyperbolas

By the end of this lesson, you will be able to identify horizontal and vertical hyperbolas, find center, vertices, foci, and asymptotes, and sketch hyperbolas using the asymptote rectangle.

The Geometric Definition

A hyperbola is the set of all points for which the difference of distances from two fixed foci is constant.

$$|PF_1 - PF_2| = 2a$$

Unlike an ellipse (sum = constant), the hyperbola uses the absolute difference — and produces two separate branches.

Standard Forms

Horizontal hyperbola (transverse axis along $x$):

$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$

Vertical hyperbola (transverse axis along $y$):

$$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$

Key Features

Feature	Horizontal	Vertical
Center	$(h, k)$	$(h, k)$
Vertices	$(h \pm a, k)$	$(h, k \pm a)$
Foci	$(h \pm c, k)$	$(h, k \pm c)$
Asymptotes	$y - k = \pm\frac{b}{a}(x-h)$	$y - k = \pm\frac{a}{b}(x-h)$

Key relationship: $c^2 = a^2 + b^2$ (note: addition, unlike ellipse!)

The Asymptote Rectangle

Draw a rectangle centered at $(h,k)$ with half-widths $a$ (horizontal) and $b$ (vertical). The diagonals of this rectangle are the asymptotes. The hyperbola opens through the vertices (ends of the transverse axis) and approaches but never crosses the asymptotes.

Example 1 — Full Analysis

$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Horizontal (x-term positive). $a^2 = 9 \to a = 3$; $b^2 = 16 \to b = 4$

$c = \sqrt{9+16} = 5$

Center: $(0,0)$; Vertices: $(\pm 3, 0)$; Foci: $(\pm 5, 0)$

Asymptotes: $y = \pm\frac{4}{3}x$

Example 2 — Vertical Hyperbola

$\dfrac{y^2}{4} - \dfrac{x^2}{12} = 1$

Vertical ($y$-term positive). $a^2 = 4 \to a = 2$; $b^2 = 12 \to b = 2\sqrt{3}$

$c = \sqrt{4+12} = 4$

Center: $(0,0)$; Vertices: $(0, \pm 2)$; Foci: $(0, \pm 4)$

Asymptotes: $y = \pm\frac{a}{b}x = \pm\frac{2}{2\sqrt{3}}x = \pm\frac{1}{\sqrt{3}}x = \pm\frac{\sqrt{3}}{3}x$

Example 3 — Translated Hyperbola

$\dfrac{(x-1)^2}{4} - \dfrac{(y+2)^2}{9} = 1$