Hyperbolas

A hyperbola is the set of all points in a plane such that the absolute difference of the distances from two fixed points (the foci) is a constant. While an ellipse uses the sum of distances, a hyperbola uses the difference — this produces two separate branches that open away from each other. Hyperbolas model the paths of some comets, the shape of cooling towers, and the mathematics behind GPS triangulation.

The Definition and Key Measurements

For any point $P$ on the hyperbola with foci $F_1$ and $F_2$:

$|d(P, F_1) - d(P, F_2)| = 2a$

Key measurements:

• $a$ = distance from center to each vertex (along the transverse axis)
• $b$ = distance from center to each co-vertex (along the conjugate axis)
• $c$ = distance from center to each focus

The crucial relationship is:

$c^2 = a^2 + b^2$

Compare this with ellipses ($c^2 = a^2 - b^2$). For hyperbolas, $c > a$ — the foci are always outside the vertices.

Standard Forms

Horizontal Transverse Axis (Opens Left and Right)

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

• Center: $(h, k)$
• Vertices: $(h \pm a, k)$
• Foci: $(h \pm c, k)$ where $c = \sqrt{a^2 + b^2}$
• Asymptotes: $y - k = \pm\frac{b}{a}(x - h)$

Vertical Transverse Axis (Opens Up and Down)

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

• Center: $(h, k)$
• Vertices: $(h, k \pm a)$
• Foci: $(h, k \pm c)$ where $c = \sqrt{a^2 + b^2}$
• Asymptotes: $y - k = \pm\frac{a}{b}(x - h)$

Key pattern: The positive term tells you the transverse axis direction. If $x^2$ is positive, the hyperbola opens left-right. If $y^2$ is positive, it opens up-down. The value under the positive term is $a^2$.

Asymptotes and the Asymptote Box

The asymptotes are the “guide rails” that the branches approach but never touch. To graph a hyperbola:

1. Plot the center
2. Draw a rectangle (the asymptote box) with width $2a$ along the transverse axis and height $2b$ along the conjugate axis
3. Draw diagonals through the box — these are the asymptotes
4. Sketch the branches curving through the vertices and approaching the asymptotes

For $\frac{x^2}{9} - \frac{y^2}{16} = 1$: the box extends $\pm 3$ horizontally and $\pm 4$ vertically. The asymptotes are $y = \pm\frac{4}{3}x$.

Worked Examples

Example 1: Finding All Parts

Analyze $\frac{(x + 1)^2}{16} - \frac{(y - 3)^2}{9} = 1$.

Step 1: Center: $(-1, 3)$. Positive term is under $x$, so the transverse axis is horizontal.

Step 2: $a^2 = 16$, $a = 4$. $b^2 = 9$, $b = 3$. $c = \sqrt{16 + 9} = \sqrt{25} = 5$.

• Vertices: $(-1 \pm 4, 3) = (-5, 3)$ and $(3, 3)$
• Foci: $(-1 \pm 5, 3) = (-6, 3)$ and $(4, 3)$
• Asymptotes: $y - 3 = \pm\frac{3}{4}(x + 1)$

Example 2: Vertical Transverse Axis

Find the vertices, foci, and asymptotes of $\frac{y^2}{25} - \frac{x^2}{4} = 1$.

Positive term is under $y$, so the transverse axis is vertical. Center: $(0, 0)$.

$a^2 = 25$, $a = 5$. $b^2 = 4$, $b = 2$. $c = \sqrt{25 + 4} = \sqrt{29}$.

• Vertices: $(0, \pm 5)$
• Foci: $(0, \pm\sqrt{29}) \approx (0, \pm 5.39)$
• Asymptotes: $y = \pm\frac{5}{2}x$

Example 3: Writing the Equation

Write the equation of a hyperbola with center $(0, 0)$, vertices at $(\pm 3, 0)$, and foci at $(\pm 5, 0)$.

Horizontal transverse axis. $a = 3$, $c = 5$.

$b^2 = c^2 - a^2 = 25 - 9 = 16$

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

Example 4: Converting from General Form

Convert $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ to standard form.

Step 1 — Group and factor:

$4(x^2 - 6x) - 9(y^2 + 4y) = 36$

Step 2 — Complete the square:

$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36$

$4(x - 3)^2 - 9(y + 2)^2 = 36$

Step 3 — Divide by 36:

$\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 1$

Center: $(3, -2)$. $a = 3$, $b = 2$, $c = \sqrt{13}$. Horizontal transverse axis.

Real-World Application: GPS and LORAN

GPS and its predecessor LORAN (Long Range Navigation) use hyperbolas. A receiver measures the time difference of signals from two transmitters. Because the signal travels at a constant speed, a fixed time difference means a fixed difference in distances — which defines a hyperbola with the transmitters as foci.

Two transmitters are 300 km apart. A ship detects that signals from transmitter $A$ arrive 0.5 milliseconds before signals from transmitter $B$. Radio signals travel at approximately 300,000 km/s, so the distance difference is:

$\Delta d = 300{,}000 \times 0.0005 = 150 \text{ km}$

This means $2a = 150$, so $a = 75$ km. The transmitters are foci with $2c = 300$, so $c = 150$ km.

$b^2 = c^2 - a^2 = 22{,}500 - 5{,}625 = 16{,}875$

The ship lies on the hyperbola $\frac{x^2}{5625} - \frac{y^2}{16875} = 1$. A second pair of transmitters produces a second hyperbola, and the intersection pinpoints the ship’s location.

Common Mistakes

1. Using $c^2 = a^2 - b^2$ (the ellipse formula). For hyperbolas: $c^2 = a^2 + b^2$.
2. Confusing which denominator is $a^2$. $a^2$ is always under the positive term, not necessarily the larger number.
3. Getting asymptote slopes backwards. For horizontal transverse axis: slope is $\pm b/a$. For vertical: slope is $\pm a/b$.
4. Subtracting instead of adding when completing the square. When you factor out a negative (for the $y^2$ group), completing the square inside the group still adds, but the negative outside flips the sign on the right side.

Practice Problems

Key Takeaways

• A hyperbola is the set of points where the absolute difference of distances to two foci equals $2a$
• The relationship $c^2 = a^2 + b^2$ means foci are farther from center than vertices (opposite of ellipses)
• The positive term in the equation determines the transverse axis direction
• Asymptotes guide graphing: use the asymptote box (width $2a$, height $2b$) and draw diagonals
• For horizontal transverse axis, asymptote slopes are $\pm b/a$; for vertical, slopes are $\pm a/b$
• Hyperbolas appear in GPS/LORAN navigation, cooling tower design, and comet trajectories

Return to College Algebra for more topics in this section.