Math Core

An ellipse keeps the sum of the distances to two foci constant. Change one word, keep the difference constant instead, and you get a hyperbola: a curve with two separate branches that bend away from each other and straighten out along a pair of lines. Hyperbolas appear in the paths of comets that pass the sun only once, in the shape of cooling towers, and in navigation systems that locate a ship from differences in signal arrival times.

Two foci and a constant difference

Definition

Hyperbola

A hyperbola is the set of all points in a plane for which the difference of the distances to two fixed points, the foci, has a constant absolute value. That constant is written $2a$ .

The vocabulary mirrors the ellipse:

The center is the midpoint of the foci, and each focus is $c$ units from it.
The two points of the hyperbola on the line through the foci are the vertices. They are $a$ units from the center, and the segment joining them is the transverse axis, of length $2a$ .
The segment of length $2b$ through the center, perpendicular to the transverse axis, is the conjugate axis. Its endpoints are not on the hyperbola, but they help you draw it.

For a hyperbola the foci are farther from the center than the vertices, so $c > a$ . The three numbers are related by

c^2 = a^2 + b^2.

Notice the plus sign. For an ellipse you subtract; for a hyperbola you add.

The standard equation and the asymptotes

With the center at the origin and foci $(\pm c, 0)$ , the definition leads (after squaring twice and setting $b^2 = c^2 - a^2$ ) to $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ .

Solve that equation for $y$ : $y = \pm\dfrac{b}{a}\sqrt{x^2 - a^2}$ . When $|x|$ is large, $\sqrt{x^2 - a^2}$ is very close to $|x|$ , so the branches hug the lines $y = \pm\dfrac{b}{a}x$ . These are the asymptotes of the hyperbola.

Standard form of a hyperbola with center (h, k)

	Opens left and right	Opens up and down
Equation	$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$	$\dfrac{(y - k)^2}{a^2} - \dfrac{(x - h)^2}{b^2} = 1$
Vertices	$(h \pm a,\ k)$	$(h,\ k \pm a)$
Foci	$(h \pm c,\ k)$	$(h,\ k \pm c)$
Asymptotes	$y - k = \pm\dfrac{b}{a}(x - h)$	$y - k = \pm\dfrac{a}{b}(x - h)$

In both cases $c^2 = a^2 + b^2$ . The positive squared term tells you the direction: $a^2$ is always the denominator of the positive term, even if it is smaller than $b^2$ .

Sketching with the central box

To sketch a hyperbola quickly:

Plot the center $(h, k)$ .
Draw a rectangle centered there that extends $a$ units along the transverse axis and $b$ units along the conjugate axis.
Draw the diagonals of the rectangle, extended. These are the asymptotes.
Draw each branch through a vertex, curving toward the asymptotes.

Worked example: A hyperbola centered at the origin

Find the vertices, foci and asymptotes of $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ .

Solution. The $x^2$ -term is positive, so the hyperbola opens left and right, with $a^2 = 9$ and $b^2 = 16$ . So $a = 3$ , $b = 4$ and

c^2 = 9 + 16 = 25, \qquad c = 5.

Vertices: $(\pm 3, 0)$
Foci: $(\pm 5, 0)$
Asymptotes: $y = \pm\dfrac{4}{3}x$

Check the definition at the vertex $(3, 0)$ : its distances to the foci are $8$ and $2$ , and $8 - 2 = 6 = 2a$ . ✓

x²/9 − y²/16 = 1 with its dashed asymptotes y = ±(4/3)x.Open in grapher →

Worked example: A shifted hyperbola that opens up and down

Find the center, vertices, foci and asymptotes of $\dfrac{(y - 1)^2}{4} - \dfrac{(x + 2)^2}{9} = 1$ .

Solution. The center is $(-2, 1)$ . The $y$ -term is positive, so the hyperbola opens up and down with $a^2 = 4$ and $b^2 = 9$ . Here $a = 2$ is smaller than $b = 3$ , and that's fine: $a$ goes with the positive term.

$c = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$ .

Vertices: $(-2, 1 \pm 2)$ , which are $(-2, 3)$ and $(-2, -1)$
Foci: $\left(-2,\ 1 \pm \sqrt{13}\right)$ , about $(-2, 4.61)$ and $(-2, -2.61)$
Asymptotes: $y - 1 = \pm\dfrac{2}{3}(x + 2)$

This hyperbola opens vertically. Its asymptotes have slopes ±2/3.Open in grapher →

Tip

You don't need to memorize which asymptote formula uses $\frac{b}{a}$ and which uses $\frac{a}{b}$ . Replace the $1$ on the right side with $0$ and solve for $y$ . For $\dfrac{(y - 1)^2}{4} - \dfrac{(x + 2)^2}{9} = 0$ , you get $y - 1 = \pm\dfrac{2}{3}(x + 2)$ .

Common mistake

Two mix-ups cause most hyperbola errors. First, for a hyperbola $c^2 = a^2 + b^2$ (add), unlike an ellipse. Second, $a^2$ is under the positive term, not the larger number. In $\dfrac{y^2}{4} - \dfrac{x^2}{9} = 1$ , $a^2 = 4$ and the hyperbola opens up and down.

Writing the equation

Worked example: From vertices and foci

A hyperbola has vertices $(1, -2)$ and $(7, -2)$ and foci $(-1, -2)$ and $(9, -2)$ . Write its equation.

Solution. The center is the midpoint of the vertices: $(4, -2)$ . The vertices lie on a horizontal line, so the hyperbola opens left and right.

From the center, the vertices are $3$ units away, so $a = 3$ . The foci are $5$ units away, so $c = 5$ . Then

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

and the equation is

\frac{(x - 4)^2}{9} - \frac{(y + 2)^2}{16} = 1.

From general form

An equation with an $x^2$ -term and a $y^2$ -term of opposite signs is a hyperbola (unless it is degenerate). Complete the square in each variable, just as for an ellipse. Be careful with the minus sign when you factor.

Worked example: Complete the square

Write $9x^2 - 4y^2 - 18x - 16y - 43 = 0$ in standard form and find its asymptotes.

Solution. Group and factor. Note that $-4y^2 - 16y = -4(y^2 + 4y)$ :

9(x^2 - 2x) - 4(y^2 + 4y) = 43.

Adding $1$ inside the first group adds $9$ . Adding $4$ inside the second group adds $-4 \cdot 4 = -16$ :

\begin{aligned} 9(x^2 - 2x + 1) - 4(y^2 + 4y + 4) &= 43 + 9 - 16 \\ 9(x - 1)^2 - 4(y + 2)^2 &= 36 \\ \frac{(x - 1)^2}{4} - \frac{(y + 2)^2}{9} &= 1. \end{aligned}

The center is $(1, -2)$ with $a = 2$ and $b = 3$ , opening left and right. The asymptotes are $y + 2 = \pm\dfrac{3}{2}(x - 1)$ .

Practice

Practice 1

The foci of $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$ lie on the $x$ -axis. Enter the $x$ -coordinates of both foci.

Separate answers with commas, e.g. 2, -5

Practice 2

Find the slopes of the two asymptotes of $\dfrac{y^2}{25} - \dfrac{x^2}{4} = 1$ .

Separate answers with commas, e.g. 2, -5

Practice 3

Which hyperbola opens up and down?

Practice 4

Find the focus with the greater $x$ -coordinate for the hyperbola $\dfrac{(x + 1)^2}{36} - \dfrac{(y - 4)^2}{64} = 1$ .

Enter a point like (2, -3)

Practice 5

Which is the equation of the hyperbola with vertices $(0, \pm 3)$ and foci $(0, \pm 5)$ ?

Practice 6

Find the center of the hyperbola $4y^2 - x^2 + 24y + 4x + 28 = 0$ .

Enter a point like (2, -3)

Practice 7

A point $P$ lies on the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ . What is the positive difference between the distances from $P$ to the two foci?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A hyperbola centered at the origin has vertices $(\pm 3, 0)$ and asymptotes $y = \pm 2x$ . How far is each focus from the center? Give an exact answer.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.