Precalculus Honors · Study guide

5.3 · Hyperbolas: branches, foci, and asymptotes

Read the positive term to find opening direction, construct the asymptotes through the center, and locate foci beyond the vertices.

Precalculus Honors · Reading: Demana 10th ed. §8.3; Stewart 8th ed. §§10.3–10.4

Learning objectives

Identify hyperbola orientation and vertices.
Find focal distance using the sum of squared parameters.
Write asymptotes with correct slopes and shifts.

The positive squared term locates the transverse axis

The equation $\displaystyle (x-h)^2/a^2-(y-k)^2/b^2=1$ opens left and right. Reversing the positive and negative terms to make y positive gives an up-down hyperbola.

For a horizontal hyperbola, the vertices are $\displaystyle (h\pm a,k)$ . The larger denominator does not determine orientation; the positive term does.

Foci lie beyond the vertices

The focal distance is $\displaystyle c=\sqrt{a^2+b^2}$ , so $\displaystyle c>a$ . For a horizontal hyperbola, the foci are $\displaystyle (h\pm c,k)$ . The absolute difference of distances to the foci is $\displaystyle 2a$ .

The plus relationship distinguishes a hyperbola from an ellipse. Inspect the equation’s signs before choosing a focal formula.

Asymptotes pass through the center

For a horizontal hyperbola, the asymptotes are $\displaystyle y-k=\pm(b/a)(x-h)$ . For a vertical hyperbola they are $\displaystyle y-k=\pm(a/b)(x-h)$ . A guiding rectangle provides a geometric check on the slope.

A complete sketch shows two branches approaching those lines, the correct vertices, and the center. Treat the asymptotes as limiting directions rather than pieces of the hyperbola.

Worked examples

An origin-centered hyperbola

Analyze $\displaystyle x^2/9-y^2/16=1$ .

The branches open left and right with $\displaystyle a=3$ , $\displaystyle b=4$ .
The focal distance is $\displaystyle 5$ .
Vertices are $\displaystyle (\pm3,0)$ , foci $\displaystyle (\pm5,0)$ , and asymptotes $\displaystyle y=\pm(4/3)x$ .

A vertical hyperbola

Give vertices of $\displaystyle y^2/4-x^2/9=1$ .

The positive y-term makes the transverse axis vertical.
The value of $\displaystyle a$ is $\displaystyle 2$ .
The vertices are $\displaystyle (0,\pm2)$ .

Common mistakes

Choosing orientation from the larger denominator instead of the positive term.
Using c squared equals a squared minus b squared.
Forgetting to shift asymptotes through the center.

Try it yourself

No calculator required.

1. Find vertices, foci, and asymptotes of $\displaystyle x^2/9-y^2/16=1$ .

The curve x squared over 9 minus y squared over 16 equals 1 has two horizontal branches. — Use the positive term to locate the transverse axis.

Show worked solution

The vertices are $\displaystyle (\pm3,0)$ , foci $\displaystyle (\pm5,0)$ , and asymptotes $\displaystyle y=\pm(4/3)x$ . The branches open horizontally.

The curve x squared over 9 minus y squared over 16 equals 1 has two horizontal branches. Vertices are (plus or minus 3, 0), foci (plus or minus 5, 0), and asymptote slopes are plus or minus 4/3. — Vertices are (plus or minus 3, 0), foci (plus or minus 5, 0), and asymptote slopes are plus or minus 4/3.

2. For $\displaystyle y^2/4-x^2/9=1$ , give the opening direction and asymptotes.

Show worked solution

It opens vertically. The asymptotes are $\displaystyle y=\pm(2/3)x$ .

3. Find the center and vertices of $\displaystyle (x-1)^2/4-(y+2)^2/9=1$ .

Show worked solution

The center is $\displaystyle (1,-2)$ and the horizontal semiaxis is $\displaystyle 2$ . The vertices are $\displaystyle (-1,-2)$ and $\displaystyle (3,-2)$ .

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