Precalculus Honors · Study guide

5.2 · Ellipses: axes, foci, and shifted equations

Identify major-axis direction, derive focal distance from semiaxes, and interpret all features relative to the center.

Precalculus Honors · Reading: Demana 10th ed. §8.2; Stewart 8th ed. §§10.2, 10.4

Learning objectives

Find ellipse vertices, covertices, and foci.
Use the larger denominator to locate the major axis.
Build an ellipse equation from geometric data.

The denominators are squared semiaxis lengths

For a horizontal ellipse, $\displaystyle (x-h)^2/a^2+(y-k)^2/b^2=1$ with $\displaystyle a>b>0$ . The center is $\displaystyle (h,k)$ and the major axis is horizontal. If the larger denominator is under the y-term, the major axis is vertical.

The major-axis length is $\displaystyle 2a$ , not $\displaystyle a$ or $\displaystyle a^2$ . Vertices and covertices are displaced from the center by the corresponding semiaxis lengths.

Foci lie inside the ellipse

The focal distance satisfies $\displaystyle c^2=a^2-b^2$ . For a horizontal ellipse the foci are $\displaystyle (h\pm c,k)$ ; for a vertical ellipse they are $\displaystyle (h,k\pm c)$ .

The sum of the distances to the two foci is $\displaystyle 2a$ . Eccentricity is $\displaystyle c/a$ , between zero and one for a noncircular ellipse. Do not use the hyperbola’s plus relationship.

Translate every feature with the center

For $\displaystyle (x-2)^2/25+(y+1)^2/9=1$ , the center is $\displaystyle (2,-1)$ , $\displaystyle a=5$ , $\displaystyle b=3$ , and $\displaystyle c=4$ . Every feature is measured from this center.

An equation with both squared terms positive can describe a real ellipse, a point, or no real locus depending on the constant after completing squares. Normalize to a positive right side before classifying.

Worked examples

An origin-centered ellipse

Analyze $\displaystyle x^2/25+y^2/9=1$ .

The major axis is horizontal with $\displaystyle a=5$ and $\displaystyle b=3$ .
The focal distance is $\displaystyle c=\sqrt{25-9}=4$ .
Vertices are $\displaystyle (\pm5,0)$ and foci are $\displaystyle (\pm4,0)$ .

A vertical shifted ellipse

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

The center is $\displaystyle (1,-2)$ .
The larger denominator lies under the y-term, so $\displaystyle a=3$ vertically.
The major vertices are $\displaystyle (1,1)$ and $\displaystyle (1,-5)$ .

Common mistakes

Using a squared denominator as an axis length.
Putting the foci on the minor axis.
Using the hyperbola focal relation instead of subtracting.

Try it yourself

No calculator required.

1. Find the vertices, covertices, and foci of $\displaystyle x^2/25+y^2/9=1$ .

The ellipse x squared over 25 plus y squared over 9 equals 1 has a horizontal major axis. — Read semiaxis lengths from the squared denominators.

Show worked solution

The vertices are $\displaystyle (\pm5,0)$ , covertices $\displaystyle (0,\pm3)$ , and foci $\displaystyle (\pm4,0)$ because $\displaystyle c^2=25-9=16$ .

The ellipse x squared over 25 plus y squared over 9 equals 1 has a horizontal major axis. Vertices are (plus or minus 5, 0), covertices (0, plus or minus 3), and foci (plus or minus 4, 0). — Vertices are (plus or minus 5, 0), covertices (0, plus or minus 3), and foci (plus or minus 4, 0).

2. Find the center and major-axis length of $\displaystyle (x-2)^2/16+(y+1)^2/4=1$ .

Show worked solution

The center is $\displaystyle (2,-1)$ and the major semiaxis is $\displaystyle 4$ , so the major-axis length is $\displaystyle 8$ .

3. An origin-centered ellipse has horizontal vertices $\displaystyle (\pm5,0)$ and foci $\displaystyle (\pm3,0)$ . Write its equation.

Show worked solution

Here $\displaystyle a=5$ , $\displaystyle c=3$ , so $\displaystyle b^2=25-9=16$ . The equation is $\displaystyle x^2/25+y^2/16=1$ .

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