Math Core

Planets travel around the sun in ellipses, and a whisper in an elliptical room can be heard clearly across the floor. A parabola has one focus; an ellipse has two, and its shape comes from a simple rule about the distances to them. In this lesson you'll learn that rule, the standard equation, and how to find the center, vertices and foci from any equation of an ellipse.

Two foci and a constant sum

Pin two tacks to a board, loop a string around them, and trace a curve while keeping the string tight. The pencil draws an ellipse, because the total length of string from the pencil to the two tacks never changes.

Definition

Ellipse

An ellipse is the set of all points in a plane whose distances to two fixed points, the foci, have a constant sum. That sum is written $2a$ .

The vocabulary for an ellipse:

The center is the midpoint of the two foci.
The major axis is the longest chord. It passes through both foci and has length $2a$ . Its endpoints are the vertices.
The minor axis is the chord through the center perpendicular to the major axis. It has length $2b$ . Its endpoints are the co-vertices.
Each focus is $c$ units from the center.

How a, b and c are related

Look at a co-vertex. It is the same distance from both foci, and the two distances add to $2a$ , so each one equals $a$ . Those two distances, together with the center, form right triangles with legs $b$ (center to co-vertex) and $c$ (center to focus) and hypotenuse $a$ . By the Pythagorean theorem,

a^2 = b^2 + c^2, \qquad\text{so}\qquad c^2 = a^2 - b^2.

This means $a$ is always the largest of the three numbers: the vertices are farther from the center than the foci and farther than the co-vertices.

The standard equation

Place the center at the origin with foci $(\pm c, 0)$ . Setting the sum of the two distances equal to $2a$ , squaring twice and using $b^2 = a^2 - c^2$ produces the equation $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ . Shifting the center to $(h, k)$ gives the general standard form.

Standard form of an ellipse with center (h, k)

	Horizontal major axis	Vertical major axis
Equation	$\dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$	$\dfrac{(x - h)^2}{b^2} + \dfrac{(y - k)^2}{a^2} = 1$
Vertices	$(h \pm a,\ k)$	$(h,\ k \pm a)$
Co-vertices	$(h,\ k \pm b)$	$(h \pm b,\ k)$
Foci	$(h \pm c,\ k)$	$(h,\ k \pm c)$

In both cases $a > b > 0$ and $c^2 = a^2 - b^2$ . The larger denominator is $a^2$ , and it sits under the variable of the major axis.

When $a = b$ , the foci merge at the center ( $c = 0$ ) and the ellipse is a circle of radius $a$ .

Worked example: An ellipse centered at the origin

Find the vertices, co-vertices and foci of $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ .

Solution. The larger denominator, $25$ , is under $x^2$ , so the major axis is horizontal with $a^2 = 25$ and $b^2 = 9$ . That gives $a = 5$ and $b = 3$ , and

c^2 = 25 - 9 = 16, \qquad c = 4.

Vertices: $(\pm 5, 0)$
Co-vertices: $(0, \pm 3)$
Foci: $(\pm 4, 0)$

Check the definition at the vertex $(5, 0)$ : its distances to the foci are $1$ and $9$ , which add to $10 = 2a$ . ✓

x²/25 + y²/9 = 1 with vertices (V), co-vertices and foci (F).Open in grapher →

Worked example: A shifted ellipse with a vertical major axis

Find the center, vertices and foci of $\dfrac{(x - 1)^2}{9} + \dfrac{(y + 2)^2}{25} = 1$ .

Solution. The center is $(1, -2)$ . The larger denominator, $25$ , is under the $y$ -term, so the major axis is vertical: $a = 5$ , $b = 3$ , and $c = \sqrt{25 - 9} = 4$ .

Move up and down from the center for the vertices and foci:

Vertices: $(1, -2 \pm 5)$ , which are $(1, 3)$ and $(1, -7)$
Co-vertices: $(1 \pm 3, -2)$ , which are $(4, -2)$ and $(-2, -2)$
Foci: $(1, -2 \pm 4)$ , which are $(1, 2)$ and $(1, -6)$

The major axis is vertical because the larger denominator is under the y-term.Open in grapher →

Common mistake

Don't assume $a^2$ is the number under $x^2$ . It is always the larger denominator. And for an ellipse, $c^2 = a^2 - b^2$ (subtract), not $a^2 + b^2$ . The foci always lie inside the ellipse, so $c$ must be smaller than $a$ .

Writing the equation

To write the equation, find the center, decide the orientation, and find $a^2$ and $b^2$ .

Worked example: From foci and major axis length

An ellipse has foci $(-2, 1)$ and $(6, 1)$ , and its major axis has length $10$ . Write its equation.

Solution. The center is the midpoint of the foci: $\left(\dfrac{-2 + 6}{2}, 1\right) = (2, 1)$ . The foci lie on a horizontal line, so the major axis is horizontal.

The foci are $4$ units from the center, so $c = 4$ . The major axis has length $2a = 10$ , so $a = 5$ . Then

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

The equation is

\frac{(x - 2)^2}{25} + \frac{(y - 1)^2}{9} = 1.

From general form

Expanded equations such as $4x^2 + 9y^2 - 16x + 18y - 11 = 0$ describe ellipses when both squared terms have positive coefficients that are different. Complete the square in each variable, then divide so the right side is $1$ .

Worked example: Complete the square twice

Write $4x^2 + 9y^2 - 16x + 18y - 11 = 0$ in standard form and find the foci.

Solution. Group and factor out the leading coefficients:

4(x^2 - 4x) + 9(y^2 + 2y) = 11.

Complete each square. Inside the first group add $4$ , which really adds $4 \cdot 4 = 16$ ; inside the second add $1$ , which adds $9 \cdot 1 = 9$ :

\begin{aligned} 4(x^2 - 4x + 4) + 9(y^2 + 2y + 1) &= 11 + 16 + 9 \\ 4(x - 2)^2 + 9(y + 1)^2 &= 36 \\ \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} &= 1. \end{aligned}

The center is $(2, -1)$ , $a = 3$ , $b = 2$ and the major axis is horizontal. Then $c = \sqrt{9 - 4} = \sqrt{5}$ , and the foci are $\left(2 \pm \sqrt{5},\ -1\right)$ , about $(4.24, -1)$ and $(-0.24, -1)$ .

(x - 2)^2/9 + (y + 1)^2/4 = 1(2 + sqrt(5), -1)(2 - sqrt(5), -1)(2, -1)Open in grapher →

Tip

When you factor out a coefficient before completing the square, remember that the number you add inside the parentheses gets multiplied by that coefficient. Add the product to the other side.

How round is it? Eccentricity

The ratio $e = \dfrac{c}{a}$ is the eccentricity of the ellipse. Since $0 \le c < a$ , it satisfies $0 \le e < 1$ . An eccentricity near $0$ means the foci are close to the center and the ellipse is nearly a circle. An eccentricity near $1$ means a long, flat ellipse. Earth's orbit has $e \approx 0.017$ , which is almost perfectly round.

Practice

Practice 1

The ellipse $\dfrac{x^2}{100} + \dfrac{y^2}{36} = 1$ has its foci on the $x$ -axis. Enter the $x$ -coordinates of both foci.

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

Practice 2

Find the vertex with the greater $y$ -coordinate of the ellipse $\dfrac{(x + 3)^2}{4} + \dfrac{(y - 1)^2}{49} = 1$ .

Enter a point like (2, -3)

Practice 3

Which is the equation of the ellipse with vertices $(0, \pm 6)$ and foci $(0, \pm 4)$ ?

Practice 4

Find the center of the ellipse $9x^2 + 4y^2 + 36x - 24y + 36 = 0$ .

Enter a point like (2, -3)

Practice 5

Find the eccentricity of $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ .

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

Practice 6

An elliptical "whispering gallery" is $50$ feet long and $30$ feet wide. Sound leaving one focus reflects off the wall to the other focus. How far from the center of the room should each person stand, in feet?

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

Practice 7

An ellipse centered at the origin has vertices $(\pm 10, 0)$ and passes through the point $(6, 4)$ . Find $b^2$ .

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