Precalculus · G.GPE.A.3

The ellipse

Pin the two ends of a loop of string, pull it tight with a pencil, and drag the pencil all the way around. What curve comes out, and what equation does it obey?

You need

Equation of a circleGeometry

Unlocks

The hyperbolaPrecalculus

Where you are 6 in Conic sections

Geometry11

Conic sections6

Conic sections from a double cone Focus and directrix

The ellipse

The hyperbola The parabola as a locus Identifying and translating conics

Why this exists

Cutting a double cone with a tilted plane gives a closed, stretched curve, the ellipse. That description is a picture, not something you can compute with. A circle became computable once it was defined by distance: every point at distance r from a center, which turned into the equation of a circle. The ellipse needs the same treatment. The idea that works uses two fixed points instead of one, and it turns the stretched oval into an equation from which you can read its center, its size and its shape.

The lesson runs over two days. Day 1 is the definition, the equation of an ellipse centered at the origin, and its parts. Day 2 moves the center to (h, k), graphs an ellipse from its equation, writes the equation from its features, measures flatness, and uses the result in a room built on an ellipse.

Definition

Fix two points F_1 and F_2, the foci (one is a focus), and a length 2a greater than the distance F_1F_2. An ellipse is the set of all points P in the plane with

PF_1 + PF_2 = 2a.

The parts that go with it:

The center is the midpoint of F_1F_2. Write c for the distance from the center to each focus, so F_1F_2 = 2c and the condition says a > c \ge 0.
The major axis is the chord through both foci. Its ends are the vertices, and its length is 2a.
The minor axis is the chord through the center perpendicular to the major axis. Its ends are the co-vertices, and its length is 2b, where b^2 = a^2 - c^2.

When c = 0 the two foci are one point, every point of the curve is at distance a from it, and the ellipse is a circle of radius a.

Figure 1For every point P on the ellipse, d_1 + d_2 is the same number, 2a.

Intuition

A loop of string around two pins, pulled tight by a pencil, keeps PF_1 + PF_2 fixed: the pencil draws exactly this curve. Pins far apart use up most of the string between them, so the curve comes out long and flat. Pins close together leave the curve nearly round.

Two special points explain the letters. Say the vertex on the right of Figure 2 is v from the center. Its distances to the foci are v - c and v + c, which add to 2v. That sum must be 2a, so v = a: the major axis has length 2a. At a co-vertex the two distances are equal by symmetry, and they add to 2a, so each is a. That gives the right triangle in Figure 2, with legs b and c and hypotenuse a:

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

The hypotenuse is the longest side, so a > b: the major axis really is the longer one.

Figure 2The parts of an ellipse. The triangle gives b^2 + c^2 = a^2.

Results

Center at the origin. With the foci on the x-axis at (\pm c, 0), the ellipse is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad a > b > 0,

with vertices (\pm a, 0), co-vertices (0, \pm b) and foci (\pm c, 0), where c^2 = a^2 - b^2. With the foci on the y-axis the roles of x and y swap:

\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, \qquad a > b > 0,

with vertices (0, \pm a), co-vertices (\pm b, 0) and foci (0, \pm c). In both, a^2 is the larger denominator, and the variable over it names the direction of the major axis.

Center at (h, k). Replacing x by x - h and y by y - k slides every point by h across and k up, so

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

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

is the ellipse with center (h, k) and a horizontal or vertical major axis. Its vertices, co-vertices and foci are those of the centered ellipse, each moved by (h, k).

Eccentricity. The number

e = \frac{c}{a}, \qquad 0 \le e < 1,

measures flatness. It is 0 for a circle and creeps toward 1 as the foci slide out toward the vertices. It does not depend on size: doubling a and c keeps e. The same letter returns in focus and directrix, where it becomes the ratio that sorts every conic.

Reflection. At any point P of the ellipse, the segments PF_1 and PF_2 make equal angles with the tangent line at P. So a ray leaving one focus bounces off the ellipse straight toward the other focus.

Why it is true

The equation. Put F_1 = (-c, 0) and F_2 = (c, 0). By the distance formula, P = (x, y) is on the ellipse exactly when

\sqrt{(x + c)^2 + y^2} = 2a - \sqrt{(x - c)^2 + y^2}.

Squaring both sides and cancelling x^2, y^2 and c^2 from each side leaves \displaystyle 4cx = 4a^2 - 4a\sqrt{(x - c)^2 + y^2}, that is

a\sqrt{(x - c)^2 + y^2} = a^2 - cx.

Squaring again and collecting terms:

\begin{aligned} a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2 &= a^4 - 2a^2cx + c^2x^2 \\ (a^2 - c^2)x^2 + a^2y^2 &= a^2(a^2 - c^2). \end{aligned}

Since a > c, the number a^2 - c^2 is positive; call it b^2. Dividing by a^2b^2 gives \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

Squaring can add solutions, so run it backward. A point of the final equation has |x| \le a, so a^2 - cx > 0, and the step before the last holds with a positive right side. That step gives \displaystyle \sqrt{(x - c)^2 + y^2} = a - \frac{cx}{a} \le a + c < 2a, so the right side of the first equation is positive too, and every step reverses. The equation and the distance condition describe the same points.

Reflection. This argument takes two facts as given, without proof here: the tangent line at P touches the ellipse only at P and otherwise runs outside it, and every point outside the ellipse has a distance sum larger than 2a. Granting them, let Q be any point of the tangent line other than P; then QF_1 + QF_2 > 2a = PF_1 + PF_2. So among the points of that line, P has the smallest value of QF_1 + QF_2. The shortest path from F_1 to a line and on to F_2 is the one that meets the line at equal angles (reflect F_2 in the line and join it to F_1 with a straight segment). So the angles at P are equal.

Worked examples

Example 1

Day 1: reading the parts

Find the vertices, co-vertices and foci of \displaystyle \frac{x^2}{169} + \frac{y^2}{25} = 1, and the lengths of its axes.

Solution

The larger denominator, 169, is under x^2, so the major axis lies on the x-axis with a^2 = 169, a = 13, and b^2 = 25, b = 5. Then c^2 = a^2 - b^2 = 144, so c = 12.

Vertices (\pm 13, 0), co-vertices (0, \pm 5), foci (\pm 12, 0). The major axis has length 2a = 26 and the minor axis 2b = 10.

Check with the definition at the co-vertex (0, 5): its distance to (12, 0) is \displaystyle \sqrt{144 + 25} = 13, and the same to (-12, 0). The sum is 26 = 2a.

Example 2

Day 1: an equation to rearrange first

Find the vertices, co-vertices and foci of 16x^2 + 7y^2 = 112.

Solution

Standard form has 1 on the right, so divide both sides by 112:

\frac{x^2}{7} + \frac{y^2}{16} = 1.

Now 16 > 7 and 16 sits under y^2, so the major axis is vertical: a = 4 and \displaystyle b = \sqrt{7}. Then c^2 = 16 - 7 = 9 and c = 3.

Vertices (0, \pm 4), co-vertices \displaystyle (\pm\sqrt{7}, 0), foci (0, \pm 3). Check at the vertex (0, 4): its distances to the foci are 1 and 7, which add to 8 = 2a.

Example 3

Day 1: the equation from the definition

Write an equation for the set of points whose distances to (0, -2) and (0, 2) add to 10.

Solution

This is an ellipse with foci (0, \pm 2), so its center is the midpoint (0, 0), its major axis is vertical, and c = 2. The constant sum is 2a = 10, so a = 5. Then

b^2 = a^2 - c^2 = 25 - 4 = 21,

and a^2 goes under y^2:

\frac{x^2}{21} + \frac{y^2}{25} = 1.

Check at the co-vertex \displaystyle (\sqrt{21}, 0): its distance to each focus is \displaystyle \sqrt{21 + 4} = 5, and the two add to 10.

Example 4

Day 2: graphing a shifted ellipse

Graph \displaystyle \frac{(x - 1)^2}{16} + \frac{(y + 2)^2}{36} = 1, marking its center, vertices, co-vertices and foci.

Solution

Reading y + 2 as y - (-2), the center is (1, -2). The larger denominator, 36, is under the y term, so the major axis is vertical with a = 6; then b = 4 and \displaystyle c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47.

From the center, go 6 up and down: vertices (1, 4) and (1, -8). Go 4 left and right: co-vertices (-3, -2) and (5, -2). The foci are \displaystyle 2\sqrt{5} up and down, at \displaystyle (1, -2 + 2\sqrt{5}) and \displaystyle (1, -2 - 2\sqrt{5}). Sketch a smooth oval through the four ends of the axes.

Figure 3Example 4. The major axis is vertical because the larger denominator sits under the y term.

Example 5

Day 2: completing the square first

Find the center, vertices and foci of 4x^2 + 25y^2 + 24x - 100y + 36 = 0.

Solution

Group the x terms and the y terms, move the constant, and factor out each leading coefficient, as in completing the square for a circle:

4(x^2 + 6x) + 25(y^2 - 4y) = -36.

Complete each square inside its parentheses. Adding 9 inside the first adds 4 \cdot 9 = 36 to the left, and adding 4 inside the second adds 25 \cdot 4 = 100, so the right side gains the same:

4(x + 3)^2 + 25(y - 2)^2 = -36 + 36 + 100 = 100.

Dividing by 100:

\frac{(x + 3)^2}{25} + \frac{(y - 2)^2}{4} = 1.

The center is (-3, 2), the major axis is horizontal with a = 5, b = 2, and \displaystyle c = \sqrt{25 - 4} = \sqrt{21}. Vertices (2, 2) and (-8, 2); foci \displaystyle (-3 + \sqrt{21}, 2) and \displaystyle (-3 - \sqrt{21}, 2).

Example 6

Day 2: the equation from features, and flatness

An ellipse has vertices (-1, 4) and (7, 4) and one focus at (6, 4). Write its equation, find its eccentricity, and compare its flatness with the ellipse in Example 5.

Solution

The center is the midpoint of the vertices, (3, 4). The vertices differ only in x, so the major axis is horizontal with a = 4. The focus is 3 from the center, so c = 3, and b^2 = 16 - 9 = 7:

\frac{(x - 3)^2}{16} + \frac{(y - 4)^2}{7} = 1.

Its eccentricity is \displaystyle e = \frac{3}{4} = 0.75. The ellipse in Example 5 has \displaystyle e = \frac{\sqrt{21}}{5} \approx 0.92, closer to 1, so Example 5’s ellipse is the flatter one. Check against the axes: Example 5 has \displaystyle \frac{b}{a} = \frac{2}{5} = 0.4, and this one has \displaystyle \frac{b}{a} = \frac{\sqrt{7}}{4} \approx 0.66, so this one is rounder.

Example 7

Day 2: a whispering gallery

The floor of a room is an ellipse 100 feet long and 60 feet wide. OpenStax describes rooms of this kind, called whispering chambers: sound leaving one focus reflects off the curved wall toward the other focus. Anwar and Farhan stand at the foci. Write an equation for the floor, find how far apart they stand and how far each is from the nearest wall, and find the length of every path from Anwar to the wall and on to Farhan.

Solution

Put the center at the origin with the long direction on the x-axis. Then 2a = 100 and 2b = 60, so a = 50, b = 30, and

\frac{x^2}{2500} + \frac{y^2}{900} = 1.

Then c^2 = 2500 - 900 = 1600, so c = 40: Anwar and Farhan stand at (\pm 40, 0), 80 feet apart, and each is a - c = 10 feet from the end wall nearest to them.

A path from Anwar to a wall point P and on to Farhan has length PF_1 + PF_2, and every wall point is on the ellipse, so every such path is 2a = 100 feet long. For example, through the co-vertex (0, 30) it is \displaystyle 2\sqrt{40^2 + 30^2} = 2 \cdot 50 = 100 feet. By the reflection result, each of these paths is one that sound bouncing off the wall at P can follow, so sound from Anwar leaving in every direction reaches Farhan after the same distance.

Figure 4Example 7. Every path from one focus to the wall and on to the other focus has the same length.

Boundaries

The sum must beat the gap. The definition needs 2a > F_1F_2. If 2a = F_1F_2, the only points with that distance sum lie on the segment F_1F_2; if 2a < F_1F_2, no point qualifies, by the triangle inequality.
a is the larger semi-axis. In standard form a^2 is whichever denominator is larger, and the foci lie along that variable’s direction. Equal denominators make a circle, with c = 0 and no separate foci.
Axes parallel to the coordinate axes. These equations describe only ellipses whose axes are horizontal and vertical. A tilted ellipse has an xy term, which these forms do not have.
Not every such equation is an ellipse. After completing the square, the right side must be positive. A right side of 0 gives a single point, as in 4(x + 3)^2 + 25(y - 2)^2 = 0, and a negative one gives no points at all.
The reflection result is about the ideal curve. It is exact for a smooth ellipse. A real room’s wall, doors and people change how sound travels, so the mathematics says where the reflections point, not how loud anything is.

Where this goes next

Change the sum of distances to a difference and you get the next curve, the hyperbola, whose equation differs from this one by a single sign. The eccentricity \displaystyle e = \frac{c}{a} reappears in focus and directrix, which describes every conic by one ratio, with the ellipses as exactly the cases below 1. The parabola is the boundary case between the two. Reading the squared terms of an equation to tell these curves apart is identifying a conic.

Exercises

Day 1: reading an ellipse

Find the vertices, co-vertices and foci.

\displaystyle \frac{x^2}{100} + \frac{y^2}{36} = 1
\displaystyle \frac{x^2}{9} + \frac{y^2}{25} = 1
\displaystyle \frac{x^2}{49} + \frac{y^2}{24} = 1
4x^2 + y^2 = 16
x^2 + 9y^2 = 9

Day 1: writing the equation

Write the equation of the ellipse centered at the origin.

Vertices (\pm 8, 0) and foci (\pm 5, 0).
Foci (0, \pm 6) and co-vertices (\pm 8, 0).
The set of points whose distances to (-3, 0) and (3, 0) add to 12.
Vertical major axis of length 20 and minor axis of length 12. Give its foci too.
Show that \displaystyle \left(3, \frac{16}{5}\right) is on \displaystyle \frac{x^2}{25} + \frac{y^2}{16} = 1 in two ways: by substituting, and by adding its distances to the foci.

Day 2: shifted ellipses

Find the center, vertices, co-vertices and foci.

\displaystyle \frac{(x - 4)^2}{9} + \frac{(y - 1)^2}{25} = 1
\displaystyle \frac{(x + 5)^2}{64} + \frac{(y - 3)^2}{16} = 1
x^2 + 4y^2 - 6x + 16y + 21 = 0
25x^2 + 16y^2 + 100x - 96y - 156 = 0

Day 2: from features to equation

Center (2, -1), a vertex at (2, 5) and a co-vertex at (5, -1). Give its foci too.
Foci (-4, 2) and (6, 2), and a vertex at (8, 2).
Vertices (0, -3) and (0, 9), and eccentricity \displaystyle \frac{2}{3}.
Find the eccentricities of the ellipses in Exercises 11 and 12. Which is flatter?

Day 2: applications

The infield of Junaid’s elliptical racetrack is 340 meters long and 160 meters wide. (a) With the center at the origin and the long direction along the x-axis, write its equation. (b) How far apart are the foci? (c) How wide is the infield, measured parallel to the y-axis, at 100 meters from the center along the long axis?
Farhan and Junaid stand at the foci of an elliptical room 72 feet long, 60 feet apart. (a) How wide is the room? (b) How far is each from the nearest end wall? (c) How long is any path from Farhan to the wall and on to Junaid?

Reasoning

Explain why the definition asks for 2a to be greater than the distance between the foci, and describe what the condition PF_1 + PF_2 = 2a gives when it is not.
Explain why a circle of radius r is an ellipse, and find its a, b, c and eccentricity.
Anwar says the foci of \displaystyle \frac{x^2}{16} + \frac{y^2}{25} = 1 are (\pm 3, 0), since c^2 = 25 - 16 = 9. Say what he got right, and correct the rest.
Show that every point of \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 has |x| \le a and |y| \le b, and explain what this says about the shape of an ellipse.

Answers

Vertices (\pm 10, 0), co-vertices (0, \pm 6), foci (\pm 8, 0). Here a = 10, b = 6, and c^2 = 100 - 36 = 64.
Vertices (0, \pm 5), co-vertices (\pm 3, 0), foci (0, \pm 4). The larger denominator is under y^2, so a = 5, b = 3, c = 4.
Vertices (\pm 7, 0), co-vertices \displaystyle (0, \pm 2\sqrt{6}), foci (\pm 5, 0). Here \displaystyle b = \sqrt{24} = 2\sqrt{6} and c^2 = 49 - 24 = 25.
Vertices (0, \pm 4), co-vertices (\pm 2, 0), foci \displaystyle (0, \pm 2\sqrt{3}). Dividing by 16 gives \displaystyle \frac{x^2}{4} + \frac{y^2}{16} = 1, and c^2 = 16 - 4 = 12.
Vertices (\pm 3, 0), co-vertices (0, \pm 1), foci \displaystyle (\pm 2\sqrt{2}, 0). Dividing by 9 gives \displaystyle \frac{x^2}{9} + \frac{y^2}{1} = 1, and c^2 = 9 - 1 = 8.
\displaystyle \frac{x^2}{64} + \frac{y^2}{39} = 1. Here a = 8, c = 5, b^2 = 64 - 25 = 39.
\displaystyle \frac{x^2}{64} + \frac{y^2}{100} = 1. The foci are on the y-axis with c = 6, and b = 8, so a^2 = b^2 + c^2 = 100.
\displaystyle \frac{x^2}{36} + \frac{y^2}{27} = 1. Here 2a = 12, a = 6, c = 3, b^2 = 36 - 9 = 27.
\displaystyle \frac{x^2}{36} + \frac{y^2}{100} = 1, foci (0, \pm 8). Here a = 10, b = 6, c^2 = 100 - 36 = 64.
Substituting: \displaystyle \frac{9}{25} + \frac{256}{25} \cdot \frac{1}{16} = \frac{9}{25} + \frac{16}{25} = 1. Distances: c^2 = 25 - 16 = 9, foci (\pm 3, 0); the distance to (3, 0) is \displaystyle \frac{16}{5}, to (-3, 0) is \displaystyle \sqrt{36 + \frac{256}{25}} = \frac{34}{5}, and \displaystyle \frac{16}{5} + \frac{34}{5} = 10 = 2a.
Center (4, 1), vertices (4, 6) and (4, -4), co-vertices (1, 1) and (7, 1), foci (4, 5) and (4, -3). Vertical major axis with a = 5, b = 3, c = 4.
Center (-5, 3), vertices (3, 3) and (-13, 3), co-vertices (-5, 7) and (-5, -1), foci \displaystyle (-5 + 4\sqrt{3}, 3) and \displaystyle (-5 - 4\sqrt{3}, 3). Here a = 8, b = 4, \displaystyle c = \sqrt{48} = 4\sqrt{3}.
\displaystyle \frac{(x - 3)^2}{4} + \frac{(y + 2)^2}{1} = 1: center (3, -2), vertices (5, -2) and (1, -2), co-vertices (3, -1) and (3, -3), foci \displaystyle (3 \pm \sqrt{3}, -2). From (x^2 - 6x) + 4(y^2 + 4y) = -21, add 9 and 4 \cdot 4 = 16 to get (x - 3)^2 + 4(y + 2)^2 = 4, then divide by 4; c^2 = 4 - 1 = 3.
\displaystyle \frac{(x + 2)^2}{16} + \frac{(y - 3)^2}{25} = 1: center (-2, 3), vertices (-2, 8) and (-2, -2), co-vertices (-6, 3) and (2, 3), foci (-2, 6) and (-2, 0). From 25(x^2 + 4x) + 16(y^2 - 6y) = 156, add 100 and 144 to get 25(x + 2)^2 + 16(y - 3)^2 = 400; c^2 = 25 - 16 = 9.
\displaystyle \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{36} = 1, foci \displaystyle (2, -1 + 3\sqrt{3}) and \displaystyle (2, -1 - 3\sqrt{3}). The vertex is 6 above the center, so the major axis is vertical with a = 6; b = 3, and \displaystyle c = \sqrt{27} = 3\sqrt{3}.
\displaystyle \frac{(x - 1)^2}{49} + \frac{(y - 2)^2}{24} = 1. The center is the midpoint (1, 2), so c = 5; the vertex is 7 from the center, so a = 7, and b^2 = 49 - 25 = 24.
\displaystyle \frac{x^2}{20} + \frac{(y - 3)^2}{36} = 1. The center is (0, 3) and a = 6; \displaystyle e = \frac{c}{6} = \frac{2}{3} gives c = 4, so b^2 = 36 - 16 = 20.
Exercise 11: \displaystyle e = \frac{4}{5} = 0.8. Exercise 12: \displaystyle e = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2} \approx 0.87. The ellipse in Exercise 12 is flatter, since its e is closer to 1.
(a) \displaystyle \frac{x^2}{28{,}900} + \frac{y^2}{6400} = 1, from a = 170 and b = 80. (b) 300 meters: c^2 = 28{,}900 - 6400 = 22{,}500, so c = 150. (c) \displaystyle \frac{480\sqrt{21}}{17} \approx 129.4 meters: at x = 100, \displaystyle y^2 = 6400\left(1 - \frac{100}{289}\right) = \frac{6400 \cdot 189}{289}, so \displaystyle y = \pm\frac{240\sqrt{21}}{17}, and the width is twice that.
(a) \displaystyle 12\sqrt{11} \approx 39.8 feet: a = 36, c = 30, \displaystyle b = \sqrt{1296 - 900} = \sqrt{396} = 6\sqrt{11}, and the width is 2b. (b) a - c = 6 feet. (c) 72 feet, since every wall point has distance sum 2a; through a co-vertex it is \displaystyle 2\sqrt{30^2 + 396} = 2 \cdot 36.
For any point P, the triangle inequality gives PF_1 + PF_2 \ge F_1F_2, with equality only when P is on the segment F_1F_2. So if 2a = F_1F_2 the condition gives just that segment, and if 2a < F_1F_2 it gives no points. Only 2a > F_1F_2 gives a closed curve around the foci.
Take both foci at the center, so c = 0. Then PF_1 + PF_2 = 2 \cdot PF_1 = 2a says every point is a from the center: a circle of radius a. So a = r, b^2 = r^2 - 0 gives b = r, c = 0, and e = 0.
The value c = 3 is right. But the larger denominator, 25, is under y^2, so the major axis is vertical and the foci are (0, \pm 3). Check at the vertex (0, 5): its distances to (0, 3) and (0, -3) are 2 and 8, adding to 10 = 2a.
Both terms are squares over positive numbers, so neither is negative. Their sum is 1, so \displaystyle \frac{x^2}{a^2} \le 1, giving x^2 \le a^2 and |x| \le a; the same for y. Every point lies in the rectangle -a \le x \le a, -b \le y \le b, so an ellipse is bounded: a closed curve that never runs off to infinity, unlike a parabola or a hyperbola.