Unit 8, lesson 93

Ellipses as Geometric Loci

How an ellipse is the set of points whose distances to two foci add up to a constant, and how its center, vertices, axes and eccentricity all follow from that one rule.

  • Theory part 185
  • Practice part 186
  • Conic Sections

Drive two stakes into the ground, tie the ends of a rope to them, and pull the rope tight with a stick. Walk the stick all the way around while the rope stays taut, and it scratches an oval into the soil. The rope never changes length, so the stick’s distance to one stake plus its distance to the other stays the same the whole time. That single fact is the definition of an ellipse.

Key terms

Ellipse as a constant sum of distances

Definition Let F1F_1 and F2F_2 be points of the plane, let 2c=∣F1F2∣2c = \lvert F_1F_2 \rvert be the distance between them, and let aa be a number with a>c≥0a > c \ge 0. The ellipse with foci F1F_1 and F2F_2 and constant sum 2a2a is the set of all points PP of the plane with

∣PF1∣+∣PF2∣=2a.\lvert PF_1 \rvert + \lvert PF_2 \rvert = 2a.

Points where the sum is less than 2a2a lie inside the ellipse, and points where it is greater lie outside. The condition a>ca > c is needed. By the triangle inequality the sum is never less than 2c2c, so a sum of exactly 2c>02c > 0 gives only the segment F1F2F_1F_2, and a sum below 2c2c gives no points at all.

In plain words Stand anywhere on the curve and add your distances to the two fixed points. You always get the same total. In the stake and rope picture that total is the length of the rope.

Example Take the foci F1=(−4,0)F_1 = (-4, 0) and F2=(4,0)F_2 = (4, 0) and the constant sum 1010.

Point PP∣PF1∣\lvert PF_1 \rvert∣PF2∣\lvert PF_2 \rvertSum
(5,0)(5, 0)99111010
(0,3)(0, 3)55551010
(3,125)(3, \frac{12}{5})375\frac{37}{5}135\frac{13}{5}1010
(4,3)(4, 3)73≈8.54\sqrt{73} \approx 8.5433≈11.54\approx 11.54

The first three points lie on the ellipse. For (4,3)(4, 3) the sum is more than 1010, so that point lies outside.

Foci

Definition The foci (singular focus) of an ellipse are the two fixed points F1F_1 and F2F_2 of its definition. When they are different points, the line through them is called the focal axis. Both foci lie inside the ellipse, since at a focus the sum of distances is 0+2c<2a0 + 2c < 2a. The ellipse meets the focal axis in exactly two points, one beyond each focus.

In plain words The foci are the two stakes. They sit inside the curve along its long direction. With the rope length fixed, the closer together they are, the rounder the ellipse.

Example Take the foci (−3,0)(-3, 0) and (3,0)(3, 0) and the sum 1010, and look for points (x,0)(x, 0) of the ellipse on the xx-axis. For −3≤x≤3-3 \le x \le 3 the two distances add up to only 66. For x>3x > 3 the sum is (x+3)+(x−3)=2x(x + 3) + (x - 3) = 2x, which equals 1010 when x=5x = 5. In the same way x<−3x < -3 gives x=−5x = -5. So the ellipse crosses its focal axis at (5,0)(5, 0) and (−5,0)(-5, 0), two units beyond each focus.

Center

Definition The center of an ellipse is the midpoint of the segment F1F2F_1F_2, so each focus lies at distance cc from the center. The ellipse is symmetric about its center. If PP is on the ellipse and P′P' is the point with the center as the midpoint of PP′PP', then P′P' is on the ellipse too, because the half turn about the center swaps the foci and keeps every distance. The ellipse is also symmetric about the focal axis and about the line through the center perpendicular to it.

In plain words The center is the point halfway between the foci. Whatever the curve does on one side of the center, it also does on the opposite side.

Example Take the foci (−2,−1)(-2, -1) and (−2,5)(-2, 5) with the sum 1010. The center is

(−2+(−2)2,−1+52)=(−2,2),\left(\frac{-2 + (-2)}{2}, \frac{-1 + 5}{2}\right) = (-2, 2),

and c=3c = 3. The point (−2,7)(-2, 7) is 88 from the first focus and 22 from the second, so it lies on the ellipse. Its mirror image through the center is (−2,−3)(-2, -3), whose distances are 22 and 88, the same pair in the other order.

Semimajor and semiminor axes

Definition Suppose the foci are different points, so that the focal axis exists. The vertices of the ellipse are the two points where it meets the focal axis, and the segment between them is the major axis. For the vertex VV beyond F2F_2 we have ∣VF1∣=∣VF2∣+2c\lvert VF_1 \rvert = \lvert VF_2 \rvert + 2c, so the sum condition gives ∣VF2∣=a−c\lvert VF_2 \rvert = a - c, and VV lies c+(a−c)=ac + (a - c) = a units from the center. The same holds for the other vertex, so the major axis has length 2a2a, the constant sum. The number aa is the semimajor axis. The covertices are the two points where the ellipse meets the line through the center perpendicular to the focal axis, and the segment between them is the minor axis. A covertex is equally far from both foci, so each of those distances is aa, and the Pythagorean theorem puts it at distance

b=a2−c2b = \sqrt{a^2 - c^2}

from the center. The number bb is the semiminor axis, and 0<b<a0 < b < a. When the foci coincide, c=0c = 0 and the ellipse is a circle of radius a=ba = b, so every diameter can serve as either axis.

In plain words The semimajor axis is half the longest width of the ellipse and the semiminor axis is half the shortest width, both measured through the center.

Example Return to the foci (±3,0)(\pm 3, 0) with the sum 1010. Then a=5a = 5 and the vertices are (±5,0)(\pm 5, 0), as found above. A point (0,y)(0, y) on the yy-axis is 9+y2\sqrt{9 + y^2} from each focus, so it is on the ellipse when

29+y2=10,2\sqrt{9 + y^2} = 10,

that is when y2=16y^2 = 16. The covertices are (0,±4)(0, \pm 4) and b=4b = 4. The ellipse is 1010 units wide and 88 units tall.

Relation between a, b and c

Definition For every ellipse with semimajor axis aa, semiminor axis bb and center to focus distance cc,

c2=a2−b2,0≤c<a.c^2 = a^2 - b^2, \qquad 0 \le c < a.

This is the Pythagorean theorem in the right triangle with vertices at the center, a focus and a covertex. Its legs are cc and bb, and its hypotenuse, from the covertex to the focus, has length aa. Any two of the three numbers determine the third.

In plain words The line from the end of the short axis to a focus has length aa. That is why aa is always the largest of the three numbers and why the foci always sit on the long axis.

Example An elliptical tabletop is 130130 centimeters long and 5050 centimeters wide, so a=65a = 65 and b=25b = 25. Then

c=652−252=4225−625=3600=60.\begin{aligned} c &= \sqrt{65^2 - 25^2} \\[0.4em] &= \sqrt{4225 - 625} \\[0.4em] &= \sqrt{3600} = 60. \end{aligned}

To draw its outline with stakes and a rope, put the stakes 120120 centimeters apart on the long center line and use a rope 130130 centimeters long. Each stake is then 55 centimeters from the nearer end of the table.

Standard equation

Definition Put the center at the origin and the foci at (±c,0)(\pm c, 0), with a>c≥0a > c \ge 0 and b=a2−c2b = \sqrt{a^2 - c^2}. A point (x,y)(x, y) lies on the ellipse exactly when

x2a2+y2b2=1.\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

To derive it, write the sum condition with the distance formula, isolate one square root and square. After simplifying this leaves

a2−cx=a(x−c)2+y2.a^2 - cx = a\sqrt{(x - c)^2 + y^2}.

Squaring again and dividing by a2b2a^2b^2 gives the equation. Squaring can create extra solutions in general, but not here. Every point of the equation has ∣x∣≤a\lvert x \rvert \le a, and its distances to F1=(−c,0)F_1 = (-c, 0) and F2=(c,0)F_2 = (c, 0) are

∣PF1∣=a+caxand∣PF2∣=a−cax.\begin{aligned} \lvert PF_1 \rvert &= a + \frac{c}{a}x \quad \text{and} \\[0.8em] \lvert PF_2 \rvert &= a - \frac{c}{a}x. \end{aligned}

Both are positive, since ca∣x∣≤c<a\frac{c}{a}\lvert x \rvert \le c < a, and they add up to 2a2a. With the foci at (0,±c)(0, \pm c) the letters trade places and the equation is

x2b2+y2a2=1.\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1.

When c>0c > 0, the larger denominator in both forms is a2a^2, and the foci lie on the axis of the variable over it. Equal denominators mean c=0c = 0 and a circle.

In plain words The equation is the sum of distances written in coordinates and tidied up. Read aa and bb from the denominators, then get cc from c2=a2−b2c^2 = a^2 - b^2.

Example Dividing 9x2+4y2=369x^2 + 4y^2 = 36 by 3636 gives

x24+y29=1.\frac{x^2}{4} + \frac{y^2}{9} = 1.

The larger denominator is under y2y^2, so the major axis is vertical, with a=3a = 3 and b=2b = 2. Then c=9−4=5c = \sqrt{9 - 4} = \sqrt{5}. The vertices are (0,±3)(0, \pm 3), the covertices are (±2,0)(\pm 2, 0) and the foci are (0,±5)(0, \pm\sqrt{5}).

Eccentricity

Definition The eccentricity of an ellipse is the ratio

e=ca=a2−b2a,0≤e<1.e = \frac{c}{a} = \frac{\sqrt{a^2 - b^2}}{a}, \qquad 0 \le e < 1.

It depends only on the shape and not on the size, since enlarging the ellipse multiplies aa and cc by the same factor. It also fixes the ratio of the axes, because

ba=1−e2.\frac{b}{a} = \sqrt{1 - e^2}.

In plain words Eccentricity measures how stretched the ellipse is. Near 00 it is almost a circle, and near 11 it is long and thin. The orbit of the Earth has eccentricity about 0.0170.017, so it is very nearly round.

Example Keep a=5a = 5 and move the foci apart.

ccbbeeb/ab/a
00550011
33440.60.60.80.8
44330.80.80.60.6
4.84.81.41.40.960.960.280.28

As ee climbs toward 11 the ellipse flattens toward the segment between its vertices.

Circle as a special case

Definition When the two foci coincide at a point CC, we have c=0c = 0, so b=ab = a and e=0e = 0. The sum condition becomes 2∣PC∣=2a2\lvert PC \rvert = 2a, that is ∣PC∣=a\lvert PC \rvert = a, which is the definition of the circle with center CC and radius aa. With CC at the origin the standard equation becomes x2+y2=a2x^2 + y^2 = a^2. So every circle is an ellipse, and an ellipse with c>0c > 0 is never a circle, because its vertices are farther from the center than its covertices.

In plain words Push the two stakes together and the rope swings around a single point, so the stick draws a circle. A circle is the roundest possible ellipse. The circles lesson covers its equation in detail.

Example Dividing 4x2+4y2=364x^2 + 4y^2 = 36 by 3636 gives

x29+y29=1,\frac{x^2}{9} + \frac{y^2}{9} = 1,

so a=b=3a = b = 3 and c=0c = 0. It is the circle of radius 33 about the origin. Compare the ellipse

x29+y28=1.\frac{x^2}{9} + \frac{y^2}{8} = 1.

It has c=9−8=1c = \sqrt{9 - 8} = 1, foci (±1,0)(\pm 1, 0) and e=13e = \frac{1}{3}. It reaches 33 units from the center along the xx-axis but only 22≈2.832\sqrt{2} \approx 2.83 units along the yy-axis, so it looks nearly round without being a circle.

Check yourself

  1. Find the vertices, the foci and the eccentricity of the ellipse 25x2+9y2=22525x^2 + 9y^2 = 225.
Show the answer

Dividing by 225225 gives

x29+y225=1.\frac{x^2}{9} + \frac{y^2}{25} = 1.

The larger denominator is under y2y^2, so a=5a = 5, b=3b = 3 and the major axis is vertical. Then c=25−9=4c = \sqrt{25 - 9} = 4. The vertices are (0,±5)(0, \pm 5), the foci are (0,±4)(0, \pm 4) and e=45e = \frac{4}{5}.

  1. The distances from any point of an ellipse to the foci (±2,0)(\pm 2, 0) add up to 88. Write its equation and find its covertices.
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The sum is 2a=82a = 8, so a=4a = 4, and c=2c = 2. Then b2=16−4=12b^2 = 16 - 4 = 12 and the equation is

x216+y212=1.\frac{x^2}{16} + \frac{y^2}{12} = 1.

The covertices are (0,±23)(0, \pm 2\sqrt{3}).

  1. Is the point (2,3)(2, 3) on, inside or outside the ellipse with foci (±3,0)(\pm 3, 0) and constant sum 1010?
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Inside. Its distances to the foci are 10\sqrt{10} and 34\sqrt{34}, and

10+34≈3.16+5.83=8.99,\sqrt{10} + \sqrt{34} \approx 3.16 + 5.83 = 8.99,

which is less than 1010. The equation agrees. This ellipse is x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1, and at (2,3)(2, 3) the left side is 425+916≈0.72<1\frac{4}{25} + \frac{9}{16} \approx 0.72 < 1.

How this lesson connects