The ellipse: a constant sum of distances to two foci

Ellipse

Why does a loop of string around two pins draw an ellipse, and where does b² = a² − c² come from?

Drag a point around the curve and watch |PF₁| + |PF₂| stay at 2a. Move the foci together to get a circle, apart to flatten the ellipse, and read b² = a² − c² off the triangle at the top vertex.

Drag the ringed points on the board.

High school
Definition by two foci
Standard equation
b² = a² − c²
Eccentricity e = c/a

a

4.0

c

circle

2.4

\theta_P

1.20

|PF_1| + |PF_2| = 2a

$|PF_1|$: 4.713
$|PF_2|$: 3.287
$|PF_1| + |PF_2|$: 8.000
$b = \sqrt{a^2 - c^2}$: 3.200
$e = c / a$: 0.600

2a > 2cyes: an ellipse

Guided tour

1A string of length 2a
Pin a string of length $2a=8$ at $F_1$ and $F_2$ , $2c=4.8$ apart, and pull it tight with a pencil at $P$ . Here $|PF_1|=4.713$ and $|PF_2|=3.287$ : they add up to $8$ . Drag $P$ anywhere on the curve and the sum does not change.
2b comes from a right triangle
Move $P$ to the top of the curve. By symmetry the string is split evenly: $|PF_1|=|PF_2|=a=4$ . The height $b$ , the half-distance $c=2.4$ and the half-string $a$ form a right triangle, so $b^2=a^2-c^2$ and $b=3.2$ .
3Foci together: a circle
Slide $c$ to $0$ . The two pins coincide, every point is $a=4$ from the centre, and the ellipse is a circle: $b=a$ and $e=\frac{c}{a}=0$ .
4Foci apart: a flatter ellipse
At $c=3.8$ almost all of the string is used up spanning the foci: $b=1.249$ and $e=0.95$ . Push $c$ up to $a$ and the curve collapses onto the segment $F_1F_2$ ; beyond that the string is too short to reach round at all.

Explanation

1
The definition
An ellipse is the set of points $P$ in the plane whose distances to two fixed points $F_1$ , $F_2$ (the foci) add up to a constant $2a$ , with $2a>|F_1F_2|=2c$ .
- $2a=2c$ : the points form the segment $F_1F_2$ .
- $2a<2c$ : no point qualifies, because $|PF_1|+|PF_2|\ge|F_1F_2|$ by the triangle inequality.
2
From the definition to the equation
Put $F_1=(-c,0)$ , $F_2=(c,0)$ and write the definition out:
$\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a$
Move one root across and square, which leaves $a^2-cx=a\sqrt{(x-c)^2+y^2}$ . Square again and collect terms: $(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$ . Name $b^2=a^2-c^2>0$ and divide by $a^2b^2$ :
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\qquad(a>b>0)$
3
What a, b, c measure
- $a$ : half the string, and also the semi-major axis, since at the vertex $A_2$ the string runs $a+c$ and $a-c$ .
- $b$ : the semi-minor axis, the height of the top vertex.
- $c$ : half the distance between the foci.
The shaded triangle is the whole reason for $a^2=b^2+c^2$ : at the top vertex each half of the string is a hypotenuse of length $a$ .
4
Eccentricity measures flatness
$e=\dfrac{c}{a}$ lies between $0$ and $1$ . It is the share of the half-string spent bridging the foci: near $0$ the ellipse is round, near $1$ it is long and thin, because $\dfrac{b}{a}=\sqrt{1-e^2}$ . The eccentricity chapter shows the same $e$ from a focus and a directrix instead.

Check yourself

Q01

For $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ , what is $|PF_1|+|PF_2|$ for any point $P$ on it?

Q02

Where are the foci of $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ ?

Q03

Points with $|PF_1|+|PF_2|=6$ where $|F_1F_2|=6$ form…

Keep exploring

Turn this lab into a videoUse this lab as the script for a narrated explainer you can share. The prompt below goes to the video creator, where you can edit it first.

Make the video

View · display