A circle pulled flat — the path that keeps a constant sum of distances.
Tie a loose loop of string around two pins, pull it taut with a pencil, and sweep the pencil all the way round. The curve it traces is an ellipse — a circle that has been gently pulled flat. The two pins are the foci, and the secret the string keeps is this: as the pencil moves, the sum of its distances to the two pins never changes, because the string length never changes. That single sentence — "the sum of distances stays constant" — is the whole definition. In this lesson we turn it into the clean equation x²/a² + y²/b² = 1, read off the foci from a tidy right triangle a² = b² + c², and measure exactly how flat the ellipse is with one number, the eccentricity e = c/a.
Pick two fixed points and call them the foci, F₁ and F₂. Now collect every point P whose two distances to the foci add up to one fixed total. Write that total as 2a (the "2" will pay off in a moment). The set of all such P is the ellipse:
|PF₁| + |PF₂| = 2a (a constant, the same for every point on the curve).
This is exactly the pinned-string picture. The string has a fixed length; one end is anchored at F₁, the other at F₂, and the pencil at P keeps both segments pulled tight. The two taut pieces are |PF₁| and |PF₂|, and together they always equal the string length, 2a. Slide the pencil to a new spot and one piece grows by exactly as much as the other shrinks — the total holds.
An ellipse is the set of points the sum of whose distances to two fixed foci is a constant 2a. (Compare a circle, where a single distance — to one centre — is constant. Slide the two foci together onto one point and the ellipse becomes a circle.)
One subtlety keeps the picture honest. For a real curve to exist, the string must be longer than the gap between the pins: 2a > |F₁F₂|. If the string were exactly the gap, the only "curve" would be the line segment between the pins; shorter still, and nothing can be drawn. So throughout, the foci sit inside the ellipse.
To get an equation, set the ellipse upright and centred: put the centre at the origin and the foci on the x-axis at F₁(−c, 0) and F₂(c, 0), where c is the half-distance between them. Take a point P(x, y) on the curve. Using the distance formula from 30.1 (which is just the Pythagorean theorem), the defining property becomes
√((x + c)² + y²) + √((x − c)² + y²) = 2a.
Move one root to the other side and square; the cross-terms cancel; isolate the remaining root and square again. The dust settles into a beautifully symmetric form (the algebra is a longer cousin of completing the square):
(a² − c²)x² + a²y² = a²(a² − c²).
Now name the positive number b² = a² − c² (we will see in a moment that b is a real length — the semi-minor axis). Divide through by a²b² and the equation reaches its standard, memorable shape:
x²/a² + y²/b² = 1
Here a is the semi-major axis (half the long width) and b is the semi-minor axis (half the short width). Set y = 0 to find the curve crosses the x-axis at (±a, 0), the two vertices; set x = 0 and it crosses the y-axis at (0, ±b), the two co-vertices. The long axis (the major axis) lies along whichever of a, b is larger.
For x²/25 + y²/9 = 1 we read a = 5 (since a² = 25) and b = 3 (since b² = 9). The ellipse stretches from x = −5 to 5 and from y = −3 to 3: a wide, gently flattened oval. Its vertices are (±5, 0) and its co-vertices are (0, ±3).
The equation b² = a² − c² rearranges into the single fact you should never get backwards:
a² = b² + c²
Read it off the picture and it is just the Pythagorean theorem. Stand at a co-vertex, say the top point (0, b). Its distances to the two foci are equal by symmetry, and since they must add to 2a, each one is exactly a. That focal radius is the hypotenuse of a right triangle whose legs are b (up the y-axis) and c (across to a focus) — so a² = b² + c². The three half-lengths a, b, c are the sides of one right triangle, with a as the hypotenuse.
Because a is the hypotenuse, a is the largest of the three: a > b and a > c. So the foci, at distance c from the centre, sit between the centre and the vertices — always inside the ellipse, on the major axis.
It is a² = b² + c², so a is the largest and c < a. Do not write a² + b² = c² — that is the hyperbola's relation, where c is the largest instead. For the ellipse, c is found by c = √(a² − b²).
Here is the defining sum, made visible. As you sweep the point P around the ellipse, watch each focal radius change while their sum stays pinned at 2a. Every length below is computed from GX.ellipsePt and GX.ellipseFoci — never measured off the picture.
Two ellipses can share the same width yet look completely different — one nearly circular, one a long thin sliver. The number that captures "how flattened" is the eccentricity:
e = ca, with 0 < e < 1.
Since the foci sit inside the ellipse, c < a, so e is always between 0 and 1. Read the extremes:
So eccentricity is a flatness dial from 0 (a circle) up toward 1 (almost a line segment). For our x²/25 + y²/9 = 1: c = √(25 − 9) = 4, so e = 4/5 = 0.8 — quite flat. Earth's orbit, by contrast, has e ≈ 0.017: so close to a circle you could not tell by eye.
Fix the semi-major axis at a = 5 and shrink the semi-minor axis b. Watch the foci slide outward and the eccentricity climb from near 0 toward 1. The numbers c = √(a² − b²) and e = c/a come straight from GX.ellipseC and GX.ellipseE.
Replace x with −x in x²/a² + y²/b² = 1 and nothing changes (because x is squared); likewise for y → −y. So the ellipse is symmetric about both axes and, doing both at once, about the centre. The two axes are its axes of symmetry; the centre is a centre of symmetry.
The same squares pin down the curve's range. Since y²/b² ≥ 0, the term x²/a² can be at most 1, forcing |x| ≤ a; symmetrically |y| ≤ b. Every point of the ellipse is boxed inside the rectangle from −a to a horizontally and −b to b vertically — touching the box exactly at the four vertices and co-vertices.
The foci always lie on the major (longer) axis — not always the x-axis. If a > b the ellipse is wide and the foci are on the x-axis at (±c, 0). But if the equation gives b > a the ellipse is tall, the major axis is vertical, and the foci move to the y-axis at (0, ±c), with c = √(b² − a²). Always compare the two denominators first: the larger denominator names the major axis.
| Idea | The fact |
|---|---|
| Definition | |PF₁| + |PF₂| = 2a (constant sum of focal distances) |
| Standard equation | x²/a² + y²/b² = 1 |
| a, b, c triangle | a² = b² + c² (a is the largest; not a²+b²=c²) |
| Foci & vertices | foci (±c, 0) on the major axis; vertices (±a, 0), co-vertices (0, ±b) |
| Eccentricity | e = c/a, with 0 < e < 1 (0 = circle, →1 = thin sliver) |
| Range & symmetry | |x| ≤ a, |y| ≤ b; symmetric about both axes and the centre |
| Which axis | foci on the major axis — the one under the larger denominator |
Next, in 30.6 The Hyperbola, we keep the same two foci but fix the difference of distances instead of the sum — and the relation flips to c² = a² + b², with two branches flying apart along their asymptotes.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the ellipse as a locus — the set of points whose distances to two foci have a constant sum — and derives the standard form x²/a² + y²/b² = 1 together with the relation a² = b² + c² and the eccentricity e = c/a. It aligns with the Common Core conic-sections standards HSG-GPE.A.3 (derive the equation of an ellipse from the focus/constant-sum definition using the distance formula) and HSG-GPE.A.1 / HSA-REI.B.4 (completing the square to move between general and standard form). It builds directly on the Pythagorean theorem and the distance and midpoint work of lesson 30.1, and sets up the contrasting hyperbola (constant difference, c² = a² + b²) in 30.5–30.6. A good check of understanding: ask which denominator names the major axis, and why the foci must lie inside the curve.