Ellipse - Equation, Formula, Properties, Graphing

#Geometry
TL;DR
An ellipse is the set of all points in a plane whose distances to two fixed points (called foci) sum to a constant. Its standard equation is x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where aa is the semi-major axis and bb is the semi-minor axis.
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Madhukar VLast updated on September 30, 202611 min read

What Is an Ellipse?

An ellipse is the locus of all points in a plane whose distances to two fixed points — the foci — sum to a constant value. Mathematically, for any point PP on the ellipse:

∣PF1∣+∣PF2∣=2a|PF_1| + |PF_2| = 2a

where F1F_1 and F2F_2 are the foci and 2a2a is the length of the major axis.

An ellipse is also a conic section — formed by slicing a cone with a plane at a slight angle (not parallel to the base, not parallel to the slant side). A circle is the special case of an ellipse where both foci coincide at the centre.

The standard equation of an ellipse centred at the origin with major axis along the x-axis:

x2a2+y2b2=1,a>b>0\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0

where aa is the semi-major axis length and bb is the semi-minor axis length.

What Are the Parts of an Ellipse?

Part

Definition

Centre

The midpoint of the major axis (and minor axis)

Foci (F1,F2F_1, F_2)

Two fixed points whose distance-sum defines the curve

Major axis

Longest diameter, length 2a2a, contains both foci

Minor axis

Shortest diameter perpendicular to major axis, length 2b2b

Vertex

Endpoint of the major axis

Co-vertex

Endpoint of the minor axis

Eccentricity (ee)

Measure of how "stretched" the ellipse is, 0≤e<10 \le e < 1

What Is the Equation of an Ellipse?

Standard Form — Centred at Origin

For an ellipse centred at (0,0)(0, 0) with major axis along the x-axis:

x2a2+y2b2=1,a>b\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b

For major axis along the y-axis (taller than wide):

x2b2+y2a2=1,a>b\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, \quad a > b

Shifted Form — Centred at (h,k)(h, k)

(x−h)2a2+(y−k)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1

Relationship Between aa, bb, and cc

The distance from the centre to each focus is cc, where:

c=a2−b2c = \sqrt{a^2 - b^2}

So the foci sit at (±c,0)(\pm c, 0) for a horizontal ellipse, or (0,±c)(0, \pm c) for a vertical ellipse.

How Do You Derive the Ellipse Equation?

The standard form x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 falls straight out of the focus-pair definition + the distance formula. Here's the derivation.

Setup. Place the centre at the origin, foci at F1=(−c,0)F_1 = (-c, 0) and F2=(c,0)F_2 = (c, 0), and let P=(x,y)P = (x, y) be any point on the ellipse. By definition, the sum of distances to the foci is 2a2a:

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

Isolate one radical.

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

Square both sides.

(x+c)2+y2=4a2−4a(x−c)2+y2+(x−c)2+y2(x + c)^2 + y^2 = 4a^2 - 4a\sqrt{(x - c)^2 + y^2} + (x - c)^2 + y^2

Expand and cancel the (x±c)2+y2(x \pm c)^2 + y^2 terms (after expanding, the x2,c2,y2x^2, c^2, y^2 cancel):

4cx=4a2−4a(x−c)2+y24cx = 4a^2 - 4a\sqrt{(x - c)^2 + y^2}

Solve for the remaining radical.

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

Square again.

a2[(x−c)2+y2]=a4−2a2cx+c2x2a^2\bigl[(x - c)^2 + y^2\bigr] = a^4 - 2a^2 cx + c^2 x^2

Expand and group.

a2x2−2a2cx+a2c2+a2y2=a4−2a2cx+c2x2a^2 x^2 - 2a^2 cx + a^2 c^2 + a^2 y^2 = a^4 - 2a^2 cx + c^2 x^2

(a2−c2)x2+a2y2=a2(a2−c2)\bigl(a^2 - c^2\bigr)x^2 + a^2 y^2 = a^2\bigl(a^2 - c^2\bigr)

Substitute b2=a2−c2b^2 = a^2 - c^2 (this is the definition of bb for an ellipse — it makes the focal-distance formula come out clean):

b2x2+a2y2=a2b2b^2 x^2 + a^2 y^2 = a^2 b^2

Divide by a2b2a^2 b^2:

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

That's the standard ellipse equation — and the substitution b2=a2−c2b^2 = a^2 - c^2 (equivalently c=a2−b2c = \sqrt{a^2 - b^2}) drops out as a consequence of the derivation, not a definition you have to memorise.

What Are the Properties of an Ellipse?

Beyond the parts list, an ellipse has six geometric properties that come up in problems and proofs.

  1. Two axes of symmetry. The major and minor axes are both lines of symmetry. (A circle has infinite; a parabola has one.)

  2. Two foci, constant distance-sum. For every point PP on the curve, ∣PF1∣+∣PF2∣=2a|PF_1| + |PF_2| = 2a. This is the defining property and is often used as a problem-solving constraint.

  3. Closed, bounded curve. Unlike parabolas and hyperbolas, an ellipse is a finite closed curve. It encloses a finite area.

  4. Area = πab\pi a b. Generalises the circle's πr2\pi r^2. (When a=b=ra = b = r, this gives πr2\pi r^2 ✓.)

  5. Perimeter — no closed-form formula. Unlike area, the ellipse perimeter doesn't have a clean elementary expression. Ramanujan's approximation gives P≈π![3(a+b)−(3a+b)(a+3b)]P \approx \pi!\left[3(a + b) - \sqrt{(3a + b)(a + 3b)}\right], accurate to a few parts per million for moderate eccentricities.

  6. Reflective property. Any ray emitted from one focus reflects off the ellipse and passes through the other focus. This is the principle behind whispering galleries and lithotripsy — a sound or shock wave from one focus arrives at the other focus all at the same instant.

How Do You Draw an Ellipse? (The Gardener's Method)

The geometric definition gives a beautifully simple physical construction — sometimes called the gardener's method or string-and-pins method — used since antiquity to mark elliptical flowerbeds and arenas.

Materials. Two pins (or pegs, or thumbtacks), a piece of string of length 2a2a, and a pencil.

Steps.

  1. Decide the major-axis length 2a2a and the focal distance 2c2c (with c<ac < a).

  2. Press the two pins into the drawing surface, 2c2c apart — those are the foci F1F_1 and F2F_2.

  3. Tie the string into a loop of total length 2a+2c2a + 2c (so when stretched around both pins, the loose part forms a triangle).

  4. Place the pencil inside the loop, pull the string taut, and trace a complete circuit.

The pencil traces an ellipse because the string keeps the sum ∣PF1∣+∣PF2∣=2a|PF_1| + |PF_2| = 2a constant at every point. That's the focus-pair definition, made tangible.

Algebraic alternative — graphing from the equation. For (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, the 6-step process below (Centre → Orientation → Vertices → Co-vertices → Foci → Curve) works the same way and is what you'll do on graph paper.

What Is Eccentricity?

The eccentricity ee of an ellipse measures how elongated it is — how much it deviates from being a perfect circle.

e=ca=1−b2a2e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}

  • e=0e = 0 → perfect circle (foci coincide at centre)

  • 0<e<10 < e < 1 → ellipse (the closer to 1, the more stretched)

  • e=1e = 1 → parabola (degenerate case)

  • e>1e > 1 → hyperbola (the other side of conic sections)

Examples. Earth's orbit has eccentricity e≈0.0167e \approx 0.0167 — almost circular. Halley's Comet has e≈0.967e \approx 0.967 — extremely elongated.

How Do You Graph an Ellipse?

For an ellipse (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1:

Step 1. Identify the centre (h,k)(h, k).

Step 2. Determine orientation. If a>ba > b under xx, major axis is horizontal. If a>ba > b under yy, major axis is vertical.

Step 3. Mark the vertices at distance aa from the centre along the major axis.

Step 4. Mark the co-vertices at distance bb from the centre along the minor axis.

Step 5. Compute c=a2−b2c = \sqrt{a^2 - b^2} and mark the foci on the major axis.

Step 6. Connect the four key points with a smooth oval curve.

Worked example. Graph x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1.

  • Centre: (0,0)(0, 0).

  • a2=25a^2 = 25, b2=9b^2 = 9, so a=5a = 5, b=3b = 3.

  • Major axis horizontal (under xx). Vertices: (±5,0)(\pm 5, 0). Co-vertices: (0,±3)(0, \pm 3).

  • c=25−9=4c = \sqrt{25 - 9} = 4. Foci: (±4,0)(\pm 4, 0).

  • Eccentricity: e=4/5=0.8e = 4/5 = 0.8.

Why Are Ellipses Important? (The Real-World GROUND)

"The orbit of every planet is an ellipse with the Sun at one of the two foci." — Johannes Kepler, Astronomia Nova, 1609.

The ellipse was one of the most important geometric discoveries in the history of science — not for its mathematics directly, but for what it revealed about the universe.

In 1609, the German astronomer Johannes Kepler published Astronomia Nova, in which he proved — using Tycho Brahe's data on Mars — that the orbit of every planet is an ellipse, not a circle, with the Sun at one focus. This was Kepler's First Law, and it overturned 2,000 years of Greek astronomy (which assumed circular orbits since Aristotle).

Newton later proved (1687) that elliptical orbits are a mathematical consequence of the inverse-square law of gravity — connecting geometry, physics, and the cosmos in one formula.

Real-world applications of the ellipse:

  • Planetary orbits. Mercury, Venus, Earth, Mars, Jupiter — all elliptical. The Hubble Space Telescope and the International Space Station both orbit Earth in elliptical paths.

  • Halley's Comet. Returns every 76 years on a highly elliptical orbit (e≈0.967e \approx 0.967).

  • Whispering galleries. St. Paul's Cathedral in London and the U.S. Capitol Rotunda have elliptical domes — a whisper at one focus can be heard clearly at the other focus, because sound waves reflect off the ellipse from one focus to the other.

  • Lithotripsy. Medical machines use elliptical reflectors to focus shock waves on kidney stones — the shock generator sits at one focus, the stone at the other.

  • Architecture. The Colosseum in Rome is an ellipse (188m × 156m). Many oval running tracks are elliptical.

  • Engineering — gears. Elliptical gears produce non-uniform rotation useful in some specialised machinery.

The ellipse was first studied by Apollonius of Perga in his Conics around 200 BCE — he gave it the name ellipsis (Greek for "falling short") and proved its main geometric properties.

Worked Example of Ellipse

Find the foci of the ellipse x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1.

The intuitive (wrong) approach. A student in a hurry computes c=a2+b2c = \sqrt{a^2 + b^2} instead of a2−b2\sqrt{a^2 - b^2}:

c=?25+16=41≈6.4c \stackrel{?}{=} \sqrt{25 + 16} = \sqrt{41} \approx 6.4

Why it fails. The formula for the focal distance in an ellipse is c=a2−b2c = \sqrt{a^2 - b^2} (minus, not plus). The plus formula applies to hyperbolas, not ellipses.

The correct method.

c=a2−b2=25−16=9=3c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3

The foci are at (±3,0)(\pm 3, 0).

Check. Eccentricity e=c/a=3/5=0.6e = c/a = 3/5 = 0.6, which is between 0 and 1 ✓ (the ellipse range). The wrong answer of c=6.4>a=5c = 6.4 > a = 5 would have produced e>1e > 1, which is the hyperbola range — a contradiction.

At Bhanzu, our trainers walk through this wrong-path-first sequence intentionally — confusing the ellipse focal formula with the hyperbola version is one of the most common student archetypes. Once the student feels the sign-flip cost, the rule sticks.

What Are the Most Common Mistakes With Ellipses?

Mistake 1: Using c=a2+b2c = \sqrt{a^2 + b^2} instead of c=a2−b2c = \sqrt{a^2 - b^2}

Where it slips in: Confusing the ellipse focal-distance formula with the hyperbola one.

Don't do this: c=a2+b2c = \sqrt{a^2 + b^2} for an ellipse.

The correct way: For ellipse, c=a2−b2c = \sqrt{a^2 - b^2} (minus). For hyperbola, c=a2+b2c = \sqrt{a^2 + b^2} (plus). The rusher who pattern-matches "Pythagorean-style formula" hits this.

Mistake 2: Misidentifying major vs minor axis

Where it slips in: Assuming the xx-direction is always the major axis.

Don't do this: For x29+y225=1\frac{x^2}{9} + \frac{y^2}{25} = 1, calling the major axis horizontal.

The correct way: The larger denominator is under the major axis variable. Here, 25>925 > 9 is under yy — so the major axis is vertical, and a2=25a^2 = 25, b2=9b^2 = 9. The memorizer who pattern-matches "a is always under x" without checking which is larger hits this.

Mistake 3: Forgetting an ellipse can be a circle when a=ba = b

Where it slips in: When the two denominators are equal, students sometimes don't recognise the resulting circle.

Don't do this: Treating x216+y216=1\frac{x^2}{16} + \frac{y^2}{16} = 1 as a normal ellipse with computed cc.

The correct way: When a=ba = b, the ellipse becomes a circle of radius aa. Eccentricity is 0, foci coincide at the centre. x216+y216=1\frac{x^2}{16} + \frac{y^2}{16} = 1 simplifies to x2+y2=16x^2 + y^2 = 16 — a circle of radius 4.

The Mathematicians Who Shaped Ellipse Theory

Apollonius of Perga (c. 240–c. 190 BCE, Greece) — Wrote Conics, the foundational treatise on parabolas, ellipses, and hyperbolas. Coined the term ellipse (Greek elleipsis, "falling short") and proved its main geometric properties two millennia before algebraic notation existed.

Johannes Kepler (1571–1630, Germany) — Proved in 1609 that planetary orbits are ellipses with the Sun at one focus (Kepler's First Law). This overturned 2,000 years of assumption that orbits must be circular, and gave Newton the geometric foundation for the inverse-square law of gravity.

Tycho Brahe (1546–1601, Denmark) — Made the most precise pre-telescope astronomical observations of his era, especially of Mars. Brahe's Mars data was what Kepler used to discover elliptical orbits — the empirical bridge between classical and modern astronomy.

A Practical Next Step

Try these three before moving on to hyperbolas.

  1. Find the foci of the ellipse x216+y27=1\frac{x^2}{16} + \frac{y^2}{7} = 1.

  2. Find the area of an ellipse with semi-major axis 6 and semi-minor axis 4.

  3. The orbit of Mars has eccentricity ~0.093 and semi-major axis ~228 million km. Estimate the focal distance (cc).

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Frequently Asked Questions

What is an ellipse in simple words?
An ellipse is an oval — a stretched circle. Mathematically, it's the set of all points where the sum of distances to two fixed points (foci) is constant. The more separated the foci, the more stretched the ellipse.
What is the equation of an ellipse?
Centred at the origin: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, where aa is the semi-major axis and bb is the semi-minor axis. Centred at (h,k)(h, k): (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.
What are the foci of an ellipse?
Two fixed points inside the ellipse. The sum of the distances from any point on the curve to the two foci equals the length of the major axis (2a2a). For a centred horizontal ellipse, foci are at (±c,0)(\pm c, 0) where c=a2−b2c = \sqrt{a^2 - b^2}.
What is eccentricity?
A number between 0 and 1 that measures how stretched the ellipse is. Formula: e=c/a=1−b2/a2e = c/a = \sqrt{1 - b^2/a^2}. e=0e = 0 means a perfect circle. ee close to 1 means a very stretched ellipse. Earth's orbit: e≈0.017e \approx 0.017 (nearly circular). Halley's comet: e≈0.967e \approx 0.967 (very elongated).
How is an ellipse different from a circle?
A circle has one centre and constant distance to all points (radius). An ellipse has two foci and the sum of distances to them is constant. A circle is the special case of an ellipse where both foci coincide at the centre and eccentricity is 0.
Where are ellipses used in real life?
Planetary orbits (Kepler's First Law), satellite orbits including the ISS, whispering galleries (St. Paul's Cathedral, US Capitol Rotunda), lithotripsy kidney-stone treatment, the Colosseum's floor plan, and elliptical running tracks.
What is the area of an ellipse?
A=πabA = \pi a b, where aa is the semi-major axis and bb is the semi-minor axis. The formula generalises the circle area πr2\pi r^2 (when a=b=ra = b = r, you get πr⋅r=πr2\pi r \cdot r = \pi r^2).
✍️ Written By
Madhukar V
Madhukar V
Math Columnist & Analyst
I see mathematics as more than just numbers and formulas—it is a way of thinking, questioning, and understanding the world around us. I work at the intersection of mathematics, EdTech, technology, and AI-powered learning, I enjoy exploring how mathematical ideas can be made simple, meaningful, and relatable. Through my articles, I aim to present mathematics with curiosity and a practical perspective, helping readers see the beauty and relevance behind every concept in a Bhanzu way.
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