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Foci of an Ellipse: Definition, Formula, Examples

TL;DR

The foci of an ellipse are two fixed points on its major axis such that, for any point on the ellipse, the sum of its distances to the two foci is constant (equal to 2a). This article covers the definition, where the foci sit, the formula (c^2 = a^2 - b^2), six worked examples, and the common mistakes.

BT Last updated on June 9, 2026 | 10 min read

What Are the Foci of an Ellipse?

An ellipse is the set of all points in a plane for which the sum of the distances to two fixed points is constant. Those two fixed points are the foci (singular: focus). Every ellipse has exactly two foci, and they always lie on the major axis, the longer of the ellipse's two axes of symmetry, placed symmetrically on either side of the centre.

State the defining rule precisely. If (F_1) and (F_2) are the foci and (P) is any point on the ellipse, then:

[ PF_1 + PF_2 = 2a = \text{constant} ]

where a is the semi-major axis (half the length of the long axis). This constant sum is the same for every point on the curve, that is exactly what makes the shape an ellipse rather than some other oval.

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

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

where a is the semi-major axis and b is the semi-minor axis (half the short axis). The foci sit at ((\pm c, 0)) on the x-axis. This material appears in NCERT Class 11, Chapter 11 (Conic Sections) and under CCSS-M HSG-GPE.A.3. The study of conics traces back to Apollonius of Perga (c. 240–190 BCE, Greek), whose treatise Conics named the ellipse, parabola, and hyperbola and first studied their focal properties.

How to Find the Foci, the Formula (c^2 = a^2 - b^2)

The foci sit at ((\pm c, 0)), so finding them means finding c, the distance from the centre to each focus. The formula is:

[ c^2 = a^2 - b^2 \Rightarrow c = \sqrt{a^2 - b^2} ]

This is not a rule to memorise blindly, it falls straight out of the constant-sum definition, and seeing why fixes it for good.

Take the co-vertex B, the point at the top of the minor axis, with coordinates ((0, b)). Since B lies on the ellipse, it obeys the constant-sum rule: the sum of its distances to the two foci is 2a. By symmetry the two distances from B to (F_1) and (F_2) are equal, so each one is exactly a.

Now look at the right triangle formed by the centre O, the focus (F_1 = (-c, 0)), and the co-vertex (B = (0, b)). Its legs are b (vertical, from O up to B) and c (horizontal, from O out to the focus), and its hypotenuse is the distance from B to the focus, which we just found is a. By the Pythagorean theorem:

[ b^2 + c^2 = a^2 \Rightarrow c^2 = a^2 - b^2 ]

So the focal distance c is one leg of a right triangle whose other leg is the semi-minor axis and whose hypotenuse is the semi-major axis. That single picture is why the formula looks the way it does.

If the ellipse is taller than it is wide (major axis vertical, (b > a)), the foci sit on the y-axis at ((0, \pm c)) and the formula uses the larger denominator: (c^2 = (\text{larger}) - (\text{smaller})). The rule is always bigger minus smaller under the square root.

Eccentricity, How Stretched the Ellipse Is

The foci also fix the ellipse's eccentricity, a number that measures how far it is from being a circle:

[ e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}} ]

Eccentricity runs from 0 to 1 for an ellipse. When e = 0, the foci coincide at the centre ((c = 0,) so (a = b)) and the ellipse is a perfect circle, a circle is just an ellipse whose two foci have merged. As e approaches 1, the foci slide out toward the vertices and the ellipse stretches into a long, thin shape. So the position of the foci and the eccentricity are two ways of saying the same thing: how elongated the curve is.

Examples of the Foci of an Ellipse

With the definition, the formula, and the eccentricity in hand, here are the foci being located. The problems move from a direct read-off up to recovering the equation from given foci.

Example 1 - Find the foci of the ellipse ( \frac{x^2}{25} + \frac{y^2}{16} = 1 )

Here (a^2 = 25) (the larger denominator, so the major axis is horizontal) and (b^2 = 16). Then:

[ c^2 = a^2 - b^2 = 25 - 16 = 9 \Rightarrow c = 3 ]

The foci lie on the x-axis at ((\pm c, 0)).

Final answer: the foci are at ((3, 0)) and ((-3, 0)).

Example 2 - Find the foci of the ellipse ( \frac{x^2}{9} + \frac{y^2}{25} = 1 )

A first instinct is to assume (a^2) is always the number under (x^2), take (a^2 = 9) and (b^2 = 25), and compute (c^2 = 9 - 25 = -16), then get stuck on the square root of a negative number. That stuck feeling is the clue: (c^2) can never be negative, so the labeling must be wrong.

The fix: a² is always the larger denominator, because the major axis is the longer one. Here the bigger number, 25, is under (y^2), so the major axis is vertical, (a^2 = 25) and (b^2 = 9), and the foci sit on the y-axis:

[ c^2 = 25 - 9 = 16 \Rightarrow c = 4 \text{, foci at } (0, \pm 4) ]

Final answer: the foci are at ((0, 4)) and ((0, -4)).

Example 3 - An ellipse has a major axis of length 12 and a minor axis of length 8, centred at the origin and horizontal. Find its foci

The semi-axes are half the full axes: (a = 6), (b = 4). Then:

[ c^2 = a^2 - b^2 = 36 - 16 = 20 \Rightarrow c = \sqrt{20} = 2\sqrt{5} \approx 4.47 ]

Final answer: the foci are at approximately ((4.47, 0)) and ((-4.47, 0)).

Example 4 - Find the eccentricity of the ellipse ( \frac{x^2}{25} + \frac{y^2}{16} = 1 )

From Example 1, (a = 5) and (c = 3), so:

[ e = \frac{c}{a} = \frac{3}{5} = 0.6 ]

Final answer: the eccentricity is 0.6, moderately stretched, well short of a circle (0) or a near-line (close to 1).

Example 5 - An ellipse centred at the origin has foci at ((\pm 4, 0)) and a semi-major axis (a = 5). Find its equation

The foci give (c = 4), and (a = 5). Recover (b) from the formula:

[ c^2 = a^2 - b^2 \Rightarrow 16 = 25 - b^2 \Rightarrow b^2 = 9 ]

With (a^2 = 25) and (b^2 = 9), and the foci on the x-axis (major axis horizontal):

[ \frac{x^2}{25} + \frac{y^2}{9} = 1 ]

Final answer: ( \frac{x^2}{25} + \frac{y^2}{9} = 1 ).

Example 6 - Find the foci of the translated ellipse ( \frac{(x - 2)^2}{16} + \frac{(y + 1)^2}{7} = 1 )

The centre has shifted to ((2, -1)). With (a^2 = 16) and (b^2 = 7) (16 is larger, so the major axis is horizontal):

[ c^2 = 16 - 7 = 9 \Rightarrow c = 3 ]

The foci sit c units left and right of the centre, at ((2 \pm 3, -1)).

Final answer: the foci are at ((5, -1)) and ((-1, -1)).

Why the Foci of an Ellipse Matter

The two foci are not a textbook curiosity; the constant-sum property and the way an ellipse reflects through its foci run through astronomy, medicine, and architecture.

• Planetary orbits. Kepler's first law says every planet orbits the Sun in an ellipse with the Sun at one focus, not at the centre. The other focus is empty space. This is why a planet's distance from the Sun changes through its year, and the foci are exactly what encode that varying distance.

• The reflective property and lithotripsy. A signal sent from one focus of an ellipse reflects off the curve and converges exactly on the other focus. Medical lithotripsy machines use this: shock waves generated at one focus pass harmlessly through the body and focus their energy at the other focus, placed on a kidney stone, shattering it without surgery.

• Whispering galleries. In elliptical rooms, a whisper at one focus is heard clearly at the other focus across the room because sound waves reflect off the walls and reconverge there. The dome of St Paul's Cathedral and the U.S. Capitol's old chamber are famous examples.

• Satellite and comet paths. Eccentricity, set by the foci, tells you whether an orbit is nearly circular (low e) or a long, sweeping ellipse (high e), which is how engineers describe everything from communications-satellite orbits to the path of Halley's Comet.

For a Class 11 student, the foci are where the abstract conic-section equation suddenly becomes physical, the same two points that fix the algebra also fix where a planet speeds up and where a shock wave lands.

Where Things Go Sideways With the Foci of an Ellipse

Mistake 1: Mislabelling which axis is major

Where it slips in: A student assumes (a^2) is always under (x^2) and ends up with a negative (c^2).

Don't do this: Fix (a^2) to the x² term by habit.

The correct way: (a^2) is always the larger denominator because the major axis is the longer one. Whichever variable sits over the bigger number tells you the axis the foci lie on. If (c^2) comes out negative, you have swapped a and b.

Mistake 2: Using (c^2 = a^2 + b^2) (the hyperbola formula)

Where it slips in: A student remembers a "c² formula" but reaches for the plus version, which belongs to the hyperbola.

Don't do this: Write (c^2=a^2+b^2) for an ellipse.

The correct way: For an ellipse it is (c^2=a^2−b^2) (the foci sit inside, so c is shorter than a). The plus version, (c^2=a^2+b^2), is for the hyperbola, where the foci sit outside the vertices. The minus sign is the ellipse's signature.

Mistake 3: Forgetting to shift the foci for a translated ellipse

Where it slips in: The ellipse is centred at ((h,k)), not the origin, and the student reports the foci as ((\pm c, 0)).

Don't do this: Place the foci relative to the origin when the centre has moved.

The correct way: The foci are c units from the centre ((h,k)), so they are at ((h \pm c, k)) for a horizontal major axis. Always locate the centre first, then step out by c.

Key Takeaways

• The foci of an ellipse are two fixed points where the sum of distances to any point on the curve is the constant 2a.
• The foci lie on the major axis at ((\pm c, 0)), with (c = \sqrt{a^2 - b^2}).
• That formula comes from a right triangle linking the semi-minor axis b, the focal distance c, and the semi-major axis a.
• Eccentricity (e = \frac{c}{a}) measures how stretched the ellipse is; (e = 0) is a circle, (e) near 1 is long and thin.
• Common mistakes are mislabelling the major axis, using the hyperbola's plus formula, and forgetting to shift the foci for a translated ellipse.

Practice These Problems to Solidify Your Understanding

1. Find the foci of the ellipse ( \frac{x^2}{169} + \frac{y^2}{144} = 1 ).
2. Find the foci of the ellipse ( \frac{x^2}{16} + \frac{y^2}{36} = 1 ).
3. Find the eccentricity of the ellipse ( \frac{x^2}{100} + \frac{y^2}{64} = 1 ).

Answer to Question 1: ((\pm 5, 0)). Answer to Question 2: ((0, \pm \sqrt{20})), since 36 is the larger denominator the major axis is vertical. Answer to Question 3: e = 0.6.