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# Foci of Hyperbola — Formula, How to Find, and Examples

## TL;DR

The foci of a hyperbola are two fixed points whose difference of distances to any point on the curve is constant. They are found from \(c^2 = a^2 + b^2\), sitting at \((\pm c, 0)\) for a horizontal hyperbola centered at the origin. This article covers the focus formula, the defining difference-of-distances property, eccentricity, and worked examples on the foci of a hyperbola — note this differs from the ellipse, where \(c^2 = a^2 - b^2\).

## What Are The Foci Of A Hyperbola?

The foci of a hyperbola are two fixed points, \(F_1\) and \(F_2\), that define the curve through a difference of distances. For every point \(P\) on the hyperbola, the absolute difference between its distances to the two foci is a constant:

\[ |PF_1 - PF_2| = 2a \]

That constant equals \(2a\), twice the distance from the center to a vertex. This difference-of-distances rule is the defining feature of a hyperbola — and it is the mirror image of the [ellipse](/content/math/geometry/ellipse/index.html), where the sum of distances to the two [foci of the ellipse](/content/math/geometry/foci-of-ellipse/index.html) is constant. Sum gives an ellipse; difference gives a hyperbola.

The foci always lie inside the two branches, further from the center than the vertices, along the axis the hyperbola opens along.

## How Do You Find The Foci Of A Hyperbola?

For a hyperbola in standard form centered at the origin, the foci are found from one relationship. Take the horizontal hyperbola:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

The distance \(c\) from the center to each focus satisfies:

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

So \(c = \sqrt{a^2 + b^2}\), and the foci sit at \((\pm c, 0)\). For a vertical hyperbola, \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \], the same \(c\) applies and the foci are at \((0, \pm c)\).

Notice the plus sign. For a hyperbola, \(c\) is larger than \(a\) because \(c^2 = a^2 + b^2\). This is the detail that separates the hyperbola from the ellipse, where the relationship is \(c^2 = a^2 - b^2\) instead.

| Symbol | Meaning |
| --- | --- |
| a | Distance from center to a vertex (semi-transverse axis) |
| b | Semi-conjugate axis (sets the asymptote slope) |
| c | Distance from center to a focus |
| e | Eccentricity, \(e = \frac{c}{a}\) (always > 1 for a hyperbola) |

The steps to find the foci, in order:

1. Write the equation in standard form so the right side equals 1.
2. Read off \(a^2\) (under the positive term) and \(b^2\) (under the negative term).
3. Compute \(c = \sqrt{a^2 + b^2}\).
4. Place the foci at \((\pm c, 0)\) for a horizontal hyperbola, or \((0, \pm c)\) for a vertical one.

## How Are The Foci Related To Eccentricity?

The foci also set the hyperbola's **eccentricity**, a number measuring how "open" the curve is:

\[ e = \frac{c}{a} \]

Since \(c > a\), the eccentricity is always greater than 1. A value just above 1 gives a narrow, nearly-closed pair of branches; a large eccentricity gives wide, flat-opening branches. The foci can also be written as \((\pm ae, 0)\), which is handy when a problem gives you \(a\) and \(e\) directly.

## Examples of Foci Of Hyperbola

### Example 1

**Find the foci of the hyperbola \(\frac{x^2}{16} - \frac{y^2}{9} = 1\).**

Read off the values:

\(a^2 = 16, \quad b^2 = 9\)

Apply the focus relationship:

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

\[ c^2 = 16 + 9 = 25 \]

\[ c = 5 \]

Final answer: foci at \((5,0)\) and \((-5,0)\).

### Example 2

**Find the foci of \(\frac{x^2}{36} - \frac{y^2}{64} = 1\). A student writes \(c^2 = 36 - 64 = -28\) and says the foci do not exist. What went wrong?**

The mistake is using the ellipse relationship:

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

For hyperbolas, use addition:

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

\[ c^2 = 36 + 64 = 100 \]

\[ c = 10 \]

Final answer: foci at \((10,0)\) and \((-10,0)\).

### Example 3

**Find the foci of the vertical hyperbola \(\frac{y^2}{25} - \frac{x^2}{144} = 1\).**

Read off the values:

\(a^2 = 25, \quad b^2 = 144\)

\[ c^2 = 25 + 144 = 169 \]

\[ c = 13 \]

Final answer: foci at \((0, 13)\) and \((0, -13)\).

### Example 4

**A horizontal hyperbola has \(a = 6\) and eccentricity \(e = \frac{5}{3}\). Find its foci.**

Use \(c = ae\):

\[ c = 6 \times \frac{5}{3} = 10 \]

Final answer: foci at \((10,0)\) and \((-10,0)\).

### Example 5

**A horizontal hyperbola has foci at \((\pm 13,0)\) and a vertex at \((5,0)\). Find \(b^2\) and write its equation.**

From the points, \(c = 13\) and \(a = 5\). Solve for \(b^2\):

\[ c^2 = a^2 + b^2 \implies 169 = 25 + b^2 \implies b^2 = 144 \]

The equation is \(\frac{x^2}{25} - \frac{y^2}{144} = 1\).

### Example 6

**A hyperbola centered at \((2,-1)\) has \(a = 3\) and \(b = 4\), opening horizontally. Find the coordinates of its foci.**

Find \(c\):

\[ c^2 = a^2 + b^2 = 9 + 16 = 25 \implies c = 5 \]

The foci at \((h \pm c, k) = (2 \pm 5, -1)\) are \((7, -1)\) and \((-3, -1)\).

## Conclusion

- The **foci of a hyperbola** are two fixed points where the _difference_ of distances to any curve point is the constant \(2a\).

- They are found from \(c^2 = a^2 + b^2\), giving foci at \((\pm c, 0)\) for a horizontal hyperbola.

- Eccentricity \(e = \frac{c}{a}\) is always greater than 1.

- The hyperbola adds \(c^2 = a^2 + b^2\) where the ellipse subtracts \(c^2 = a^2 - b^2\).

- For a center at \((h,k)\), shift the foci to \((h \pm c, k)\).