# Where are the hyperbola’s foci?

Finds a hyperbola’s center, vertices, co-vertices, foci, asymptotes, eccentricity and standard-form equation from its standard form or its general form Ax² + Cy² + Dx + Ey + F = 0.

- Page: https://www.acalculator.org/math/hyperbola-calculator
- JSON spec: https://www.acalculator.org/math/hyperbola-calculator.json
- Version: c202a1242f0d

## Default answer

Example with the default inputs (Equation Standard form, Opens Left and right, a² (under the positive term) 9, b² (under the negative term) 16, Center x (h) 0, Center y (k) 0): The hyperbola x²/9 − y²/16 = 1 has foci (−5, 0), (5, 0) and asymptotes y = ±(4/3)x.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| form | Equation | Type the standard form’s numbers, or the coefficients of the general form. |
| axis | Opens | Left and right: (x − h)²/a² − (y − k)²/b² = 1. Up and down: (y − k)²/a² − (x − h)²/b² = 1. |
| a2 | a² (under the positive term) | The number under the squared term with the plus sign: a is the distance from the center to a vertex. |
| b2 | b² (under the negative term) | The number under the squared term with the minus sign: b is the distance from the center to a co-vertex. |
| h | Center x (h) | The x coordinate of the center; 0 when left empty. |
| k | Center y (k) | The y coordinate of the center; 0 when left empty. |
| A | A (x²) | The coefficient of x². |
| C | C (y²) | The coefficient of y²; it must have the opposite sign to A. |
| D | D (x) | The coefficient of x. |
| E | E (y) | The coefficient of y. |
| F | F (constant) | The constant term. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| equation | Standard form | The hyperbola’s equation in standard form. |
| center | Center (h, k) | The point midway between the vertices. |
| vertices | Vertices | The two points where the curve turns, a from the center. |
| coVertices | Co-vertices | The ends of the conjugate axis, b from the center. |
| foci | Foci | The two focus points, c from the center on the transverse axis. |
| asymptotes | Asymptotes | The two lines the curve approaches, through the center. |
| a | a | The distance from the center to each vertex. |
| b | b | The distance from the center to each co-vertex. |
| c | Focal distance c | The distance from the center to each focus: √(a² + b²). |
| e | Eccentricity e | c ÷ a, always above 1 for a hyperbola. |
| transverse | Transverse axis length | 2a: the distance between the vertices. |
| conjugate | Conjugate axis length | 2b: the distance between the co-vertices. |
| latus | Latus rectum | The chord through a focus at a right angle to the transverse axis: 2b² ÷ a. |

## Method

(x − h)²/a² − (y − k)²/b² = 1 or (y − k)²/a² − (x − h)²/b² = 1; c² = a² + b²; vertices a and foci c from the center along the transverse axis; asymptotes y = ±(b/a)(x − h) + k (or ±(a/b)); e = c ÷ a. The general form is completed to the square.

## Assumptions

- The hyperbola’s axes are parallel to the x and y axes (no xy term).
- Typed numbers are read as exact decimals; a root that is not exact is rounded once.

## Worked examples

1. form = standard, axis = y, a2 = 49, b2 = 32 gives vertices = (0, −7), (0, 7), foci = (0, −9), (0, 9), c = 9, a = 7, e = 1.285714286. Source: OpenStax, Precalculus 2e, §10.2 The Hyperbola (standard forms centered at (0, 0) and (h, k), c² = a² + b², vertices, co-vertices, foci and asymptotes y = ±(b/a)(x − h) + k; Example 1: y²/49 − x²/32 = 1 has vertices (0, ±7) and foci (0, ±9); Example 3: (x − 3)²/9 − (y + 2)²/16 = 1 has vertices (0, −2), (6, −2) and foci (−2, −2), (8, −2); Example 5: 9x² − 4y² − 36x − 40y − 388 = 0 is (x − 2)²/36 − (y + 5)²/81 = 1), https://openstax.org/books/precalculus-2e/pages/10-2-the-hyperbola (retrieved 2026-10-19).
2. form = standard, axis = x, a2 = 9, b2 = 16, h = 3, k = -2 gives equation = (x − 3)²/9 − (y + 2)²/16 = 1, center = (3, −2), vertices = (0, −2), (6, −2), coVertices = (3, −6), (3, 2), foci = (−2, −2), (8, −2), asymptotes = y = ±(4/3)(x − 3) − 2, c = 5, e = 1.666666667, latus = 10.66666667. Source: OpenStax, Precalculus 2e, §10.2 The Hyperbola (standard forms centered at (0, 0) and (h, k), c² = a² + b², vertices, co-vertices, foci and asymptotes y = ±(b/a)(x − h) + k; Example 1: y²/49 − x²/32 = 1 has vertices (0, ±7) and foci (0, ±9); Example 3: (x − 3)²/9 − (y + 2)²/16 = 1 has vertices (0, −2), (6, −2) and foci (−2, −2), (8, −2); Example 5: 9x² − 4y² − 36x − 40y − 388 = 0 is (x − 2)²/36 − (y + 5)²/81 = 1), https://openstax.org/books/precalculus-2e/pages/10-2-the-hyperbola (retrieved 2026-10-19); OpenStax, Precalculus 2e, §10.5 Conic Sections in Polar Coordinates (eccentricity e = PF ÷ PD; a hyperbola has e > 1), https://openstax.org/books/precalculus-2e/pages/10-5-conic-sections-in-polar-coordinates (retrieved 2026-10-19).
3. form = general, A = 9, C = -4, D = -36, E = -40, F = -388 gives equation = (x − 2)²/36 − (y + 5)²/81 = 1, center = (2, −5), vertices = (−4, −5), (8, −5), coVertices = (2, −14), (2, 4), foci = (−8.816653826, −5), (12.81665383, −5), asymptotes = y = ±(3/2)(x − 2) − 5, c = 10.81665383. Source: OpenStax, Precalculus 2e, §10.2 The Hyperbola (standard forms centered at (0, 0) and (h, k), c² = a² + b², vertices, co-vertices, foci and asymptotes y = ±(b/a)(x − h) + k; Example 1: y²/49 − x²/32 = 1 has vertices (0, ±7) and foci (0, ±9); Example 3: (x − 3)²/9 − (y + 2)²/16 = 1 has vertices (0, −2), (6, −2) and foci (−2, −2), (8, −2); Example 5: 9x² − 4y² − 36x − 40y − 388 = 0 is (x − 2)²/36 − (y + 5)²/81 = 1), https://openstax.org/books/precalculus-2e/pages/10-2-the-hyperbola (retrieved 2026-10-19).
4. form = standard, axis = x, a2 = 36, b2 = 4 gives foci = (−6.32455532, 0), (6.32455532, 0), asymptotes = y = ±(1/3)x, c = 6.32455532. Source: OpenStax, Precalculus 2e, §10.2 The Hyperbola (standard forms centered at (0, 0) and (h, k), c² = a² + b², vertices, co-vertices, foci and asymptotes y = ±(b/a)(x − h) + k; Example 1: y²/49 − x²/32 = 1 has vertices (0, ±7) and foci (0, ±9); Example 3: (x − 3)²/9 − (y + 2)²/16 = 1 has vertices (0, −2), (6, −2) and foci (−2, −2), (8, −2); Example 5: 9x² − 4y² − 36x − 40y − 388 = 0 is (x − 2)²/36 − (y + 5)²/81 = 1), https://openstax.org/books/precalculus-2e/pages/10-2-the-hyperbola (retrieved 2026-10-19).
5. form = general, A = -1, C = 2, D = 0, E = 0, F = -8 gives equation = y²/4 − x²/8 = 1, vertices = (0, −2), (0, 2), asymptotes = y = ±0.7071067812x, b = 2.828427125. Source: OpenStax, Precalculus 2e, §10.2 The Hyperbola (standard forms centered at (0, 0) and (h, k), c² = a² + b², vertices, co-vertices, foci and asymptotes y = ±(b/a)(x − h) + k; Example 1: y²/49 − x²/32 = 1 has vertices (0, ±7) and foci (0, ±9); Example 3: (x − 3)²/9 − (y + 2)²/16 = 1 has vertices (0, −2), (6, −2) and foci (−2, −2), (8, −2); Example 5: 9x² − 4y² − 36x − 40y − 388 = 0 is (x − 2)²/36 − (y + 5)²/81 = 1), https://openstax.org/books/precalculus-2e/pages/10-2-the-hyperbola (retrieved 2026-10-19).

## FAQ

### How do you find the foci of a hyperbola?

Find c from c² = a² + b² (for a hyperbola the squares add). The foci are c from the center along the transverse axis. For y²/49 − x²/32 = 1, c² = 49 + 32 = 81, so the foci are (0, −9) and (0, 9).

### How do you find the asymptotes of a hyperbola?

For (x − h)²/a² − (y − k)²/b² = 1 they are y = ±(b/a)(x − h) + k; for a hyperbola that opens up and down, (y − k)²/a² − (x − h)²/b² = 1, they are y = ±(a/b)(x − h) + k.

### How do I write a hyperbola from the general form in standard form?

Complete the square in x and in y. 9x² − 4y² − 36x − 40y − 388 = 0 becomes 9(x − 2)² − 4(y + 5)² = 324; dividing by 324 gives (x − 2)²/36 − (y + 5)²/81 = 1 (OpenStax Example 5).

### What is the eccentricity of a hyperbola?

e = c ÷ a, always greater than 1. For (x − 3)²/9 − (y + 2)²/16 = 1, c = 5 and a = 3, so e = 5/3 ≈ 1.667.

### How do I know which way a hyperbola opens?

It opens along the axis of the positive squared term. x²/36 − y²/4 = 1 opens left and right; y²/49 − x²/32 = 1 opens up and down. In the general form, the sign of G ÷ A decides: positive opens left and right.

### What is the difference between a hyperbola and an ellipse?

An ellipse adds its squared terms, x²/a² + y²/b² = 1, and has c² = a² − b²; a hyperbola subtracts them and has c² = a² + b². In the general form, A and C have the same sign for an ellipse and opposite signs for a hyperbola.

## Sources

- OpenStax, Precalculus 2e, §10.2 The Hyperbola: standard forms centered at (0, 0) and at (h, k), c² = a² + b², vertices, co-vertices, foci, transverse and conjugate axes (2a and 2b), asymptotes y = ±(b/a)(x − h) + k and y = ±(a/b)(x − h) + k; Examples 1, 2, 3 and 5, retrieved 2026-10-19. https://openstax.org/books/precalculus-2e/pages/10-2-the-hyperbola
- OpenStax, Precalculus 2e, §10.5 Conic Sections in Polar Coordinates: eccentricity e = PF ÷ PD, and e > 1 for a hyperbola, retrieved 2026-10-19. https://openstax.org/books/precalculus-2e/pages/10-5-conic-sections-in-polar-coordinates