acalculator

Where are the hyperbola’s foci?

Type the hyperbola equation in standard form, or the general form Ax² + Cy² + Dx + Ey + F = 0. The hyperbola calculator finds the center, vertices, co-vertices, foci, asymptotes and eccentricity.

Your numbers

Equation
Standard form
x²/9 − y²/16 = 1

The hyperbola x²/9 − y²/16 = 1 has foci (−5, 0), (5, 0) and asymptotes y = ±(4/3)x.

Center (h, k)
(0, 0)
Vertices
(−3, 0), (3, 0)
Co-vertices
(0, −4), (0, 4)
Foci
(−5, 0), (5, 0)
Asymptotes
y = ±(4/3)x
a
3
b
4
Focal distance c
5
Eccentricity e
1.666666667
Transverse axis length
6
Conjugate axis length
8
Latus rectum
10.66666667

Standard form: x²/9 − y²/16 = 1. The hyperbola x²/9 − y²/16 = 1 has foci (−5, 0), (5, 0) and asymptotes y = ±(4/3)x.

How to calculate

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.

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.

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.

  • 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.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Equation Standard form, Opens Up and down, a² (under the positive term) 49, b² (under the negative term) 32 gives Vertices (0, −7), (0, 7), Foci (0, −9), (0, 9), Focal distance c 9, a 7, Eccentricity 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. Equation Standard form, Opens Left and right, a² (under the positive term) 9, b² (under the negative term) 16, Center x (h) 3, Center y (k) -2 gives Standard form (x − 3)²/9 − (y + 2)²/16 = 1, Center (h, k) (3, −2), Vertices (0, −2), (6, −2), Co-vertices (3, −6), (3, 2), Foci (−2, −2), (8, −2), Asymptotes y = ±(4/3)(x − 3) − 2, Focal distance c 5, Eccentricity e 1.666666667, Latus rectum 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. Equation General form, A (x²) 9, C (y²) -4, D (x) -36, E (y) -40, F (constant) -388 gives Standard form (x − 2)²/36 − (y + 5)²/81 = 1, Center (h, k) (2, −5), Vertices (−4, −5), (8, −5), Co-vertices (2, −14), (2, 4), Foci (−8.816653826, −5), (12.81665383, −5), Asymptotes y = ±(3/2)(x − 2) − 5, Focal distance 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. Equation Standard form, Opens Left and right, a² (under the positive term) 36, b² (under the negative term) 4 gives Foci (−6.32455532, 0), (6.32455532, 0), Asymptotes y = ±(1/3)x, Focal distance 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. Equation General form, A (x²) -1, C (y²) 2, D (x) 0, E (y) 0, F (constant) -8 gives Standard form 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)

How it works

A hyperbola that opens left and right has the standard form (x − h)²/a² − (y − k)²/b² = 1; one that opens up and down has (y − k)²/a² − (x − h)²/b² = 1. In both, a² is under the positive term, (h, k) is the center, and

  • c² = a² + b², so c = √(a² + b²);
  • the vertices are a from the center along the transverse axis: (h ± a, k), or (h, k ± a);
  • the co-vertices are b from the center along the conjugate axis: (h, k ± b), or (h ± b, k);
  • the foci are c from the center along the transverse axis: (h ± c, k), or (h, k ± c);
  • the asymptotes are y = ±(b/a)(x − h) + k for a hyperbola that opens left and right, and y = ±(a/b)(x − h) + k for one that opens up and down;
  • the transverse axis is 2a long and the conjugate axis 2b.

Eccentricity. e = c ÷ a. It is the ratio PF ÷ PD of OpenStax §10.5 for the focus F = (h + c, k) and the directrix x = h + a²/c: a point P on the right branch has PF = (c/a)(x − h) − a and PD = (x − h) − a²/c, and PF ÷ PD = c/a. Since c > a, e > 1.

Latus rectum. The chord through a focus at a right angle to the transverse axis. At x − h = c the equation gives (y − k)² = b²(c²/a² − 1) = b⁴/a², so the chord is 2b² ÷ a long.

General form Ax² + Cy² + Dx + Ey + F = 0 (no xy term). Completing the square, A(x − h)² + C(y − k)² = G with h = −D ÷ (2A), k = −E ÷ (2C) and G = D² ÷ (4A) + E² ÷ (4C) − F. Dividing by G: (x − h)²/(G/A) + (y − k)²/(G/C) = 1. When G/A > 0 the hyperbola opens left and right with a² = G/A and b² = −G/C; otherwise it opens up and down with a² = G/C and b² = −G/A.

Rules

  • Standard form: a² and b² from 10⁻¹² to 10¹²; h and k from −10¹² to 10¹², 0 when left empty.
  • General form: A, C, D, E and F from −10¹² to 10¹². No answer when A or C is 0 ("A and C must both be nonzero for a hyperbola: with one of them 0 it is a parabola or a line."), when A and C have the same sign ("A and C have the same sign, which makes an ellipse or a circle, not a hyperbola."), or when G = 0 ("The equation is two crossing lines (a degenerate hyperbola), not a hyperbola.").
  • A number output too large or too small to show as a number gives no answer ("A result is too large or too small to show as a number.").

Output format

  • a², b², c², h and k are exact fractions of the typed decimals. a, b, c, e = √(c²/a²), 2a = √(4a²), 2b = √(4b²), the latus rectum √(4b⁴/a²) and the asymptote slope √(b²/a²) (or √(a²/b²)) are exact when the fraction under the root is the square of a fraction, and otherwise the true root rounded once to a 64-bit float.
  • Standard form writes each square as x², (x − h)² or (x + |h|)² (and the same for y and k), and a² and b² as decimals when they have at most 10 significant digits, otherwise as a fraction p/q in lowest terms in brackets: (x − 1)²/(100/3).
  • Points are written (x, y). An exact coordinate (h, k, and h ± a when a is exact) is rounded once to 10 significant figures; a coordinate with a root that is not exact is h + c or h − c computed in 64-bit floats, then shown to 10 significant figures. Numbers use the true minus sign (−) and e-notation below 10⁻⁶ or from 10²¹ up; vertices, co-vertices and foci list the smaller coordinate first.
  • Asymptotes read y = ±m X K: m is left out when the slope is exactly 1, a whole number when it is one, a fraction (p/q) in brackets when it is a fraction, or a decimal to 10 significant figures when it is not exact; X is x when h = 0, otherwise (x − h) or (x + |h|); K is left out when k = 0, otherwise + k or − |k|.
  • The number outputs show to 10 significant figures.
  • The answer sentence reads The hyperbola E has foci P and asymptotes Q., with the standard form, the foci and the asymptotes as above.

Worked examples by hand

y²/49 − x²/32 = 1 (OpenStax Example 1). It opens up and down with a² = 49 and b² = 32, so a = 7 and c² = 49 + 32 = 81, c = 9. Vertices (0, −7), (0, 7), foci (0, −9), (0, 9), e = 9 ÷ 7 = 1.285714286.

(x − 3)²/9 − (y + 2)²/16 = 1 (OpenStax Example 3). Center (3, −2), a = 3, b = 4, c² = 9 + 16 = 25, c = 5. Vertices (3 ± 3, −2) = (0, −2), (6, −2); co-vertices (3, −2 ± 4) = (3, −6), (3, 2); foci (3 ± 5, −2) = (−2, −2), (8, −2); asymptotes y = ±(4/3)(x − 3) − 2; e = 5/3 = 1.666666667; latus rectum 2 × 16 ÷ 3 = 10.66666667.

9x² − 4y² − 36x − 40y − 388 = 0 (OpenStax Example 5). h = 36 ÷ 18 = 2, k = 40 ÷ (−8) = −5, G = 1,296 ÷ 36 + 1,600 ÷ (−16) + 388 = 36 − 100 + 388 = 324. G/A = 36 > 0, so it opens left and right: (x − 2)²/36 − (y + 5)²/81 = 1. a = 6, b = 9, c = √117 = 3√13 = 10.81665383. Vertices (−4, −5), (8, −5); co-vertices (2, −14), (2, 4); foci (−8.816653826, −5), (12.81665383, −5); asymptotes y = ±(3/2)(x − 2) − 5.

x²/36 − y²/4 = 1 (OpenStax Example 2). c = √40 = 2√10 = 6.32455532, so the foci are (−6.32455532, 0), (6.32455532, 0); the asymptote slope is √(4/36) = 1/3: y = ±(1/3)x.

2y² − x² = 8. A = −1, C = 2, F = −8: G = 8, G/A = −8 < 0, so it opens up and down with a² = 8 ÷ 2 = 4 and b² = 8: y²/4 − x²/8 = 1. Vertices (0, −2), (0, 2); b = √8 = 2.828427125; the slope √(4/8) = √½ is not exact: y = ±0.7071067812x.

Other questions people ask

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.