About the Hyperbola Calculator
This hyperbola calculator works from the standard form of a hyperbola: (x − h)²/a² − (y − k)²/b² = 1 when it opens left and right, or (y − k)²/a² − (x − h)²/b² = 1 when it opens up and down. Enter the centre (h, k), the semi-transverse axis a and the semi-conjugate axis b, and it returns every key feature you are asked for in algebra, precalculus and analytic geometry courses.
You get the vertices, co-vertices, foci, the equations of both asymptotes, the eccentricity, the directrices, the focal distance c, the lengths of the transverse and conjugate axes and the length of the latus rectum. A table lists the points so you can sketch the curve quickly: plot the centre, draw the box through the vertices and co-vertices, extend its diagonals as asymptotes, then draw each branch through its vertex.
If your equation is not yet in standard form, complete the square on x and y first and divide so the right-hand side equals 1; the denominator under the positive term is a².
How to use the hyperbola calculator
- 1Write your equation in standard form so the right-hand side is 1.
- 2Choose horizontal if the x term is positive, vertical if the y term is positive.
- 3Enter the centre (h, k) and the square roots of the two denominators as a and b.
- 4Read the vertices, foci, asymptotes and eccentricity.
- 5Use the key-points table to sketch the box, asymptotes and both branches.
Formula and method
For a horizontal hyperbola centred at (h, k), the vertices sit a units left and right of the centre, the co-vertices b units above and below, and the foci c units from the centre along the transverse axis, where c² = a² + b² (note the plus sign — for an ellipse it is a minus). The eccentricity c/a is always greater than 1.
The asymptotes pass through the centre with slopes ±b/a (horizontal) or ±a/b (vertical); they are the diagonals of the box formed by the vertices and co-vertices. The directrices lie a²/c from the centre, and the latus rectum — the chord through a focus perpendicular to the transverse axis — has length 2b²/a. A vertical hyperbola swaps the roles of x and y.
- (h, k)
- Centre of the hyperbola
- a
- Distance from centre to each vertex
- b
- Distance from centre to each co-vertex
- c
- Distance from centre to each focus
- e
- Eccentricity
Worked examples
x²/9 − y²/16 = 1
With a = 3 and b = 4, c = √(9 + 16) = 5, so the foci are at (±5, 0). The asymptotes are y = ±(4/3)x, e = 5/3 ≈ 1.667, and the latus rectum is 2 × 16 ÷ 3 ≈ 10.67.
Vertical hyperbola centred at (2, −1) with a = 5, b = 12
c = √(25 + 144) = 13, so the foci are 13 units above and below the centre: (2, 12) and (2, −14). The asymptotes have slope ±5/12 ≈ ±0.417 and e = 13/5 = 2.6.
Rectangular hyperbola x²/4 − y²/4 = 1
When a = b the asymptotes are y = ±x, which are perpendicular. c = √8 ≈ 2.828 and the eccentricity is √2 ≈ 1.414.
Frequently asked questions
How do you find the foci of a hyperbola?+
Compute c = √(a² + b²) and move c units from the centre along the transverse axis. For x²/9 − y²/16 = 1, c = 5 and the foci are (5, 0) and (−5, 0).
How do you find the asymptotes of a hyperbola?+
For a horizontal hyperbola they are y − k = ±(b/a)(x − h); for a vertical one, y − k = ±(a/b)(x − h). They pass through the centre and the corners of the central box.
How can I tell which way a hyperbola opens?+
Look at which squared term is positive in standard form. If the x term is positive it opens left and right; if the y term is positive it opens up and down.
What is the eccentricity of a hyperbola?+
e = c/a, and it is always greater than 1. Values just above 1 give narrow branches; larger values give wide, flat branches. A rectangular hyperbola has e = √2.
What is the difference between an ellipse and a hyperbola?+
An ellipse is the set of points whose distances to two foci sum to 2a, with c² = a² − b². A hyperbola uses the difference of the distances, giving c² = a² + b² and two separate branches.