Hyperbola Calculator
Graph a hyperbola with its asymptotes and fundamental rectangle, and find the center, vertices, foci, eccentricity and asymptote equations step by step.
Try an example
Hyperbola
(x − 2)²/16 − (y + 1)²/9 = 1
Center (2, −1), branches open left and right, vertices (−2, −1) and (6, −1), foci (−3, −1) and (7, −1), asymptote slopes ±3/4 ≈ ±0.75, e = 1.25.
Hyperbola properties
- Standard equation
- (x − 2)²/16 − (y + 1)²/9 = 1
- Orientation
- ↔ Horizontal transverse axis (opens left/right)
- Center
- (2, −1)
- Semi-transverse axis a
- 4
- Semi-conjugate axis b
- 3
- Transverse axis length 2a
- 8
- Conjugate axis length 2b
- 6
- Vertices (on the curve)
- (−2, −1), (6, −1)
- Conjugate-axis endpoints (not on the curve)
- (2, −4), (2, 2)
- Foci
- (−3, −1), (7, −1)
- Focal distance c
- 5
- Eccentricity e
- 1.25
- Asymptote slopes
- ±3/4 ≈ ±0.75
- Asymptotes
- y + 1 = (3/4)(x − 2), y + 1 = −(3/4)(x − 2) (y = (3/4)x − 2.5, y = −(3/4)x + 0.5)
- Fundamental rectangle
- 8 wide × 6 tall, area 4ab = 48
The center is the midpoint of the two vertices and the point where the asymptotes cross. Every hyperbola has eccentricity e > 1; here e = 1.25.
Step-by-Step Calculation
Step 1 — Read the orientation from the positive term
The x-term is positive, so the transverse axis is horizontal and the branches open left and right.
(x − 2)²/16 − (y + 1)²/9 = 1
Center (h, k) = (2, −1)
a² = 16 → a = 4 (under the positive term)
b² = 9 → b = 3
Step 2 — Focal distance c
For a hyperbola the squares add: c² = a² + b² (an ellipse subtracts). So c is always larger than a.
c = √(a² + b²)
= √(16 + 9)
= √25
= 5
Step 3 — Eccentricity
e = c/a. Because c > a, every hyperbola has e > 1.
e = c / a
= 5 / 4
= 1.25
Step 4 — Vertices, conjugate-axis endpoints and foci
Vertices and foci lie on the transverse axis; the conjugate-axis endpoints are not on the curve, but they fix the fundamental rectangle.
Vertices: (h ± a, k) = (2 ± 4, −1) → (−2, −1) and (6, −1)
Conjugate-axis endpoints: (h, k ± b) = (2, −1 ± 3) → (2, −4) and (2, 2)
Foci: (h ± c, k) = (2 ± 5, −1) → (−3, −1) and (7, −1)
Step 5 — Asymptotes
The asymptotes pass through the center with slope ±b/a: the diagonals of the fundamental rectangle.
y − k = ±(b/a)(x − h)
slope = ±3/4 = ±3/4 ≈ ±0.75
y + 1 = (3/4)(x − 2)
y + 1 = −(3/4)(x − 2)
Step 6 — Fundamental rectangle
The rectangle centered at (h, k) with half-width and half-height a and b (swapped for a vertical hyperbola).
width 2a = 8, height 2b = 6
Corners: (−2, −4), (6, −4), (6, 2), (−2, 2)
Advanced hyperbola properties
Focal property. For every point P on the hyperbola, the difference of its distances to the foci is constant: |PF₁ − PF₂| = 2a = 8. (An ellipse uses the sum instead.)
Directrices (distance a/e from the center, between the center and each vertex): x = −1.2 and x = 5.2
Standard forms of a hyperbola
Horizontal hyperbola ↔ ← current
(x − h)²/a² − (y − k)²/b² = 1
- x-term positive: transverse axis horizontal
- Branches open left and right
- Vertices (h ± a, k), foci (h ± c, k)
- Asymptote slopes ±b/a
Vertical hyperbola ↕
(y − k)²/a² − (x − h)²/b² = 1
- y-term positive: transverse axis vertical
- Branches open up and down
- Vertices (h, k ± a), foci (h, k ± c)
- Asymptote slopes ±a/b
a is always under the positive term (it can be smaller than b), c² = a² + b², and e = c/a. The fundamental rectangle is centered at (h, k), 2a wide and 2b tall here; its diagonals are the asymptotes, and its sides touch the vertices.
Hyperbola vs ellipse
| Hyperbola | Ellipse | |
|---|---|---|
| Equation | (x − h)²/a² − (y − k)²/b² = 1 (minus) | (x − h)²/a² + (y − k)²/b² = 1 (plus) |
| Shape | Two separate open branches | One closed curve |
| Asymptotes | Two, through the center | None |
| Foci | c² = a² + b², so c > a | c² = a² − b², so c < a |
| Eccentricity | e > 1 | 0 ≤ e < 1 |
| Focal property | |PF₁ − PF₂| = 2a | PF₁ + PF₂ = 2a |
Why the sign flips: on a hyperbola, the vertex is between the center and the focus, so c > a, and the Pythagorean relation in the fundamental rectangle gives c² = a² + b² (c is the distance from the center to a corner). On an ellipse the focus is inside, so c < a and c² = a² − b².
Common mistakes
- Using c² = a² − b². That is the ellipse relation. For a hyperbola, c² = a² + b², so the foci are farther out than the vertices.
- Calling the conjugate-axis endpoints vertices. The points b from the center are not on the hyperbola. Only the two vertices, a from the center, are.
- Opening the branches the wrong way. The positive term decides: positive x-term → left and right; positive y-term → up and down.
- Drawing only one branch. Every hyperbola has two separate, mirror-image branches.
- Wrong asymptote slope. Horizontal: ±b/a. Vertical: ±a/b. The slope is rise over run of the rectangle’s diagonal.
- Treating a as the full axis. a is the semi-transverse axis; the transverse axis (vertex to vertex) is 2a.
- Assuming a > b. Unlike an ellipse, b can be larger than a. a is simply the term under the positive fraction.
- Drawing a closed curve. A hyperbola never closes; its branches run off to infinity along the asymptotes.
- Letting the curve cross its asymptotes. In standard position the branches approach the asymptotes but never touch or cross them.
Compare with the other conic sections in the Ellipse Calculator, the Parabola Calculator and the Circle Calculator. The Distance Formula Calculator checks distances such as PF₁ and PF₂, and the Slope Calculator works with lines like the asymptotes.
This hyperbola calculator analyses and graphs any hyperbola in standard position. Enter the center (h, k), the semi-transverse axis a and the semi-conjugate axis b, and choose whether the branches open left and right (horizontal) or up and down (vertical). The calculator writes the standard equation, for example (x − 2)²/16 − (y + 1)²/9 = 1, and draws both branches together with the asymptotes and the fundamental rectangle that defines them.
It also finds the vertices, the conjugate-axis endpoints, the foci, the focal distance c = √(a² + b²), the eccentricity, the asymptote equations and the axis lengths, and shows the working for each step. Values stay exact where possible, such as c = √29 or slopes ±3/4.
Worked Calculation Examples
| Scenario | Result | Calculation Step |
|---|---|---|
| Horizontal: (x − 2)²/16 − (y + 1)²/9 = 1 | Foci (−3, −1) and (7, −1), e = 1.25 | Center (2, −1); the x-term is positive, so the branches open left and right with a = 4, b = 3. Vertices (−2, −1), (6, −1). c = √(16 + 9) = 5, e = 5/4 = 1.25. Asymptotes y + 1 = ±(3/4)(x − 2). |
| Vertical: (y − 3)²/25 − (x + 2)²/9 = 1 | Vertices (−2, −2) and (−2, 8) | The y-term is positive, so the branches open up and down, with a = 5, b = 3 and center (−2, 3). c = √(25 + 9) = √34 ≈ 5.831, so the foci are (−2, 3 ± √34). Asymptotes y − 3 = ±(5/3)(x + 2); e = √34/5 ≈ 1.1662. |
| Negative center: (x + 3)²/4 − (y + 4)²/1 = 1 | Center (−3, −4), foci (−3 ± √5, −4) | Vertices (−5, −4) and (−1, −4). c = √(4 + 1) = √5 ≈ 2.2361, e = √5/2 ≈ 1.118. Asymptotes y + 4 = ±(1/2)(x + 3). |
| Decimal parameters: a = 1.5, b = 2 | c = 2.5, e = 5/3 | x²/2.25 − y²/4 = 1. c = √(2.25 + 4) = √6.25 = 2.5, so the foci are (±2.5, 0). e = 2.5/1.5 = 5/3 ≈ 1.6667. Asymptotes y = ±(4/3)x. |
| Different aspect ratio: a = 1, b = 4 | Asymptote slopes ±4, e = √17 | x²/1 − y²/16 = 1. The steep asymptotes y = ±4x make the branches open very widely. c = √17 ≈ 4.1231 and e = √17 ≈ 4.1231, compared with e = √17/4 ≈ 1.0308 for a = 4, b = 1, whose branches are narrow. |
What Is a Hyperbola?
A hyperbola is the set of points whose distances to two fixed points, the foci, have a constant difference: |PF₁ − PF₂| = 2a. It has two separate, mirror-image branches that open away from each other, and two asymptotes, straight lines through the center that the branches approach more and more closely without touching.
The line through the two vertices is the transverse axis, of length 2a. The perpendicular line through the center is the conjugate axis, of length 2b. Its endpoints are not on the hyperbola, but together with the vertices they fix the fundamental rectangle.
Reading the Standard Equation
In a standard-form hyperbola, one squared term is positive and one is negative. The positive term decides the direction: if the x-term is positive, as in (x − 2)²/16 − (y + 1)²/9 = 1, the transverse axis is horizontal and the branches open left and right; if the y-term is positive, as in (y − 3)²/25 − (x + 2)²/9 = 1, they open up and down.
a² is always the denominator of the positive term, even when it is smaller than b². The signs inside the brackets are opposite to the center: (x − 2) and (y + 1) give the center (2, −1).
The Fundamental Rectangle and the Asymptotes
Draw a rectangle centered at (h, k) extending a units along the transverse axis and b units along the conjugate axis in each direction: 2a by 2b for a horizontal hyperbola, 2b by 2a for a vertical one. The vertices touch the middle of two of its sides, and its diagonals, extended, are the asymptotes.
That gives the asymptote slopes directly as rise over run: ±b/a for a horizontal hyperbola and ±a/b for a vertical one. For (x − 2)²/16 − (y + 1)²/9 = 1, the asymptotes are y + 1 = ±(3/4)(x − 2). Sketching the rectangle and its diagonals first is the quickest way to draw a hyperbola by hand.
Foci and Eccentricity: Why c² = a² + b²
The foci lie on the transverse axis, c units from the center, where c² = a² + b²; geometrically, c is the distance from the center to a corner of the fundamental rectangle. For a = 4 and b = 3, c = √25 = 5. Because c > a, the foci lie beyond the vertices, outside the curve's turning points.
This is the opposite of an ellipse, where c² = a² − b² and the foci sit inside. The eccentricity e = c/a is therefore greater than 1 for every hyperbola (1.25 in the example); the larger e is, the more widely the branches open.
How to Use the Hyperbola Calculator
- Enter the center (h, k), the semi-transverse axis a (center to each vertex) and the semi-conjugate axis b (half the other side of the fundamental rectangle). Both a and b must be positive.
- Choose the orientation: Horizontal if the x-term is positive (branches open left and right), Vertical if the y-term is positive (branches open up and down). The standard equation updates as you type.
- Read the graph: both branches, the dashed asymptotes, the fundamental rectangle, the center C, vertices V, conjugate-axis endpoints B and foci F. Zoom in or out if needed.
- Check the Hyperbola Properties list for the axis lengths, foci, focal distance c, eccentricity e, asymptote equations and rectangle size.
- Follow the step-by-step working: c = √(a² + b²), e = c/a, the key points, the asymptote slopes and the rectangle.
- Try the examples to compare a horizontal and a vertical hyperbola, or wide and narrow branches.
Frequently Asked Questions
What is a hyperbola?
A curve with two separate branches, made of all points whose distances to two fixed points (the foci) differ by a constant 2a. It has two asymptotes through its center.
What is the standard equation of a hyperbola?
(x − h)²/a² − (y − k)²/b² = 1 for a horizontal transverse axis, and (y − k)²/a² − (x − h)²/b² = 1 for a vertical one, with center (h, k).
How do you find the center of a hyperbola?
Read h and k from the brackets with their signs reversed: (x − 2)² and (y + 1)² give the center (2, −1). The center is also the midpoint of the vertices and the point where the asymptotes cross.
How do you find the vertices?
Move a units from the center along the transverse axis: (h ± a, k) for a horizontal hyperbola, (h, k ± a) for a vertical one. For (x − 2)²/16 − (y + 1)²/9 = 1, the vertices are (−2, −1) and (6, −1).
How do you find the foci?
Compute c = √(a² + b²), then move c units from the center along the transverse axis. For a = 4, b = 3 and center (2, −1), c = 5 and the foci are (−3, −1) and (7, −1).
How do you find the asymptotes?
They pass through the center with slope ±b/a (horizontal) or ±a/b (vertical): y − k = ±(b/a)(x − h) or y − k = ±(a/b)(x − h). They are the diagonals of the fundamental rectangle.
What is the eccentricity of a hyperbola?
e = c/a. Since c² = a² + b² > a², it is always greater than 1. Larger values mean the branches open more widely.
Why is c² = a² + b² for a hyperbola?
For a hyperbola the foci lie beyond the vertices, and c equals the distance from the center to a corner of the fundamental rectangle, whose sides are a and b, so c² = a² + b². An ellipse’s foci lie inside, giving c² = a² − b² instead.
What is the difference between a horizontal and vertical hyperbola?
A horizontal hyperbola has a positive x-term and branches opening left and right, with asymptote slopes ±b/a. A vertical hyperbola has a positive y-term and branches opening up and down, with slopes ±a/b.
What is the difference between a hyperbola and an ellipse?
An ellipse’s equation has a plus sign and gives one closed curve with c² = a² − b² and e < 1. A hyperbola’s equation has a minus sign and gives two open branches with asymptotes, c² = a² + b² and e > 1.
What are the transverse and conjugate axes?
The transverse axis joins the two vertices and has length 2a. The conjugate axis is perpendicular to it through the center and has length 2b; its endpoints are not on the hyperbola.
Do hyperbolas intersect their asymptotes?
Not in standard position: the branches get arbitrarily close to the asymptotes as they extend outward but never touch or cross them.
Related Calculators
Ellipse Calculator
Graph an ellipse from its center and semi-axes or its equation, and find the vertices, co-vertices, foci, eccentricity, area and perimeter step by step.
Parabola Calculator
Graph y = ax² + bx + c and find the vertex, axis of symmetry, intercepts, discriminant, vertex and factored forms, focus and directrix, step by step.
Circle Calculator
Find radius, diameter, circumference, and area from any one of them, plus arc length, sector area, chords, central angles, and segments.
Distance Formula Calculator
Find the distance between two points in 2D or 3D with the distance formula, showing Δx, Δy, Δz, every substitution step, the exact answer and a graph.
Slope Calculator
Find the slope, equation, intercepts, angle, distance, and midpoint of a line from two points, an equation, or a point and slope, with a graph.
Midpoint Calculator
Find the midpoint between two points with step-by-step working for each coordinate and a coordinate plane showing A, B and M.
Last updated: September 27, 2026.