Hyperbola Properties
Enter the center (h, k), the semi-transverse axis a and the semi-conjugate axis b for the standard horizontal-transverse hyperbola (x − h)²/a² − (y − k)²/b² = 1. The calculator returns the vertices, the foci with c = √(a² + b²), the asymptote equations and the eccentricity.
The asymptotes are the point
A hyperbola is the conic with two separate branches. Where a closed conic keeps the distances to its two foci summing to a constant, a hyperbola keeps their difference constant — and that switch splits the curve in half and sends both halves off to infinity.
This calculator takes the centre and the two semi-axes of a hyperbola whose branches open left and right, and reports the vertices, the foci, the asymptote equations and the eccentricity.
Every quantity here except the asymptotes has a counterpart in the closed conics. The asymptotes do not: they are the two lines the branches approach ever more closely without meeting, and they govern the curve far from the centre more than any point on it does.
They also explain the second semi-axis. Nothing on the curve is b units from the centre vertically — the conjugate axis touches no point of the hyperbola. Its only job is to set the asymptote slope, and through it the steepness of the branches.
How to use this calculator
- Enter the centre as h and k The midpoint between the two vertices, which is also where the asymptotes cross. It is not a point on the curve — the centre of a hyperbola lies in the gap between the branches.
- Enter a, the semi-transverse axis The distance from the centre to either vertex along the horizontal direction. Strictly positive; zero and negative values are rejected.
- Enter b, the semi-conjugate axis Also strictly positive. It does not reach the curve. It sets the asymptote slope as the ratio b over a, and nothing else.
- Read the asymptotes alongside the vertices Two points and two lines together determine the sketch. The vertices fix where the branches start; the asymptotes fix where they go.
The formula, and where it comes from
(x − h)²/a² − (y − k)²/b² = 1 c = √(a² + b²) y − k = ±(b/a)(x − h) e = c/a
The minus sign between the squared terms is the whole difference from the closed-conic equation, and it changes everything downstream. Because the y-term is subtracted, no real point exists for x between the vertices: the equation would need a negative quantity to equal a positive one. That is why the curve has a gap in the middle.
The focal distance adds the squared semi-axes rather than subtracting, so c always exceeds a. The foci sit beyond the vertices, outside the curve, whereas in a closed conic they sit inside. That sign is the most common error when the two are studied together.
The asymptotes come from asking what the equation approaches when x is enormous. The constant on the right becomes negligible against two large squares, so the relation collapses to a difference of squares equalling zero — which factors into two lines through the centre with slopes ±b/a.
Eccentricity is the same ratio c/a as in the closed case, but since c exceeds a it is always above one. Just above one means narrow branches hugging the transverse axis; a large value means branches that open out almost immediately.
What each input means
- h Centre x-coordinate — form field “Center x (h)”
- Horizontal position of the centre, midway between the vertices and on neither branch.
- k Centre y-coordinate — form field “Center y (k)”
- Vertical position of the centre. The asymptotes intersect here.
- a Semi-transverse axis — form field “Semi-axis a (transverse)”
- Distance from the centre to each vertex. Must be positive. It is the eccentricity denominator, so it sets the scale everything else is judged against.
- b Semi-conjugate axis — form field “Semi-axis b (conjugate)”
- Must be positive, but reaches no point of the curve. It fixes the asymptote slope, and through it how quickly the branches spread.
- c Focal distance
- Computed as the root of the sum of the squared semi-axes, and reported with that addition shown so it is not confused with the closed-conic subtraction.
Worked examples
Every number below is produced by the same calculation engine the tool above runs. Nothing here is typed by hand, so the walkthrough cannot drift from what you get when you enter the same values yourself.
The 3-4-5 hyperbola
Centred at the origin with a semi-transverse axis of 3 and a semi-conjugate axis of 4. The two squares add to 25, so the focal distance is a whole number.
Inputs Center x (h) = 0, Center y (k) = 0, Semi-axis a (transverse) = 3, Semi-axis b (conjugate) = 4
- Equation (x − 0)²/9 − (y − 0)²/16 = 1
- Center (0, 0)
- Orientation horizontal transverse axis (x-term positive)
- Vertices (-3, 0), (3, 0)
- Foci c = √(9 + 16) = 5 → (-5, 0), (5, 0)
- Asymptotes y − 0 = ±(1.33333)·(x − 0)
- Eccentricity e = c/a = 1.66667
Result center (0, 0), vertices (-3, 0) & (3, 0), foci (-5, 0) & (5, 0), e = 1.66667
The focal distance is 5, comfortably beyond the vertices at 3. That ordering is guaranteed here: adding a positive square to a² can only produce something larger, so the foci always lie outside the branches.
The asymptotes have slopes of four thirds and minus four thirds, steeper than the diagonal. The eccentricity of five thirds says the same thing in one number — well above one, so the branches open rather than hug their axis.
Equal semi-axes
Centre at the origin with a and b both 2. Equal axes give the equilateral case, where the asymptotes are the two diagonals.
Inputs Center x (h) = 0, Center y (k) = 0, Semi-axis a (transverse) = 2, Semi-axis b (conjugate) = 2
- Equation (x − 0)²/4 − (y − 0)²/4 = 1
- Center (0, 0)
- Orientation horizontal transverse axis (x-term positive)
- Vertices (-2, 0), (2, 0)
- Foci c = √(4 + 4) = 2.82843 → (-2.82843, 0), (2.82843, 0)
- Asymptotes y − 0 = ±(1)·(x − 0)
- Eccentricity e = c/a = 1.41421
Result center (0, 0), vertices (-2, 0) & (2, 0), foci (-2.82843, 0) & (2.82843, 0), e = 1.41421
The asymptote slopes become plus and minus one, so the branches approach the two forty-five-degree lines. This is the hyperbola analogue of a circle among the closed conics — the maximally symmetric member of the family.
The eccentricity comes out as the square root of two for every equilateral hyperbola, whatever the size of the axes. It depends only on their ratio, so scaling the curve leaves it untouched.
Reading the result
The branches never reach their asymptotes
The gap narrows without limit as you move from the centre but never closes. A sketch that lets a branch touch or cross its asymptote is drawing a different curve, and the error shows most near the vertices.
Eccentricity is always above one
Since c exceeds a by construction, the ratio cannot be one or less. A value near one means the two foci sit close to the vertices and the branches are narrow; a large value means they flare open quickly.
When you would use this
Sketching from an equation in standard form
The denominators under the squared terms are a² and b², so their roots feed straight in. Vertices and asymptotes together draw a recognisable pair of branches without plotting points.
Navigation and ranging by time difference
A constant difference in arrival time between two transmitters means a constant difference in distance, and therefore a hyperbola with the transmitters at the foci. Intersecting two such curves fixes a position — the principle behind hyperbolic navigation.
Assumptions and limitations
What this calculator assumes
- The transverse axis is horizontal, so the branches open left and right and the x-term carries the positive sign.
- Both semi-axes are strictly positive; either may be the larger.
- The inputs are semi-axes measured from the centre, not full axis lengths.
- The curve is axis-aligned rather than rotated.
Where it stops being the right tool
- Vertical-transverse hyperbolas are not accepted: an equation with the y-term positive needs the coordinates swapped by hand.
- Rotated hyperbolas, and general conic equations with an xy term, are out of scope.
- No area is reported, since a hyperbola encloses no finite region.
Common mistakes
Subtracting the squares to find the foci
Why it happens. The closed-conic relation uses a subtraction, and the two formulas are taught together. Subtracting here gives a focal distance smaller than a, which would put the foci between the vertices — inside a region the curve does not occupy.
How to avoid it. Add. Check the result against a: for a hyperbola the focal distance must exceed the semi-transverse axis, and a value below it means the wrong sign was used.
Looking for the curve on the conjugate axis
Why it happens. The second semi-axis is measured from the centre like the first, so it is natural to expect a point of the curve b units away vertically. Nothing is there.
How to avoid it. Use b only for the asymptote slope. The conjugate axis is a construction line, drawn to build the box the asymptotes run through diagonally.
Assuming the larger semi-axis is the transverse one
Why it happens. For a closed conic the longer axis determines the orientation, so the same habit gets applied here. A hyperbola opens along whichever term is positive, regardless of size.
How to avoid it. Look at the sign, not the size. The variable whose squared term is added is the one the branches open along, and here that is always x.
Frequently asked questions
What are the asymptotes of a hyperbola?
Two straight lines through the centre that the branches approach without ever touching. For the horizontal case they are y − k = ±(b/a)(x − h), and they come from dropping the constant term in the equation when x and y are both large.
How does the focal distance differ from the closed-conic case?
By a sign. Here c is the root of a² + b², so it always exceeds a and the foci lie beyond the vertices, outside the curve. In a closed conic the squares are subtracted instead, placing the foci inside.
Is the eccentricity always greater than 1?
Yes, necessarily. It is c/a and c is always larger than a, so the ratio exceeds one. That is the numerical statement of the foci lying beyond the vertices.
Can I enter a hyperbola that opens up and down?
Not directly. This solver assumes the transverse axis is horizontal, with the x-term positive. For a vertical one, swap the roles of the two coordinates and read the results with x and y interchanged.