Hyperbola eccentricity
Measure how widely a hyperbola opens from its two semi-axes: 3 and 4 give a focal distance of 5 and an eccentricity of 5/3. Every true hyperbola lands above 1, unlike an ellipse. Free in your browser.
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Worked example
- A
- 3
- B
- 4
- C
- 5
- Eccentricity
- 1.6666666667
The eccentricity of a hyperbola measures how sharply the curve opens compared with a circle: for every standard hyperbola it is strictly greater than one. Given the semi-transverse axis a and the semi-conjugate axis b, the closed formula e equals the square root of one plus b squared over a squared, which is the same as the linear eccentricity c divided by a. This capability accepts those two positive semi-axis lengths and returns the eccentricity together with c, so homework checks, CAD helpers, and geometry pipelines can reuse one deterministic result instead of re-deriving the radical by hand.
What hyperbola eccentricity means
In conic-section geometry a hyperbola is the set of points whose absolute difference of distances to two fixed foci is constant. That constant is 2a, the full transverse axis length, while the distance from center to each focus is c. Eccentricity is defined as e equals c over a.
Because the foci always lie outside the vertices of a hyperbola, c is strictly larger than a, so e is always greater than one—unlike an ellipse, where e stays between zero and one, or a parabola, where e equals one. The conjugate semi-axis b is related by the identity c squared equals a squared plus b squared. Substituting that identity into the definition yields the convenient working formula e equals the square root of one plus b squared over a squared.
Textbooks, contest problems, and orbital-analogy exercises all use this dimensionless number to compare hyperbolas of different sizes: two hyperbolas with the same e are similar; larger e means a more open, sharper angle between the asymptotes. Computing e by hand is short algebra, but students often mix the ellipse formula with the hyperbola one, flip a and b, or forget that a must be the transverse semi-axis.
How the capability computes the answer
Both must be finite and strictly positive; zero is rejected because division by a would be undefined or degenerate, and any negative length is rejected because a semi-axis is a geometric length, not a signed coordinate. Both expressions are algebraically equivalent; the dual output lets you verify e against c divided by a in client code if you wish. The response reports a, b, c, and eccentricity as numbers.
The algorithm is a handful of multiplications and one square root per path, so it stays well inside the analytic cost model used for pricing this tier-A parse capability.
Where to use it and what it costs
Typical placements include verifying student answers that ask for eccentricity given a and b, generating catalog rows of similar hyperbolas that share e while scaling a and b, documenting asymptote geometry for technical drawings, and feeding e into larger conic classifiers that already know the semi-axes from a fit. The browser path runs locally and needs no account.
Keep a and b positive and finite so the radical is well defined, treat the returned c as the linear eccentricity to the foci, and prefer the server-computed eccentricity field rather than re-deriving the square root in client code that might use a different floating-point path.
What you can do with it
Homework and exam checks
Confirm a student’s eccentricity for given semi-axes a and b against one canonical e and c instead of re-grading hand square roots.
Similar-hyperbola catalogs
Scale a and b while holding e fixed, or compute e once per shape family when building tables of conics for design or teaching materials.
FAQ
What happens if a or b is negative?
Semi-axis lengths must be strictly positive geometric lengths; negative values are rejected.
Is eccentricity always greater than one?
Yes for a non-degenerate hyperbola with positive a and b. The formula sqrt(1 + b²/a²) is always greater than 1 when b and a are positive.
What is the formula for the eccentricity of a hyperbola?
Take the square root of one plus b squared over a squared. It is the same as the focal distance c divided by a, where c squared equals a squared plus b squared.
How does it differ from the eccentricity of an ellipse?
An ellipse stays between zero and one and a parabola sits at exactly one, while a hyperbola is always above one. Mixing up the ellipse and hyperbola formulas is one of the most frequent slips.
What does a larger eccentricity tell me?
That the hyperbola is more open, with a wider angle between its asymptotes. Two hyperbolas with the same eccentricity are similar, even when their sizes are different.
Related tools
For developers — API access
Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.
API endpoint
Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.
Call it from your stack
curl -X POST https://api.kit.forhosting.com/geom/hyperbola-eccentricity \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"a":3,"b":4}'const res = await fetch("https://api.kit.forhosting.com/geom/hyperbola-eccentricity", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"a": 3,
"b": 4
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/geom/hyperbola-eccentricity",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"a": 3,
"b": 4
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/geom/hyperbola-eccentricity", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"a":3,"b":4}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"a":3,"b":4}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/geom/hyperbola-eccentricity", body)
req.Header.Set("Authorization", "Bearer "+os.Getenv("KIT_KEY"))
req.Header.Set("Content-Type", "application/json")
res, _ := http.DefaultClient.Do(req)Example request
{
"a": 3,
"b": 4
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "geom.hyperbola_eccentricity",
"status": "queued",
"_links": {
"result": "/tasks/tsk_…/result"
}
}The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.
Pricing
Published price — no tokens, no invented credits. A failed task is never charged.
Errors
| HTTP | Code | Meaning |
|---|---|---|
401 | unauthorized | Missing or invalid API key. |
402 | insufficient_balance | Your balance doesn't cover the task price. |
404 | unknown_type | That task type doesn't exist. |
429 | rate_limited | Too many requests. Use the webhook instead of polling. |
Technical notes
What hyperbola eccentricity means
Automating the calculation with validated positive inputs removes those slips while still exposing a, b, c, and eccentricity in one JSON object for the next step of a solution or pipeline.
How the capability computes the answer
You send a JSON body with two required numeric fields: a, the semi-transverse axis, and b, the semi-conjugate axis. The engine parses numbers from JSON numbers or plain numeric strings, then computes c as the square root of a squared plus b squared and eccentricity as the square root of one plus b squared over a squared. Missing fields, non-numeric values, zero, or negative semi-axes fail with invalid_input and are never charged. There is no randomness, no clock dependency, and no network call; two identical requests always produce the same JSON.
Where to use it and what it costs
Use the free browser tool when you want a quick check for a single pair of semi-axes, or call the API when a script, gradebook, or design job must resolve many hyperbolas the same way every time. API usage draws from prepaid KIT balance at the published $0.002 base price per successful request; failed validation does not consume balance.
Integration use cases
CAD and asymptote prep
Pass measured semi-axes from a drawing into the API and store eccentricity next to focus distance c for documentation or further geometry steps.
Technical questions
How do I calculate hyperbola eccentricity with this API?
POST a JSON body with positive numbers a (semi-transverse) and b (semi-conjugate). The response includes eccentricity and the linear eccentricity c.
Is the hyperbola eccentricity calculator free?
The browser widget is free and runs locally. API calls use prepaid KIT balance at the published $0.002 base price; invalid input is not charged.
Is the result always the same for the same a and b?
Yes. The algorithm is pure analytic math with no randomness or timestamps, so identical inputs produce identical JSON.