Line and parabola intersection
The line y = x + 2 meets the parabola y = x^2 at (-1, 1) and (2, 4): the calculator sets both equal, solves the quadratic and reads the discriminant, here 9, to tell a secant, a tangent or no meeting point. Free in your browser.
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Worked example
- A
- 1
- B
- 0
- C
- 0
- M
- 1
- K
- 2
- Parabola
- y=1x^2+0x+0
- Line
- y=1x+2
- Equation
- 1x^2-1x-2=0
- A
- 1
Finding where a straight line meets a parabola is a core algebra skill: set the expressions equal, reduce to a quadratic, and read the discriminant. Students need both ordered pairs when the line is a secant, the single tangent point when it touches once, and a clear empty result when the graphs miss each other in the real plane. This line and parabola intersection calculator accepts the five finite coefficients of y equals a x squared plus b x plus c and y equals m x plus k, substitutes the line into the parabola, solves A x squared plus B x plus C equals zero with A equals a, B equals b minus m, and C equals c minus k, and returns the real points with y recovered from the line.
Substitution: from two graphs to one quadratic
A vertical-opening parabola written as y equals a x squared plus b x plus c and a non-vertical line written as y equals m x plus k intersect where both expressions give the same y for the same x. Equating them yields a x squared plus b x plus c equals m x plus k.
Moving every term to one side produces the monic-looking but fully general quadratic A x squared plus B x plus C equals zero, where the three coefficients are assembled without iteration: A is simply a, B is b minus m, and C is c minus k. The discriminant Delta equals B squared minus four A C then decides the geometry.
If Delta is positive the line is a secant and two distinct real x values appear from the quadratic formula; each x is paired with y equals m x plus k so the answer is two Cartesian points ordered by ascending x. If Delta is exactly zero the line is tangent and the double root collapses to a single point of contact.
If Delta is negative there is no real intersection: the returned solution count is zero and the points array is empty, which matches classroom language that the system has no real solution without treating that outcome as a hard input error. Coefficient a must be non-zero because a equals zero would turn the first equation into another line and the problem would no longer be a nonlinear line-and-parabola system.
Use the echoed parabola string, line string, reduced equation, and formula field to show the algebraic steps next to the numeric points in a worksheet or LMS feedback panel.
Where line and parabola intersections help
Classrooms and tutoring platforms use a line and parabola intersection calculator so learners can check ordered pairs for arbitrary coefficients without sign errors when expanding b minus m or c minus k by hand. Exam-prep generators need a deterministic oracle that returns zero, one, or two points from the discriminant instead of silently inventing a complex pair or treating an empty success as a network failure. Failed validation does not consume balance.
Keep requests small, pin the rounded points in tests, and you get reproducible intersections for homework, LMS plugins, and edge workers without standing up symbolic math infrastructure of your own.
What you can do with it
Classroom and homework checks
Let students submit parabola coefficients a, b, c and line m, k, then verify zero, one, or two ordered intersection points against textbook substitution.
Notebooks and simple curve–target models
Solve y=ax²+bx+c against y=mx+k in scripts for projectiles, cost curves, or design checks without importing a full CAS.
FAQ
How do I find where a line meets a parabola?
Send finite a (non-zero), b, c for y=ax²+bx+c and m, k for y=mx+k.
What does a zero, one, or two solution count mean?
Discriminant positive → two secant points; zero → one tangent point; negative → no real intersection (empty points array, not an error).
Why must a be non-zero?
If a is zero, y=ax²+bx+c is a line, not a parabola. The solver requires a≠0 so the system stays a genuine line–parabola problem.
What if the line only touches the parabola?
Then the line is a tangent: the discriminant is zero and there is a single point of contact. For the parabola y = x^2 and the line 2x − 1, that point is (1, 1).
Does the calculator give complex intersections?
No, it works only in the real plane. When the discriminant is negative the graphs do not meet, so the solution count is zero and the list of points is empty, which is a valid result rather than an error.
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/algebra/system-nonlinear-line-parabola \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"a":1,"b":0,"c":0,"m":1,"k":2}'const res = await fetch("https://api.kit.forhosting.com/algebra/system-nonlinear-line-parabola", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"a": 1,
"b": 0,
"c": 0,
"m": 1,
"k": 2
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/algebra/system-nonlinear-line-parabola",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"a": 1,
"b": 0,
"c": 0,
"m": 1,
"k": 2
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/algebra/system-nonlinear-line-parabola", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"a":1,"b":0,"c":0,"m":1,"k":2}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"a":1,"b":0,"c":0,"m":1,"k":2}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/algebra/system-nonlinear-line-parabola", 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": 1,
"b": 0,
"c": 0,
"m": 1,
"k": 2
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "algebra.system_nonlinear_line_parabola",
"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
When a is zero the curve is not a parabola and the task fails with invalid_input.
Substitution: from two graphs to one quadratic
This endpoint solves only the real plane; complex roots are not emitted.
Input contract, response shape, and edge cases
Example: a equals one, b equals zero, c equals zero, m equals one, k equals two describes y equals x squared and y equals x plus two. Substitution gives x squared minus x minus two equals zero, roots x equals negative one and x equals two, and points negative one comma one and two comma four. A tangent example is y equals x squared with y equals two x minus one: discriminant zero and the single point one comma one. Pair the numeric response with your own plotter if you need a sketch; this capability returns coordinates and algebra only. POST a JSON body with five required finite numbers: a, the non-zero quadratic coefficient of the parabola; b and c, the remaining parabola coefficients; m, the slope of the line; and k, the y-intercept of the line. Each field may arrive as a JSON number or as a numeric string. The slope alias is accepted for m, and intercept or y_intercept are accepted for k, so clients that already speak slope-intercept language do not need a rename layer. Missing fields, empty strings, non-finite values, and a equal to zero fail with invalid_input naming the problem; those failures are not billed. On success the body echoes a, b, c, m, and k rounded to twelve decimal places for stable goldens; parabola and line display strings; the reduced equation string; the intermediate A, B, and C; the discriminant; solution_count in zero, one, or two; points as an array of objects with x and y; method set to substitution; and a short formula restating the pipeline. A miss is y equals x squared with y equals negative one: discriminant negative and solution_count zero.
Where a line–parabola intersection API earns its keep
Engineering notebooks often reduce projectile height versus a straight target line, cable sag approximations, or simple cost–revenue curves to y equals a x squared plus b x plus c against y equals m x plus k; calling a tiny parse API avoids pulling a full computer-algebra system for a five-coefficient problem.
Integration use cases
Exam and LMS auto-grading
Mark nonlinear system items with a deterministic oracle that grades empty intersections as solution_count zero and never invents extra real roots.
Technical questions
Does the calculator return complex intersections?
No. Only real points are returned. A negative discriminant yields solution_count 0 and an empty points list.
What does the line and parabola intersection calculator cost?
The browser tool is free. API requests use prepaid KIT balance at $0.002 per successful call; invalid input is not billed.