This Java Script calculator finds the general conic equation corresponding to five points in the Cartesian (x-y) plane. Press the Compute button to obtain the solution. The Test button loads a test case to demonstrate how the calculator works. On invalid entries, a popup window will display an error message.
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The Java Script source code for this program can be viewed by using the View|Source command of your web browser.
You may use or modify this source code in any way you find useful, provided that you agree that the author has no warranty, obligations or liability. You must determine the suitability of this source code for your use.
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The general conic equation
A conic equation represents the intersection of an extended cone and a plane. The general form of the equation of a conic section is:
A·x2 + B·x·y + C·y2 + D·x + E·y + F = 0
An example
The equation below represents an ellipse with axes aligned with the coordinate frame and with its center at the origin.
This equation can be changed by a translation that moves the center of the ellipse to a point {p, q} by replacing x, y using:x2 y2 --- + --- = 1 a2 b2
It can also be changed by a rotation by angle θ around the coordinate origin by again replacing x, y using:x := x - p y := y - q
If the ellipse is translated or rotated or both, the resulting equation will have terms in x2, x·y, y2, x, y, and a constant. Transformation of the normal equations for a parabola or a hyperbola would also produce a new equation with these terms.x := x·cos(θ) + y·sin(θ) y := -x·sin(θ) + y·cos(θ)
Fitting a conic section through five points
The general conic section has five independent coefficients (the equation can be divided by any of the six coefficients without changing the equality). Therefore, five points are necessary and sufficient to define a unique conic section. One and only one conic section can be drawn through five points.
One method to determine the five coefficients is to substitute each of the {x, y} points into the equation to obtain five linear equations using five independent coefficients as variables. For example, to find the conic section that fits these five points:
{0.5, 8.0}, {1.0, 4.0}, {2.0, 2.0}, {4.0, 1.0}, {8.0, 0.5}
Solve the system of equations (and arbitrarily choosing F = 4):
to obtain the solution:0.25·A + 4.00·B + 64.00·C + 0.50·D + 8.00·E + 4.00 = 0 1.00·A + 4.00·B + 16.00·C + 1.00·D + 4.00·E + 4.00 = 0 4.00·A + 4.00·B + 4.00·C + 2.00·D + 2.00·E + 4.00 = 0 16.00·A + 4.00·B + 1.00·C + 4.00·D + 1.00·E + 4.00 = 0 64.00·A + 4.00·B + 0.25·C + 8.00·D + 0.50·E + 4.00 = 0
The calculator performs these computations using Gaussian elimination.- x·y + 4.0000 = 0
Elementary Mathematical Analysis
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