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Well, the formula for a conic section has six variables (with minor restrictions), and the parabolic case is a single additional equation. To specify the remaining five variables, you'd need five points, yes.
Black Carrot (
talk)
09:36, 29 January 2008 (UTC)reply
(after edit conflict) It is somewhat implicit, but from the general equation of a conic section
Ax2 + Bxy + Cy2 +Dx + Ey + F = 0
of a conic section with 6 parameters you have 5 degrees of freedom, since you can multiply all coefficients by a common multiplier so as to have A2 + B2 + C2 = 1. It gives you a parabola if B2 − 4AC = 0, which leaves 4 degrees of freedom. That means that you need, "in general", 4 points to fix it. I write "in general" because not all choices of 4 points will do (for example, you can't make a parabola go through the corners of a square), and in some cases two parabolas will go through 4 given points; for example, the set
I haven't looked at how common it is that a given set will have multiple solutions, but suspect it will always be a finite number. --
Lambiam10:07, 29 January 2008 (UTC)reply
Another way to get to same result is to define the parabola by two points that determine its
directrix, a third non-colinear point that is its focus, and a fourth point on the parabola itself to select a unique curve from the family of parabolas that have the given directrix and
focus. Incidentally, you can have a parabola that passes through the 4 corners of a square if you allow degenerate cases such as x2=1, when the "parabola" becomes two parallel lines.
Gandalf61 (
talk)
10:35, 29 January 2008 (UTC)reply
Yet another way to look at it: We all know that for a parabola of a known orientation we need 3 points. If we also need to find the orientation, that's another degree of freedom for which we need one more point. --
Meni Rosenfeld (
talk)
13:33, 29 January 2008 (UTC)reply
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the
current reference desk pages.
Well, the formula for a conic section has six variables (with minor restrictions), and the parabolic case is a single additional equation. To specify the remaining five variables, you'd need five points, yes.
Black Carrot (
talk)
09:36, 29 January 2008 (UTC)reply
(after edit conflict) It is somewhat implicit, but from the general equation of a conic section
Ax2 + Bxy + Cy2 +Dx + Ey + F = 0
of a conic section with 6 parameters you have 5 degrees of freedom, since you can multiply all coefficients by a common multiplier so as to have A2 + B2 + C2 = 1. It gives you a parabola if B2 − 4AC = 0, which leaves 4 degrees of freedom. That means that you need, "in general", 4 points to fix it. I write "in general" because not all choices of 4 points will do (for example, you can't make a parabola go through the corners of a square), and in some cases two parabolas will go through 4 given points; for example, the set
I haven't looked at how common it is that a given set will have multiple solutions, but suspect it will always be a finite number. --
Lambiam10:07, 29 January 2008 (UTC)reply
Another way to get to same result is to define the parabola by two points that determine its
directrix, a third non-colinear point that is its focus, and a fourth point on the parabola itself to select a unique curve from the family of parabolas that have the given directrix and
focus. Incidentally, you can have a parabola that passes through the 4 corners of a square if you allow degenerate cases such as x2=1, when the "parabola" becomes two parallel lines.
Gandalf61 (
talk)
10:35, 29 January 2008 (UTC)reply
Yet another way to look at it: We all know that for a parabola of a known orientation we need 3 points. If we also need to find the orientation, that's another degree of freedom for which we need one more point. --
Meni Rosenfeld (
talk)
13:33, 29 January 2008 (UTC)reply