Wikipedia:Reference desk/Archives/Mathematics/2008 January 29
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January 29
edithow many points to determine a general parabola?
editSurprising that neither parabola nor conic section gives this. How many points do you need to fix a parabola if you don't know its orientation? Is it one less than the number for a general conic? —Tamfang (talk) 07:09, 29 January 2008 (UTC)
- 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)
- (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
- {(0, −1), (−1, 0), ((1+√5)/2, (1+√5)/2), ((1−√5)/2, (1−√5)/2)}
- can be fitted both to y = x2 − 1 and x = y2 − 1.
- 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. --Lambiam 10:07, 29 January 2008 (UTC)
- 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)
- 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)
- 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)