Coefficients from five points
- Ryan Hirst
Let me start here: A simple method for obtaining coefficients from five points.
The meat of this worksheet is in the Algebra view. The standard form: Five points determine the conic section, so why six parameters? Depending on the arrangement of the points in space, we may have to choose a different free parameter. For example, if the curve passes through (0,0), F is zero, irrespective of the kind of curve. So if we were to choose, say, F=1, our solution would be undefined whenever the curve passed through the coordinate origin. So suppose we choose one of the points to be the local origin, throw it out of the list, and set F to zero? The same problem arises: there is one free parameter among A, B, C, D, E, and if we decide, for example, to set E=1, the solution will be undefined whenever the curve is tangent to the y-axis at our chosen local origin. The conditions become more complex the more parameters we try to resolve in advance. A much simpler soltuion, is to let the given points tell us which of the coefficients to choose as the free parameter. Let the system be exactly as given in the standard form: Then, using Gaussian Elimination, reduce the matrix to Row Echelon Form. Count down the diagonal entries a11, a22 .... one of them will be zero. This is the free parameter we are looking for. If we let it be 1, we can read the other coefficients directly out of the column. In this way, we neatly sidestep the difficulties above.