Eileen M. Klimick Schoaff wrote: > > A parabola, on the other hand, is determined by any 4 of its points. > John Conway> > > Am I missing something here? Don't 3 points determine a parabola if the axis > of symmetry is either vertical or horizontal? But if we consider any axes, > then there are an infinite number of parabolas passing through three points. > The generic equation is ax^2 + bxy +cy^2 + dx + ey + f = 0. If Jon Roberts is > considering parabolas of the form y = ax^2 + bx + c, then knowing 3 points > gives you three equations with three unknowns which can easily be solved -- > unless there is no solution. > > In the April 1997 issue of the Mathematics Teacher, a colleague of mine, Dr. > Ellie Johnson, wrote an article "A Look at Parabolas with a Graphing > Calculator". In this article she using the calculator to generate many > solutions to the generic equation. Of course this just shows that given three > points and restricting yourself to a parabola of the form y = ax^2 + bx + c, > you can derive the equation. That does not, of course, construct it. > > Does the fourth point determine whether the axis of symmetry is vertical, > horizontal, or rotated? In ax^2 + bxy +cy^2 + dx + ey + f = 0, it looks like > you need more than 4 points to determine a, b, c, d, e, f. > > Then again, I am only a math education person and do not have a PhD in math so > I am probably far in the dark. > > Eileen Schoaff > Buffalo State College
This might be easier to see if you look at the *definition* of the parabola and the other conics. A parabola is determined by its focus (a,b) and its directrix. The directrix can be specified by two parameters as well, say r and theta, where r is the distance of the line from the origin and theta is a reference angle. If you know four points on the parabola, you get four equations in four unknowns.
Similarly, a circle is also a conic, but it is completely determined by three parameters: its center (a,b) and radius r, so three points determine a circle.
An ellipse requires five points, as it is determined by two foci (a,b) and (c,d) and one additional parameter, say the sum of the distances from the foci to a point on the ellipse. Similarly, a hyperbola requires five points as well.
I have a question on a related topic which I'd like to put to the group. Suppose I have 5 points on a branch of a hyperbola. Then we know that there is a unique hyperbola through these five points. I would like to know a geometric method of finding the foci of this hyperbola. I seem to remember reading the solution to this problem somewhere. (When I say geometric, I mean co-ordinate free, not "solve 5 equations in five unknowns". It doesn't have to be a ruler and compass construction, although I seem to recall that the solution that I saw was.) I think the solution may be fairly complex, and I'd be happy just to have a reference.