Date: Nov 11, 1997 10:36 PM
Author: Peter Ash
Subject: Re: parabola through 4 points
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
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
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
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
a geometric method of finding the foci of this hyperbola. I seem to
remember reading the solution to this problem somewhere. (When I say
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.