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

Eileen,

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.

--Peter Ash