Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Math Topics » geometry.pre-college.independent

Topic: construction of triangle of given perimeter, given point and angle
Replies: 16   Last Post: Jun 10, 2011 12:58 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ] Topics: [ Previous | Next ]
Peter Ash

Posts: 13
Registered: 12/6/04
Re: parabola through 4 points
Posted: Nov 11, 1997 10:36 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.