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

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Eileen M. Klimick Schoaff

Posts: 42
Registered: 12/6/04
Re: construction of triangle of given perimeter, given point and angle
Posted: Nov 11, 1997 12:19 PM
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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

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