Defining a Parabola with Three Points

Date: 05/09/2009 at 01:45:31
From: Junichi 
Subject: Do 3 points determining a parabola if 1 point is the vertex

If we have 3 distinct points, and we know that one point is the 
vertex (of a parabola), is there only one parabola (any direction)
that passes through the 3 points?  I think there is one and only one 
parabola that passes through the 3 points, but I'm not sure.  This is
not homework by the way.  I just came up with this crazy thought while
learning parabolas in school.

Note: We know one point is the vertex, it's not like the "5 point 
locate one parabola" theorem. 

Since the parabola can be going towards any direction, hence it 
doesn't have to apply to y=ax^2+bx+c.  Therefore the only way to prove
this is by non-coordinate geometry?

Note: If we know that the parabola satisfies y=ax^2+bx+c (or y=a(x-
h)^2+k) then we only need 2 points.  A vertex (h,k) and a point (x,y).

I tried to find the location of the axis of symmetry first.  If we can
locate it, and if it's unique, then we can reflect the points besides
the vertex, then we'll have 5 points.  And we know that 5 point
determine a parabola (2 parabolas intersect at most 4 points).

Date: 05/10/2009 at 05:17:49
From: Doctor Jacques
Subject: Re: Do 3 points determining a parabola if 1 point is the vertex

Hi Junichi,

First, note that, although you need 5 points to determine a general
conic, 4 points will suffice if you know that the conic is a parabola.
This is because a parabola has only one asymptotic direction, which
means that the discriminant of the second degree terms is 0.  This
gives you an additional equation, which removes one "degree of
freedom" (this is strictly correct, you will see below what I mean).

If, in addition to that, you know that one (identified) point is a
vertex, you have another equation, because the tangent at the vertex
is perpendicular to the asymptotic direction, and you need one point less.

We may choose the vertex (A) as the origin of the axes.  The general
equation of a parabola (not vertical) through the origin is:

  (y - mx)^2 + ay + bx = 0         [1]

where m is the slope of the asymptotic direction.  The slope of the
tangent at the origin is (-b/a).  If the origin is the vertex, this
slope must be perpendicular with m, and we have:

  (-b/a) = (-1/m)
  a = mb                           [2]

and, by substituting into [1], you get the equation:

  (y - mx)^2 + b(my + x) = 0       [3]

Each of the other two points (B and C) will give you an equation, and
you can solve those two equations for the unknowns m and b.

The point is that you will get a cubic equation in m (you can see
this, for example, by choosing one of the coordinate axes as the line
through AB).  This equation may have up to three real roots for m, and
each such root will give you one value for b.  This means that you can
actually have up to 3 parabolas through the given points.

For example, assume that the points are:

  A (0,0)
  B (0,1)
  C (1,3)

If you work through the substitutions, you will find that m is given
by the equation:

  m^3 - 6m^2 + 6m - 1 = 0           [4]

and this equation has three real roots: {0.20871215252208, 1.0,
4.79128784747792}.  Each of these roots gives you the equation of a
parabola through A, B, C :

  (y - 0.20871215252208x)^2 - y - 4.79128784747792x = 0
  (x - y)^2 - y - x = 0
  (y - 4.79128784747792x)^2 - y - 0.20871215252208x = 0

(The symmetry between the first and third equations is accidental, it
is related to the fact that the equation [4] is self-reciprocal; this
is not typical of the general case).

You can see the three parabolas on:

     

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/20/2009 at 01:47:40
From: Junichi 
Subject: Do 3 points determining a parabola if 1 point is the vertex

Hi,

First of all thank you for helping me.  But can you clarify what's 
the asymptotic direction you're referring to?  Is it the direction of 
the axis of symmetry?  Or the directrix? 

Thanks again,

Junichi

Date: 05/20/2009 at 04:55:54
From: Doctor Jacques
Subject: Re: Do 3 points determining a parabola if 1 point is the vertex

Hi again Junichi,

An asymptotic direction on a curve is the limit of the ratio y/x (the
slope of a line through the origin) as the point on the curve goes to
infinity.

If the curve has an asymptote, then the asymptotic direction is the
slope of the asymptote.  For example, a hyperbola has two asymtoptes,
and the asymptotic directions are the slopes of those asymptotes.

In a parabola, there is only one asymptotic direction, and there is no
corresponding asymptote.  The asymptotic direction is the direction of
the axis of symmetry.

For a general conic, you can compute the asymptotic directions by
keeping only the second degree terms, and solving the corresponding
homogeneous equation for y/x.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/