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Parabola


Date: 12/23/97 at 17:34:41
From: jen
Subject: Parabola

How can you, from an ordered pair, find the vertex, axis of symmetry, 
directrix, the focus, the distance from the vertex to the focus, and 
the latus rectum?  I use the following two equations in class: 

  (x-h)^2 = 4p(y-k) or (y-k)^2 = 4p(x-h).

Also, how can you tell if your numbers go in the x squared or 
y squared equation? 

Thank you for your time.


Date: 12/24/97 at 11:48:00
From: Doctor Rob
Subject: Re: Parabola

You can't find these things from a single ordered pair (x,y) of 
coordinates of a point on the curve. You need three pairs to do that, 
since there are three unknowns in the equation: h, k, and p. You also 
cannot tell which of the two equations you listed to use, even with 
three pairs. A fourth pair would let you figure out which of the two 
equations is the right one.

With both equations, the vertex is at (h,k), its distance from the 
focus is |p|, and the length of the latus rectum is 4*|p|.

With the first equation, the axis of symmetry is the line x = h, the
directrix is the line y = k - p, and the focus is at (h,k+p).

With the second equation, the axis of symmetry is the line y = k, the
directrix is the line x = h - p, and the focus is at (h+p,k).

Example:  The points (-4,10), (-2,-3), and (3,7) lie on a parabola.
If the first equation is used, then

   (-4-h)^2 = 4*p*(10-k),
   (-2-h)^2 = 4*p*(-3-k),
    (3-h)^2 = 4*p*(7-k).

We can use the first equation to solve for k, and substitute in the 
other two equations:

         k = 10 - (4+h)^2/(4*p)
  (-2-h)^2 = 4*p*[-3 - 10 + (4+h)^2/(4*p)] = 4*p*(-13) + (4+h)^2,
   (3-h)^2 = 4*p*[7 - 10 + (4+h)^2/(4*p)]  = 4*p*(-3) + (4*h)^2,

Now we can use the second equation to solve for p, and substitute in 
the other equations:

   p = (h+3)/13,
   k = -(13*h^2+64*h+88)/(4*[3+h]) = -(13*h+25)/4 - 13/(4*[h+3]),
   0 = 5*(34*h+11)/13.

Using the last equation to solve for h, and substituting in the other
equations, we get:

   h = -11/34,
   k = -6105/952,
   p = 7/34.

This allows us to answer all the questions posed.

If the second equation is used, we get instead,

   (10-k)^2 = 4*p*(-4-h),
   (-3-k)^2 = 4*p*(-2-h),
    (7-k)^2 = 4*p*(3-h),

which gives the solution

   h = -13633/2856,
   k = -107/6,
   p = 119/6.

Again, this allows us to answer all the questions posed.

If you know five points, you can determine the conic section on which
they lie by using the equation

   A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0,

substituting in the five pairs of values (x,y), and getting five 
linear equations in the six unknowns A, B, C, D, E, and F. You can 
solve them for five of the unknowns in terms of the sixth. If not all 
of A, B, and C are zero, and B^2 - 4*A*C = 0, this is a parabola. You 
can determine the answers to all your questions for this parabola, 
too. It is very possible that its axis may be oblique, that is, not 
parallel to either coordinate axis. In fact the slope of the axis is 
B/(2*C) if C is nonzero. If C is zero, then B must also be zero, and 
then the axis is vertical, and parallel to the y-axis.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
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High School Equations, Graphs, Translations

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