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Equation of a Parabola


Date: 03/04/97 at 16:05:40
From: croman family
Subject: Parabolas as quadratic functions

Dr. Math,

I'm a 10th grade student and I'm having some trouble in my honors 
algebra II class with parabolas as quadratic functions. Could you tell 
me how to do these problems, please:

1. Determine the equation of a parabola with a vertex at the origin 
   and a focus of (10,0).

2. Determine the equation of a parabola with a vertex at the origin 
   and a directrice of y = -2

If you know any way to make this stuff easier, please tell me. If it 
is at all possible, please show your work. The entire class will be 
grateful.

Sincerely,
April Croman


Date: 03/06/97 at 16:54:12
From: Doctor Keith
Subject: Re: Parabolas as quadratic functions

Hi!

First let's show how we got that a parabola is a quadratic equation.  
Remember that a quadratic equation is of the form:

           y = f(x) = Ax^2 + Bx + C

A parabola is the set of points in the plane that are equally distant
from a point (focus) and a line (the directrix). We want to write 
this as an equation, so let's define our notation and assume for the 
moment that our directrix is a horizontal line (i.e., the parabola 
opens up or down):

focus      is the point         (a,b)
point      of parabola          (x,y)
directrix  is the line          y = d
distance   euclidean distance d(pt1,pt2) = sqrt((x1-x2)^2 + (y1-y2)^2)

With these definitions, we know that the distance from the focus to 
any point on the parabola is:

    d((a,b),(x,y)) = sqrt((a-x)^2 + (b-y)^2)

The distance from the directrix to a point on the parabola is:

    d((x,d),(x,y))= sqrt((x-x)^2 + (d-y)^2) = | d-y |

Now we set them equal to each other and square both sides to clean 
things up:

          (a-x)^2 + (b-y)^2 = (d-y)^2

We need to collect terms so we will expand the ones that deal with y 
first:

       (a-x)^2 + y^2 - 2by + b^2 = y^2 - 2dy + d^2

Take the expanded terms all to the right side to cancel y^2 terms:

    (a-x)^2 = 2(b-d)y + d^2 - b^2

Now expand the left side:

    x^2 -2ax + a^2 =  2(b-d)y + d^2 - b^2

Now solve for y:

     x^2 -2ax + a^2 + b^2 - d^2 = 2(b-d)y
    (1/2(b-d))x^2 + (a/(d-b))x +((a^2)/2(b-d) + (b+d)/2) = y

Let's make the following definitions:

    A = 1/2(b-d)
    B = a/(d-b)
    C = (a^2)/2(b-d) + (b+d)/2

Then we have our quadratic equation: y = Ax^2 + Bx + C

I will leave the analogous derivation for parabolas opening left or 
right to you.

Now we can solve your problems quickly by using this derivation and 
the fact that the vertex is a point on the parabola and that the 
directrix is perpendicular to the line traveling through the focus and 
the vertex:

1) We know (a,b) = (10,0) and vertex = (0,0). Thus the directrix is 
   avertical line (perpendicular to x axis) and ten units away from 
   the origin in the negative direction (for the equal distance part).  
   So the directrix is x = -10.  Then substitute these values in the 
   equation you were to derive above for A, B, and C.

2) We know y = -2 is the directrix so the parabola opens up or down
   (perpendicular requirement) and since the vertix is the origin, it 
   must open up and the focus is two units up on the y-axis (equal 
   distance on opposite side) so (a,b) = (0,2).  You can now 
   substitute these values into the equation I derived above for 
   A, B, and C and you are done.

Hope this helps,  If you would like more explanation, or something is
confusing, write back and I will give you more.  I left you some work 
so you could test yourself.

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

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