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Definition of a Parabola

Date: 02/23/97 at 19:00:06
From: Cindy Smith
Subject: Re:  Parabola Problems

This problem is long, and I can't draw the diagram, but here it is:

A parabola can be defined as the set of all points in a plane equally 
distant from a given point and a given line not containing the point.  

(a) Suppose that (x, y) is to be on the parabola. Suppose that the 
line mentioned in the definition is given by x = -p.  Find the 
distance between (x, y) and the line.  (The distance from a point to a 
line is the length of the perpendicular from the point to the line.)

(b) If x = -p is the line mentioned in the definition, why is it 
reasonable to use (p, 0) as the given point?
(c) Find the distance from (x, y) to (p, 0).

(d) Find an equation for the parabola of the figure. (Hint: Use the 
results of parts (a) and (c) and the fact that (x, y) is equally 
distant from the point and the line.)

Could you please point me in the right direction so I can figure it 
out?  Thanks.

Cindy Smith

Date: 03/08/97 at 02:23:22
From: Doctor Luis
Subject: Re:  Parabola Problems


Let's hope you understand my diagram!  :)

               y  |
           |      |
           |      |     
           |      | 
           |      |     
           |      |                    * 
           |      |     *    /  A(x,y)
           |      | *      /  
   --------|------*------@---------------------> x
           |      | *     F(p,0)
           |      |     *
     [line x=-p]  |            *
                  |                    *

(a) Using the distance formula:

         AD = sqrt( (x-(-p))^2 + (y-y)^2 )
            = sqrt( (x+p)^2 )
            = x + p

(b) Remember the definition? "the set of all points in a plane EQUALLY 
DISTANT from a given point and a GIVEN LINE not containing the point." 
This should be more than enough to answer this question.

(c) Using the distance formula again:

         AF = sqrt( (x-p)^2 + (y-0)^2 )
            = sqrt( (x-p)^2 + y^2 )

(d) Ahh. This is the fun part. From the definition, AD = AF (right?).

                  (x+p) = sqrt( (x-p)^2 + y^2 )
                (x+p)^2 = (x-p)^2 + y^2
      (x^2 + 2px + p^2) = (x^2 - 2px + p^2) + y^2
The x^2's and p^2's cancel, leaving:

                    2px = -2px + y^2
                    4px = y^2 

So, the equation of the parabola is y^2 = 4px
You did notice that this was not a function (there are two possible
values of y for each value of x in the domain). If you interchange x 
and y, however, you will have a function (the parabola will open
about the y axis, instead of the x-axis):

                    4py = x^2

Conic sections in general are very interesting, and have many
applications in astronomy and optics. (The planets follow elliptical
paths around the sun, comets follow hyperbolic or even parabolic paths
around the sun. All of this is a result of the inverse square nature 
of the gravitational force.)

-Doctor Luis,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Equations, Graphs, Translations

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