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Derivation of the Equation of an Ellipse


Date: 06/23/99 at 19:04:52
From: Rob Walsman
Subject: Derivation of equation of an ellipse

I have been trying to derive the equation of an ellipse from its 
definition (the locus of points having a fixed sum of distances to the 
two foci - sorry for the loose definition). My algebra gets very 
complex trying to eliminate radicals. Could you point me to a source 
where I can see a good derivation?

Thanks,
Rob Walsman


Date: 06/24/99 at 15:43:52
From: Doctor Peterson
Subject: Re: Derivation of equation of an ellipse

Hi, Rob.

The derivation really isn't too hard, but you have to know how to 
handle an equation containing the sum of two radicals. The trick is to 
move one radical to the other side and square; this leaves you with 
only one radical.

Then you can rearrange the equation so this radical is alone on one 
side, then square again to eliminate the radical entirely.

Here's how it goes:


             (x,y)********+********
            ****+*___     |  \_    ******
         ***   /     \____|    \__a      ***
       **     /           \b___   \_        **
      *      /            |    \____\_        *
     *(-f,0)/      f      |      f  \_\_(f,0)  *
     *-----+--------------+-------------+------*
     *         a          |          a         *
      *                   |                   *
       **                 |b                **
         ***              |              ***
            ******        |        ******
                  *****************


We want the sum of the distances from (-f,0) to (x,y) and from (x,y) 
to (f,0) to be 2a, where a is the horizontal semiaxis; the vertical 
semiaxis will be b, where b^2 + f^2 = a^2. The equation is

     sqrt[(x+f)^2 + y^2] + sqrt[(x-f)^2 + y^2] = 2a

Move the second root to the right and square:

   sqrt[(x+f)^2 + y^2] = 2a - sqrt[(x-f)^2 + y^2]

         (x+f)^2 + y^2 = 4a^2 - 4a sqrt[(x-f)^2 + y^2] + (x-f)^2 + y^2

 x^2 + 2fx + f^2 + y^2 = 4a^2 + x^2 - 2fx + f^2 + y^2
                               - 4a sqrt[(x-f)^2 + y^2]

Move the root to the left side by itself, and divide by 4:

     4a sqrt[(x-f)^2 + y^2] = 4a^2 - 4fx

      a sqrt[(x-f)^2 + y^2] = a^2 - fx

Square both sides again:

               a^2 (x^2 - 2fx + f^2 + y^2) = a^4 - 2a^2 fx + f^2 x^2

     a^2 x^2 - 2a^2 fx + a^2 f^2 + a^2 y^2 = a^4 - 2a^2 fx + f^2 x^2

                  (a^2 - f^2)x^2 + a^2 y^2 = a^4 - a^2 f^2

Divide by a^2(a^2 - f^2):

     x^2      y^2
     --- + --------- = 1
     a^2   a^2 - f^2

Write b^2 = a^2 - f^2, and we're done:

     x^2   y^2
     --- + --- = 1
     a^2   b^2

It's always fun to see how much cancels out when you're on the right 
track!

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Exponents
High School Geometry

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