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Derivation of the cartesian formula for an ellipse

The derivation of the cartesian form for an ellipse is simple and instructive. An ellipse is defined as a the loci of points equidistant to two fixed points called the foci. Assuming that the foci are located at (-c,0) and (c,0) (ie. the ellipse is centered at (0,0)) then the sum of the distance between any point (x,y) and the two foci is constant.

If (x,y) is any point on the ellipse and if $d 1$ is the distance between (x,y) and (-c,0) and $d 2$ is the distance between (x,y) and (c,0), ie

$d_1 = \sqrt {(x+c)^2+y^2}$

$d_2 = \sqrt {(x-c)^2+y^2}$

then

d 1 + d 2 = 2 a

where a is the semimajor axis. From this we can derive the cartesian equation. Substituting:

$\sqrt {(x+c)^2+y^2} + \sqrt {(x-c)^2+y^2} = 2a$

To simplify we rearrange and square both sides.

$\sqrt {(x+c)^2+y^2} = 2a - \sqrt {(x-c)^2+y^2}$

$(x+c)^2 + y^2 = \left ( 2a - \sqrt{(x-c)^2+y^2} \right )^2$

$(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2 +y^2$

Solving for the root and simplifying:

$\sqrt{(x-c)^2+y^2} = -{1 \over 4a} ((x+c)^2+y^2-4a^2-(x-c)^2-y^2)$

$\sqrt{(x-c)^2+y^2} = -{1 \over 4a} (x^2 + 2xc + c^2 -4a^2 -x^2 +2xc -c^2)$

$\sqrt{(x-c)^2+y^2} = -{1 \over 4a} (4xc - 4a^2)$

$\sqrt{(x-c)^2+y^2} = a - {c \over a}x$

A final squaring

$(x-c)^2+y^2 = a^2 - 2cx + {c^2 \over a^2}x^2$

$x^2 - 2xc + c^2 + y^2 = a^2 -2xc + {c^2 \over a^2}x^2$

$x^2 + c^2 + y^2 = a^2 + {c^2 \over a^2}x^2$

Grouping the x-terms and dividing with $a 2 - c 2$

$x^2 \left( 1 - {c^2 \over a^2} \right) + y^2 = a^2 - c^2$

$x^2 \left( {a^2 - c^2 \over a^2} \right) + y^2 = a^2 - c^2$

${x^2 \over a^2} + {y^2 \over a^2-c^2} = 1$

It is easy to see that by the Pythagorean theorem $a 2 - c 2$ is the square of the semiminor axis.

Therefore we can substitute

b 2 = a 2 - c 2

And we have our desired equation:

${x^2 \over a^2} + {y^2 \over b^2} = 1$

Last updated: 05-16-2005 16:51:54

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