NationMaster - Encyclopedia: 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. The ellipse and some of its mathematical properties. ... In mathematics, a locus (Latin for place, plural loci) is a collection of points which share a common property. ... personal space, proxemics. ... The word focus (pl. ... The word focus (pl. ...

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), i.e.

Image File history File links Ellipse_derivation_1. ...

then we can define $a$

d 1 + d 2 = 2 a

(a here is the semi-major axis, although this is irrelevant for the sake of the proof). From this simple definition we can derive the cartesian equation. Substituting:

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

To simplify we isolate the radical and square both sides.

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

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

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

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 $a 2 - c 2$

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

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

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

If x = 0 then

$d_1 = d_2 = a = sqrt {c^2+b^2}$

(where b is the semi-minor axis)

Therefore we can substitute

b 2 = a 2 - c 2

And we have our desired equation:

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

Categories: Article proofs | René Descartes | Conic sections

Results from FactBites:

Britain.tv Wikipedia - Cartesian coordinate system (1674 words)
	Cartesian coordinate systems are also used in space (where three coordinates are used) and in higher dimensions.
	Using the Cartesian coordinate system geometric shapes (such as curves) can be described by algebraic equations, namely equations satisfied by the coordinates of the points lying on the shape.
	An example of a point P on the system is indicated in Figure 3, using the coordinate (3,5).

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