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Falling short

The polar form of the equation for the ellipse  can be derived in a number of ways.

If we consider the ellipse the locus of  points which maintain the sum of distances between from point to foci constant, then consider the following diagram:

The red point and the cross are the two foci. 2a is the length of the major axis.

From the definition: r+d =2a.

From Pythagras theorem,

(2c-x)^2 +y^2 =(2a -r)^2


4c^2 -4cr\cos\theta + r^2=4a^2-4ar +r^2

using x^2+y^2=r^2

Simplifying and solving for r yields:

r =\frac{a^2-c^2}{a-c\cos\theta}

Eccentricity, e=c/a,

r =\frac{a^2(1-e^2)}{a(1-e\cos\theta)} =\frac{a(1-e^2)}{1-e\cos\theta}

Note that when \cos\theta =e, r =a. r \sin\theta =a\sin\theta at this point is the semi-minor axis. Let b represent the semi-minor axis. As a\cos\theta =ae, b=a\sin\theta=a\sqrt{1-e^2}

From the above graphic, the periapsis occurs when \theta =\pi, yielding r=a(1-e). Similarly, the apoapsis occurs when \theta =0 and r =a(1+e).

The ellipse can be defined (as can all conic sections) as the locus of points that  maintain a constant ratio of distance from a fixed line (directrix) to distance to a fixed point (focus). Using the same ellipse as above and the same angle convention (as illustrated in the following graphic) the derivation is straightforward. Let k be the distance between the focus and the directrix, and e emerges as illustrated.

r/e + r\cos (\pi-\theta) = k

Solving for r,

r = \frac {ek}{1 -e\cos\theta}

This is consistent with the above derivation.


  • The graphics were made using Mathematica and the drawing tool.
  • The specific ellipse used was r =\frac{0.75}{1-0.5\cos\theta} where a =1, e =0.5. The arguments are, however, general.
Categories: Mathematica, Mathematics
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  1. July 10, 2011 at 5:40 am

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