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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,

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

Therefore,

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

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.

Comment

• 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.
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Categories: Mathematica, Mathematics
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1. July 10, 2011 at 5:40 am