THE ELLIPSE

To draw an ellipse, stick two pins A and B upright into a sheet of paper on a drawing board.  Tie the ends of a thin nylon thread together to form a loop that fits loosely around the two pins.  Place the loop around the pins and pull the loop tight by means of a pencil point.  Move the pencil point right around the pins, thus tracing the ellipse on the paper. While the pencil pulls the nylon thread tight, the first property of an ellipse is being complied with: the sum of the distances from the two fixed points, the foci, is constant.

Thus AC + BC = 6 + 4 = 10; AD + BD = 5 + 5 = 10 and AE + BE = 3 + 7 = 10, and so on.
 

Please see ellipdgm.doc for diagram.
 

Another property of the ellipse, is that the areas of sectors 1, 2 and 3, that are traced out in equal times t, by a revolving body, between F and G, H and I, and J and K, under the force of attraction of a massive body at B (one of the foci), are equal.  This is so because the revolving body moves faster between F and G than between H and I, and J to K even slower.

In the case of a planet’s ellipse, angle FBG is less than 10, BF and BG are thus perpendicular to the ellipse.  The area of triangle BFG is ½ FG x FB.  Multiply by 1 or by FG/FG. Therefore the area of the triangle BFG is ½ FG 2.  FB/FG = ½ FG 2 / FG/FB = ½ FG 2 / sinFBG. The eccentricity of the ellipse is AB divided by LM = 6/10 = 0.6.

In the case of Halley’s comet the eccentricity of the orbit is 0.967 and the perihelion distance LB = 0.587 AU.

The distance between the foci BA = major axis LM - 2(0.587).
This distance divided by the major axis is equal to the eccentricity.
 

ie                   Major axis - 2(0.587) / Major axis = 0.967

                      Major axis - 2(0.587) = Major axis x 0.967

so that             0.033 x Major axis = 2 x 0.587 = 1.174.

Therefore        Major axis = 1.174/ 0.033 = 35.576 AU,

so that             the aphelion distance = 35.576 - 0.587 = 35 AU (very nearly)
 

This is half way between the orbits of Neptune and Pluto.

Eben van Zyl