Up: Explore: Projective Conics

An Explanation of the Form and Program

Your first six inputs should be angle measures (in degrees) specifying the siz points on the unit circle that will form the hexagon; an angle theta will specify the point (cos(theta), sin(theta)) on the circle. These will lead to a graph showing the unit circle and the hexagon through those six points. Each side of the hexagon will have a different color. The graph will also show the Pascal line of the hexagon: one gray line through the intersections of the red and light blue lines, the orange and dark blue lines, and the green and purple lines. We call all of this a Pascal configuration; it is the first output of the program.

We can think of this Pascal configuration as lying in the plane P: z=1. Now we can use the origin to project this plane onto any other plane P'. The projection will take points in the P to points in the P', and lines in P to lines in P'. The same projection will take the circle from P to some conic in P'. However, in P' the line at infinity may intersect or be tangent to the conic; in other words, the conic may become a hyperbola or a parabola.

Let P' have the line at infinity l(inf)'. This line corresponds to some line l in P. Now consider the reverse process: specifying the line l in P and allowing it to determine the plane P'. This does not determine the plane P' uniquely; it only restricts the possibilities for P' to a family of parallel planes. However, this does determine uniquely the projection of the Pascal configuration from P. To be more precise: Say we take two planes from this family and consider the projections of the Pascal configuration in both of them. Then the two projections can differ only in size; they are identical up to a dilation. Thus, once we specify l, we have specified what the Pascal configuration looks like in P'.

The nature of this specification becomes much more clear when it is displayed graphically; this is the point of the program called below. This program produces two graphs, one of which is the already-described Pascal configuration in P. The second graph shows the same Pascal configuration in P'; we specify P' by specifying l in P, and we specify l using the bottom half of the form.

If the projection in P' of our unit circle is a hyperbola, it will contain two points on the line at infinity. These will correspond to two points in P which lie both on the unit circle and on l, and we may specify these two points by means of angles. So, when the hyperbola option is selected on the form, the last two inputs are taken as angles specifying these points, the line l, and the plane P'.

If the projection in P' of our unit circle is an ellipse, it will not intersect the line at infinity at all. So the line l in P will not intersect the unit circle at all; we can not specify l as we did in the hyperbolic case. Instead, we imagine dropping a perpendicular from the origin to l; we specify the line l by specifying this perpendicular. So, when the elliptic option is selected on the form, the second- to-last input is taken as the angle from the x-axis to the perpendicular, and the last input is taken as the length of that perpendicular.

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Created: Nov 30 1995 --- Last modified: Thu Nov 30 15:58:53 1995