is itself a theorem. That is to say, given five points no four of which are co-linear, we may construct the conic through these five points as follows. From a mathematical point of view, if we take five points no four of which are co-linear then there exists a unique conic section through them. This may be a special case such as a circle or a degenerate case as two intersecting straight lines. It is not immediately clear that this follows from the usual definition as a planar section through a cone. Consider the interpolation problem of fitting the equation
ax2 + by2 + cxy + dx + ey = k, with k = 0 or k = 1
through points (xi, yi). Since we have an equation in five unknowns, a, b, c, d and e it follows that five equations with (xi, yi), i = 1,...,5 provide the prospect of a unique solution.
Since five points are sufficient to define a conic one might expect six points on a conic to contain some redundancy and hence to generate interesting theorems. This is the case, and our key to understanding the conics is Pascal's Theorem, named for Blaise Pascal.
Consider six points A, B, C, A′, B′ and C′ on a conic section. Assume they are joined, as in Figure 3, with lines AB′, AC′ and BA′, BC′ and CA′, CB′. Find the intersections of the lines from A with those from A′, etc. Pascal's Theorem claims that the resulting three intersections of lines, at D, E and F say, are co-linear.
Figure 3. Pascal's Theorem
While Figure 3 illustrates the case of an ellipse, this holds for the other conic sections, including a circle. Depending on the configuration of the points, the Pascal line, dashed in Figure 3, may not intersect the ellipse. Furthermore, if two joined points, eg A and B′, coincide then the line AB′ is the tangent to the conic through A. Hence, given a point A on the conic by taking other points B, C, A′ and C′ it is possible to construct the Pascal line and from this the line DA must be a tangent to the conic.
If, in Pascal's Theorem, two pairs of points coincide then we have two tangent lines to a conic, and this provides the key to constructing the tangents to a conic from a point not on the conic. You are encouraged to fill in the missing details. If you would like the details then please email me, but for now I would prefer not to post all the answers with this article.
Rather surprisingly the button is itself a theorem. That is to say, given five points no four of which are co-linear, we may construct the conic through these five points as follows.
Take five points A, B, C, A′ and B′, so named to be consistent with Pascal's Theorem shown in Figure 3. Join these points as if we were going to show Pascal's Theorem, noting that the point C′ is missing. Mark D as the intersection of A′B and AB′. Draw a circle, centered at A through D and place a point P to be constrained on this circle. The line PA will eventually go through C′, but for now label E as the intersection of PA and CA′. The line DE is the Pascal line, and hence the intersection of this with B′C is the point F. The intersection of AE and BF is the point C′, which by Pascal's Theorem must be constrained to lie on the conic. As P moves, C′ so defined, moves around the entire conic section. See Figure 4
.Figure 4. Constructing a conic from five points