If we determine in S1 (fig. 16) the ray corresponding to the ray S2S1 in S2, we get the tangent at S1. Similarly, we can determine the point of contact of the tangents u1 or u2 in fig. 17.
| Fig. 17. |
| Fig. 18. |
§ 51. If five points are given, of which not three are in a line, then we can, as has just been shown, always draw a curve of the second order through them; we select two of the points as centres of projective pencils, and then one such curve is determined. It will be presently shown that we get always the same curve if two other points are taken as centres of pencils, that therefore five points determine one curve of the second order, and reciprocally, that five tangents determine one curve of the second class. Six points taken at random will therefore not lie on a curve of the second order. In order that this may be the case a certain condition has to be satisfied, and this condition is easily obtained from the construction in § 49, fig. 16. If we consider the conic determined by the five points A, S1, S2, K, L, then the point D will be on the curve if, and only if, the points on D1, S, D2 be in a line.
This may be stated differently if we take AKS1DS2L (figs. 16 and 18) as a hexagon inscribed in the conic, then AK and DS2 will be opposite sides, so will be KS1 and S2L, as well as S1D and LA. The first two meet in D2, the others in S and D1 respectively. We may therefore state the required condition, together with the reciprocal one, as follows:—
| Pascal’s Theorem.—If a hexagon be inscribed in a curve of the second order, then the intersections of opposite sides are three points in a line. | Brianchon’s Theorem.—If a hexagon be circumscribed about a curve of the second class, then the lines joining opposite vertices are three lines meeting in a point. |
These celebrated theorems, which are known by the names of their discoverers, are perhaps the most fruitful in the whole theory of conics. Before we go over to their applications we have to show that we obtain the same curve if we take, instead of S1, S2, any two other points on the curve as centres of projective pencils.
§ 52. We know that the curve depends only upon the correspondence between the pencils S1 and S2, and not upon the special construction used for finding new points on the curve. The point A (fig. 16 or 18), through which the two auxiliary rows u1, u2 were drawn, may therefore be changed to any other point on the curve. Let us now suppose the curve drawn, and keep the points S1, S2, K, L and D, and hence also the point S fixed, whilst we move A along the curve. Then the line AL will describe a pencil about L as centre, and the point D1 a row on S1D perspective to the pencil L. At the same time AK describes a pencil about K and D2 a row perspective to it on S2D. But by Pascal’s theorem D1 and D2 will always lie in a line with S, so that the rows described by D1 and D2 are perspective. It follows that the pencils K and L will themselves be projective, corresponding rays meeting on the curve. This proves that we get the same curve whatever pair of the five given points we take as centres of projective pencils. Hence—
| Only one curve of the second order can be drawn which passes through five given points. | Only one curve of the second class can be drawn which touches five given lines. |