Page:EB1911 - Volume 11.djvu/716

Rh §48.

It follows that every line in one of the two pencils cuts the curve in two points, viz. once at the centre S of the pencil, and once where it cuts its corresponding ray in the other pencil. These two points, however, coincide, if the line is cut by its corresponding line at S itself. The line p in S, which corresponds to the line SS′ in S′, is therefore the only line through S which has but one point in common with the curve, or which cuts the curve in two coincident points. Such a line is called a tangent to the curve, touching the latter at the point S, which is called the “point of contact.”

In the same manner we get in the reciprocal investigation the result that through every point in one of the rows, say in s, two tangents may be drawn to the curve, the one being s, the other the line joining the point to its corresponding point in s′. There is, however, one point P in s for which these two lines coincide. Such a point in one of the tangents is called the “point of contact” of the tangent. We thus get—

§ 49. Two projective pencils are determined if three pairs of corresponding lines are given. Hence if a1, b1, c1 are three lines in a pencil S1, and a2, b2, c2 the corresponding lines in a projective pencil S2, the correspondence and therefore the curve of the second order generated by the points of intersection of corresponding rays is determined. Of this curve we know the two centres S1 and S2, and the three points a1a2, b1b2, c1c2, hence five points in all. This and the reciprocal considerations enable us to solve the following two problems:

In order to solve the left-hand problem, we take two of the given points, say S1 and S2, as centres of pencils. These we make projective by taking the rays a1, b1, c1, which join S1 to A, B, C respectively, as corresponding to the rays a2, b2, c2, which join S2 to A, B, C respectively, so that three rays meet their corresponding rays at the given points A, B, C. This determines the correspondence of the pencils which will generate a curve of the second order passing through A, B, C and through the centres S1 and S2, hence through the five given points. To find more points on the curve we have to construct for any ray in S1 the corresponding ray in S2. This has been done in § 36. But we repeat the construction in order to deduce further properties from it. We also solve the right-hand problem. Here we select two, viz. u1, u2 of the five given lines, u1, u2, a, b, c, as bases of two rows, and the points A1, B1, C1 where a, b, c cut u1 as corresponding to the points A2, B2, C2 where a, b, c cut u2.

We get then the following solutions of the two problems:

§ 50. These constructions prove, when rightly interpreted, very important properties of the curves in question.

If in fig. 16 we draw in the pencil S1 the ray k1 which passes through the auxiliary centre S, it will be found that the corresponding ray k2 cuts it on u2. Hence—

As A is any given point on the curve, and u1 any line through it, we have solved the problems:

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.

§ 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:—