§48.
| Theorem.—The curve of second order which is generated by two projective flat pencils passes through the centres of the two pencils. | Theorem.—The envelope of second class which is generated by two projective rows contains the bases of these rows as enveloping lines or tangents. |
| Proof.—If S and S′ are the two pencils, then to the ray SS′ or p′ in the pencil S′ corresponds in the pencil S a ray p, which is different from p′, for the pencils are not perspective. But p and p′ meet at S, so that S is a point on the curve, and similarly S′. | Proof.—If s and s′ are the two rows, then to the point ss′ or P′ as a point in s′ corresponds in s a point P, which is not coincident with P′, for the rows are not perspective. But P and P′ are joined by s, so that s is one of the enveloping lines, and similarly s′. |
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—
| Theorem.—To the line joining the centres of the projective pencils as a line in one pencil corresponds in the other the tangent at its centre. | Theorem.—To the point of intersection of the bases of two projective rows as a point in one row corresponds in the other the point of contact of its base. |
§ 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:
| Problem.—To construct a curve of the second order, of which five points S1, S2, A, B, C are given. | Problem.—To construct a curve of the second class, of which five tangents u1, u2, a, b, c are given. |
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:
| Solution.—Through the point A draw any two lines, u1 and u2 (fig. 16), the first u1 to cut the pencil S1 in a row AB1C1, the other u2 to cut the pencil S2 in a row AB2C2. These two rows will be perspective, as the point A corresponds to itself, and the centre of projection will be the point S, where the lines B1B2 and C1C2 meet. To find now for any ray d1 in S1 its corresponding ray d2 in S2, we determine the point D1 where d1 cuts u1, project this point from S to D2 on u2 and join S2 to D2. This will be the required ray d2 which cuts d1 at some point D on the curve. | Solution.—In the line a take any two points S1 and S2 as centres of pencils (fig. 17), the first S1 (A1B1C1) to project the row u1, the other S2 (A2B2C2) to project the row u2. These two pencils will be perspective, the line S1A1 being the same as the corresponding line S2A2, and the axis of projection will be the line u, which joins the intersection B of S1B1 and S2B2 to the intersection C of S1C1 and S2C2. To find now for any point D1 in u1 the corresponding point D2 in u2, we draw S1D1 and project the point D where this line cuts u from S2 to u2. This will give the required point D2, and the line d joining D1 to D2 will be a new tangent to the curve. |
§ 50. These constructions prove, when rightly interpreted, very important properties of the curves in question.
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| Fig. 16. |
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—
| Theorem.—In the above construction the bases of the auxiliary rows u1 and u2 cut the curve where they cut the rays S2S and S1S respectively. | Theorem.—In the above construction (fig. 17) the tangents to the curve from the centres of the auxiliary pencils S1 and S2 are the lines which pass through u2u and u1u respectively. |
As A is any given point on the curve, and u1 any line through it, we have solved the problems:
| Problem.—To find the second point in which any line through a known point on the curve cuts the curve. | Problem.—To find the second tangent which can be drawn from any point in a given tangent to the curve. |
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.
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| Fig. 17. |
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| 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:—
