| 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.
| 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. |