GENERATION OF CURVES AND CONES OF SECOND ORDER
OR SECOND CLASS
§ 45. Conics.—If we have two projective pencils in a plane, corresponding rays will meet, and their point of intersection will constitute some locus which we have to investigate. Reciprocally, if two projective rows in a plane are given, then the lines which join corresponding points will envelope some curve. We prove first
Theorem.—If two projective Theorem.—If two projective flat pencils lie in a plane, but
are neither in perspective nor concentric, then the locus of intersections of corresponding rays is a curve of the second order, that is, no line contains more than two points of the locus.
Proof.—We draw any line t. This cuts each of the pencils in a row, so that we have on t two rows, and these are projective because the pencils are projective. If corresponding rays of the two pencils meet on the line t, their intersection will be a point in the one row which coincides with its corresponding point in the other. But two projective rows on the same base cannot have more than two points of one coincident with their corresponding points in the
other (§ 34)• (§34)•
It will be seen that the proofs are reciprocal, so that the one may be copied from the other by simply interchanging the words point and line, locus and envelope, row and pencil, and so on. We shall therefore in future prove seldom more than one of two reciprocal theorems, and often state one theorem only, the reader being recommended to go through the reciprocal proof by himself, and to supply the reciprocal theorems when not given.
§ 46. We state the theorems in the pencil reciprocal to the last, without proving them:
Theorem.—If two projective Theorem.—If two projective flat pencils are concentric, but axial pencils lie in the same are neither perspective nor co pencil (their axes meet in a planar, then the envelope of the point), but are neither perspecplanes joining corresponding rays tive nor coaxial, then the locus is a cone of the second class; of lines joining corresponding that is, no line through the planes is a cone of the second common centre contains more order; that is, no plane in the than two of the enveloping planes. pencil contains more than two
of these lines. W
§ 47. Of theorems about cones of second order andcones of second class we shall state only very few. We point out, however, the following connexion between the curves and cones under consideration :
The lines which join any point Every plane section of a cone in space to the points on a curve of the second order is a curve of of the second order form a cone the second order.
of the second order.
The planes which join any Every plane section of a cone point in space to the lines en of the second class is a curve of 'veloping a curve of the second the second class.
class envelope themselves a cone
of the second class.
By its aid, or by the principle of duality, it will be easy to obtain theorems about them from the theorems about the curves.
We prove the first. A curve of the second order is generated by two projective pencils. These pencils, when joined to the point in space, give rise to two projective axial pencils, which generate the cone in question as the locus of the lines where corresponding planes meet.
rows lie in a plane, but are neither in perspective nor on a common base, then the envelope of lines joining corresponding points is a curve of the second class, that is, through no point pass more than two of the enveloping lines.
Proof.—We take any point T and join it to all points in each row. This gives two concentric pencils, which are projective because the rows are projective. If a line joining corresponding points in the two rows passes through T, it will be a line in the one pencil which coincides with its corresponding line in the other. But two projective concentric flat pencils in the same plane cannot have more than two lines of one coincident with their corresponding line in the other
II
§ 48.
Theorem.—The curve of second order which is generated by two projective flat pencils passes through the centres of the two pencils.
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'.
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 beings, 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 Theorem.—To the point of the centres of the projective intersection of the bases of two pencils as a line in one pencil projective rows as a point in one corresponds in the other the row corresponds in the other the
tangent at its centre. point of contact of its base.
§ 49. Two projective pencils are determined if three pairs of corresponding lines are given. Hence if al, b1, cl are three lines in a pencil Si, 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 SI and S2, and the three points ala2, b'b2; c'c2, hence five points in all. This and the reciprocal considerations enable us to solve the following two problems:
Problem.—To construct a curve Problem.—To construct a curve
of the second order, of which five of the second class, of which five
points SI, S2, A, B, C are given. tangents uI, u2, a, b, c are given.
In order to solve the lefthand problem, we take two of the given points, say SI and S2, as centres of pencils. These we make projective by taking the rays a', bI, c', which join Si 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 Si and S2, hence through the five given points. To find more points on the curve we have to construct for any ray in SI 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 righthand problem. Here we select two, viz. ul, u2 of the five given lines, u2, a, b, c, as bases of two rows, and the points Ai, Bi, CI where a, b, c cut u' 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 Solution.—In the line a take A draw any two lines, ul and u2 any two points Si and S2 as (fig. 16), the first ul to cut the centres of pencils (fig. 17), the pencil SI in a row ABICI, the first Si (A'BICI) to project the other u2 to cut the pencil S2 in a row uI, the other S2 (A2B2C2) to row AB2C2. These two rows will project the row u2. These two be perspective, as the point A pencils will be perspective, the corresponds to itself, and the line SIAI being the same as the centre of projection will be the corresponding line S2A2, and the point S, where the lines BIB2 axis of projection will be the line and CIC2 meet. To find now for u, which joins the intersection B any ray di in SI its corresponding of SIBI and S2B2 to the intersecray d2 in S2, we determine the tion C of SIC' and S2C2. To find
point DI where di cuts project now for any point DI in u' the this point from S to D2 on 14.2 and corresponding point D2 in u2, we join S2 to D2. This will be the draw SIDI and project the point required ray d2 which cuts di at D where this line cuts u from S2
some point D on the curve. to u2. This will give the required
point D2, and the lined joining D'
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.
If in fig. 16 we draw in the pencil SI the ray k1 which passes
ui
through the auxiliary centre S, it will be found that the corresponding ray k2 cuts it on u2. Hence
Theorem.—In the above con Theorem.—In the above construction the bases of the auxil struction (fig. 17) the tangents to iary rows ul and u2 cut the curve the curve from the centres of the where they cut the rays S2S and auxiliary pencils Si and S2 are the
End of Article: GENERATION OF CURVES AND CONES OF SECOND 

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