C Is Proper. Clearly, C ∩ C Is Also the Intersection of C with Any
208 Chapter VI. Conics and Quadrics C is proper. Clearly, C ∩ C is also the intersection of C with any other conic of the pencil generated by C and C: Q(x, y, z)=Q(x, y, z)=0 if and only if Q(x, y, z)=λQ(x, y, z)+λQ(x, y, z)=0forλ =0 . Hence C ∩ C is the intersection of C with a degenerate conic C in the pencil. But C consists of two secant lines or of a double line. Figure 15, in which the “heavy” lines represent double lines, shows the “four” expected intersection points (notice, however, that the pairs of conics shown on this figure do not define distinct pencils (see Exercise VI.43). 4. The cross-ratio of four points on a conic and Pascal’s theorem AconicC and a point m are given in a plane P .Weconsiderthe pencil of lines through m.Recall that this is the line m of P . Any point m of C defines a mapping πm : C −−−−→ m which, with any point n of C,associates the line mn,withtheconvention that πm(m)isthetangent to C at m.Letusprove that πm,ormoreexactly −1 C πm is a parametrization of the conic by the projective line m (this is the first assertion in the next proposition). Proposition 4.1. For any point m of C,the mapping πm is a bijection. Moreover, if m and n are two points of C,the composed mapping ◦ −1 −−−−→ πn πm : m n is a homography. Remark 4.2. There is a converse to this proposition: if m and n are two points in the plane and if f : m → n is a homography, there exists a conic C through m and n such that Im(C)={D ∩ f(D) | D ∈ m } (see [Ber77]or[Sid93]).
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