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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]). The set consisting of the proposition and its con- verse is often called the Chasles–Steiner theorem. The transformation m → n is shown in Figure 16. Corollary 4.3. Let C be a proper conic with nonempty image and let m1,...,m4 be four points of C.Thecross-ratio [mm1,mm2,mm3,mm4] does not depend on the choice of the point m on C. 4. The cross-ratio of four points on a conic and Pascal’s theorem 209 n a m n 1 m a Fig. 16 Fig. 17 This cross-ratio is called the cross-ratio of the four points m1,...,m4 on the conic C. Proof of the proposition.Toavoidusing technical results on homographies, let us use coordinates. All we have to do is to choose these cleverly. We choose the line mn as the line at infinity (assuming m = n,otherwisethere is nothing to prove). In the remaining affine plane, the conic C is a hyperbola. We choose the origin at its center and the two basis vectors as directing vectors of the asymptotes, scaling them so that the equation of the affine conic is xy =1.Thepointsm and n arethe points at infinity of the x-and y-axes respectively, so that the pencil m consists of the lines y = a parallel to the x-axis and the pencil n of the lines x = b. ◦ −1 Thus the mapping πn πm is (see Figure 17) K ∪{∞}= m −−−−→ n = K ∪{∞} 1 a −−−−→ a aprojective transformation indeed. We are now going to prove Pascal’s theorem, special cases of which we have already encountered (see Exercise III.51). Pascal has proved his theorem in acircle(the general case of conics being a consequence of the special case of circles) probably using Menelaus’¨ theorem (Exercise I.37). Theorem 4.4 (Pascal’s theorem). Let C be (the image of) a proper conic and let a, b, c, d, e and f be six points of C.Thenthe intersection points of ab and de, bc and ef, cd and fa are collinear. Proof.Letusgivenames to some of the intersection points: x = bc ∩ ed y = cd ∩ ef ∩ z = ab de t = af ∩ cd 210 Chapter VI. Conics and Quadrics b e z x d y a t f c Fig. 18. The “mystic hexagram” of Pascal (see Figure 18). We want to prove that the intersection point of the lines bc and ef lies on the line zt.Todothis,letuscalculate the cross-ratio [z,x,d,e] (these four points are collinear on the line ed): [z,x,d,e]=[bz, bx, bd, be] =[ba, bc, bd, be] (these are other names for the same lines) =[fa,fc,fd,fe](because f is another point of C) =[t, c, d, y](usingthe secant cd). We thus have the equality [z,x,d,e]=[t, c, d, y]. Let m be the intersection point of zt and bc.Theperspectivity(13) of center m (from the line ed to the line cd)mapsz to t, x to c, d to d and thus, as it preserves the cross-ratio, it must map e to y,hence e, m and y arecollinear. Therefore m,whichbelongs to bc and zt,alsobelongs to ef... andthisis what we wanted to prove. 5. Affine quadrics, via projective geometry Let us consider now the real projective plane as the completion of some affine plane. We have said that any proper projective conic defines an affine conic. We have also seen that there is only one type of (nonempty) real proper conic over R and over C.Thequestionof the affine classification is thus reduced to the question of the relative positions of the conic and the line at infinity. We have already noticed that the intersection of the conic and the line at infinity is the set of isotropic vectors of the dehomogenized quadratic form q. Figure 19 shows the affine conics, from top to bottom: (13)See if necessary Exercise V.12. 5. Affine quadrics, via projective geometry 211 – in the projective completion of the affine plane P,together with the line at infinity, – in the affine plane P itself, – in the 3-dimensional vector space P × R,wheretheaffineplaneisthe z =1-plane. ellipse parabola hyperbola Fig. 19 For those of our readers who prefer equations to very clear pictures as those shown here, here are some equations. Let us choose the line of equation z =0as line at infinity and describe the conic by a homogeneous equation Q(x, y, z)=0forsomequadratic form Q of signature (2, 1). There are three possibilities: – The conic does not intersect the line at infinity, it is thus contained in the affine plane, this is an ellipse in this plane. In equations: the conic has an equation x2 + y2 − z2 =0,its affine part is the ellipse of equation x2 + y2 =1. – The conic is tangent to the line at infinity, its affine part is a parabola. In equations: take x2 − yz =0fortheconic; an equation of the affine part is y = x2. – The conic intersects the line at infinity at two points; its affine part is a hyperbola, the asymptotes are the tangents at the two points at infinity. In equations: x2 − y2 − z2 =0fortheprojective conic, x2 − y2 =1foritsaffine part. What we have just described can be rephrased, in a more pedantic way, as “two proper affine conics are (affinely) equivalent if and only if their points 212 Chapter VI. Conics and Quadrics at infinity are two projectively equivalent 0-dimensional quadrics”. We have used both the fact that there is only onetypeofprojectiveconicsandthe affine classification: we knew, e.g., that the affine conicswithnopoint at infinity were all affinely equivalent. but the affine classification can also be deduced from the projective classification, as I show now, in any dimension. Proposition 5.1. To give the affine type of a proper affine quadric is equi- valent to giving the types of the projectivized quadric and of its intersection with the hyperplane at infinity. Proof.LetC and C be two proper quadrics in anaffinespaceE.Wechoose an origin O,onC to make life easier. We will use, as ever, the vector space E ⊕K in which E is the affinehyperplanez =1andO is the point (0, 1). The projective completion of E is P (E ⊕ K), the hyperplane at infinity is P (E). −−→ −−→ An equation for C is f(m)=q(OM)+L(OM)orf(u, 1) = q(u)+L(u). An equation for C is f (u, 1) = q(u)+L(u)+c.Theprojective quadrics under consideration are the completions Q(u, z)=q(u)+zL(u),Q(u, z)=q(u)+zL(u)+z2c in P (E ⊕ K)and the quadrics at infinity q(u), q(u)inP (E). Let us assume first that the two affine quadrics are equivalent. That is, there is an affine transformation ψ : E −−−−→ E such that f ◦ ψ(M)=λf(M) (for some nonzero scalar λ). In our notation, −→ −−−−→ ψ(u, 1) = (v + ψ (u), 1) where v = Oψ(O) so that the condition is equivalent to −→ −→ −→ q◦ ψ (u)=λq(u),L◦ ψ (u)+2ϕ(v, ψ (u)) = λL(u),q(v)+L(v)+c =0. The first equation gives the equivalence of the quadrics at infinity, the last one just says that ψ(O) ∈ C. Recall (from Proposition V-5.7) that ψ extends to a linear isomorphism Ψ:E ⊕ K −−−−→ E ⊕ K −→ (u, z) −−−−→ (zv + ψ (u),z). Hence −→ −→ Q(Ψ(u, z)) = q(zv + ψ (u)) + zL(zv + ψ (u)) + z2c −→ −→ −→ = q( ψ (u)) + z(L( ψ (u)) + 2ϕ(v, ψ (u))) + z2(q(v)+L(v)+c). Eventually, Q ◦ Ψ(u, z)=λQ(u, z), so that the two projective quadrics are indeed equivalent. 5. Affine quadrics, via projective geometry 213 Conversely, assume that the projective quadrics are equivalent and the quadrics at infinity too.
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