Notes on Advanced Geometry Dr. John Sarli Approximate Schedule of Topics: Week 1 Affi ne transformations and conics Week 2 Introduction to projective space Week 3 Projective transformations; Fundamental Theorem of Projective Geometry Week 4 Theorems of Desargues, Pappus, Brianchon; Cross-ratio Week 5 Projective conics; tangents and secants; midterm exam Week 6 Affi ne conics revisited Week 7 Poles and polars; LaHire’sTheorem Week 8 Standard forms; determination of conics Week 9 Theorems on tangents and secants Week 10 Pascal’sTheorem, its dual and converse 4 Conics Defined by Collineations Let T be a collineation of the Euclidean plane. A basic theorem tells us that T is an affi ne transformation, which means it can be represented as a transformation of R2 in the form a b x p T (x, y) = + , ad bc = 0 c d y q 6 The condition that = ad bc = 0 ensures that T is invertible, as every trans- formation must be. In fact, 6 1 1 a b x p T (x, y) = c d y q 1 1 = (dx by + bq dp) , (ay cx aq + cp) Let P be any point and consider the pencil of lines L concurrent at P . We define the conic at P afforded by T to be (T,P ) = L T (L): P L E f \ 2 g To understand this definition we should make the connection with the familiar description of conics by Cartesian curves. The fundamental principle that applies is as follows: Suppose is a figure in the plane described by a Cartesian equa- tion f(x, y) =F 0. For any transformation T the image figure T ( ) is 1 F described by the Cartesian equation f (T (x, y)) = 0. Now suppose, for example,that the point P in the definition is O = (0, 0), the origin of the Cartesian plane. Any line L through O is a figure we want to transform by T . Since L is either the line x = 0 or is described by a Cartesian equation of the form y mx = 0 5 the image line T (L) is either 1 (dx by + bq dp) = 0 dx by = dp bq or is described by (ay cx aq + cp) = m (dx by + bq dp) ay cx aq + cp m = dx by + bq dp For any given m we can find the intersection of L with T (L) and the collection of all such intersection points must satisfy ay cx aq + cp y = mx = x dx by + bq dp and so = (T,O) is the Cartesian curve E E cx2 + (d a) xy by2 = (cp aq)x + (dp bq)y We say is an ellipse if > 0, a parabola if = 0, a hyperbola if < 0, whereE = 4 2 = a + d This conic might be degenerate. For example, if 1 1 x 1 T (x, y) = + 0 1 y 0 then is the curve y2 y = 0, which consists of the the two parallel lines y = 0 and yE= 1. Since = 0, this is an example of a degenerate parabola. First Graded Assignment 6 We say that T preserves orientation if > 0 and reverses orientation if < 0. If = 1 then T is an isometry (preserves distance between points). For any isometry± there is an angle such that a b cos sin = , if = 1 c d sin cos cos sin = , if = 1 sin cos a) If T reverses orientation show that is a hyperbola. E b) If T preserves orientation then can be any affi ne type. Find a transformation T for each affi ne type and sketch theE resulting conic. c) If T is an isometry that reverses orientation show that is a rectangular hyperbola (asymptotes are perpendicular, so eccentricity is p2E). Then provide a non-degenerate example and sketch the hyperbola with its asymptotes. d) If T is an isometry that preserves orientation show that is a circle unless = 0 (T is a translation) or = (T is a half-turn). Find theE center and radius of the circle in the general case. What happens if T is a translation or half-turn? e) Find a collineation T such that the conic E 4x2 + 4xy + y2 + x + y = 0 is (T,O). Sketch the conic. Let L be the line y = x. Show where L and T (L) intersectE on . E 7 Degenerate Affi ne Conics Degenerate affi ne conics are described qualitatively by examining the extreme cases of a plane intersecting a cone. For example, if the plane intersects the cone only at its vertex we obtain a single point. Since such an intersection is the limiting case of a plane that intersects the cone in an ellipse, a single point could be considered a degenerate ellipse. By similar reasoning, a single line could be a degenerate parabola since a plane tangent to the cone is the limiting case of a plane parallel to a generator of the cone; and a pair of intersecting lines could be a degenerate hyperbola, the limiting case of a plane that intersects both branches of the cone. What about a pair of parallel lines? To understand how this and the other cases occur we need to describe affi ne conics algebraically. We know from algebraic geometry that any affi ne conic satisfies a Cartesian rela- tion of the form Ax2 + Bxy + Cy2 + Dx + Ey + K = 0 The set of points (x, y) that satisfy this relation may be empty, for example, x2 + y2 + 1 = 0. We will assume that the conic consists of at least one point. Then, without loss of generality we can assume K = 0, by applying a translation if necessary. Working with the equation F (x, y) = Ax2 + Bxy + Cy2 + Dx + Ey = 0 where we assume not every coeffi cient is zero, we have at least the point (0, 0) on the conic. When will (0, 0) be the only point on the conic? Writing (Ax + D) x + (Cy + E) y = Bxy we see that if Ax + D = 0 has a non- zero solution x0 then (x0, 0) is on the conic; similarly, (0, y0) is on the conic if Cy +D = 0 has a non-zero solution y0. If Ax+D = 0 has only x = 0 as a solution then D = 0, whereby A = 0. Finally, if Ax + D = 0 has no solution then A = 0 and D = 0; but then we6 can solve for x as a rational function of y and obtain 6 8 infinitely many points on the conic. Applying a similar analysis to the solutions of Cy + E = 0 we conclude that (0, 0) must be the only solution to Ax2 + Bxy + Cy2 = 0 in other words, Ax2 + Bxy + Cy2 is a definite quadratic form. Theorem. The conic F (x, y) = 0 consists of a single point provided D = E = 0 and 4AC > B2 Corollary. A single point is a degenerate ellipse. What other degenerate geometric figures are possible? First, if A = B = C = 0 then Dx + Ey = 0 is a single line, which we would interpret to be a degenerate parabola since 4AC = B2. Otherwise, if the equation represents a degenerate form that is not a single point then F (x, y) must factor in one of the following ways: i) the square of a linear polynomial; ii) the product of distinct linear polynomials. Case i) again gives us a single line. Case ii) gives us a pair of lines but there are two possibilities: The two lines could be parallel or they could intersect. So there are three possibilities which we need to analyze to determine which type of conic each is a degenerate form of. Case i): F (x, y) = ( x + y)2, for some choice of , not both 0. Then A = 2, B = 2 ±, C = 2 (and D = E = 0) so 4AC = B2, which is consistent± with F±(x, y) = 0 being± a degenerate parabola. Case iia): F (x, y) = ( x + y)( x + y + ), for some choice of , not both 0 and = 0. This± is a pair of parallel lines, and expanding the product we see that the6 triple (A, B, C) is a non-zero multiple of (D2, 2DE, E2) with at least one of D, E non-zero. Again we have 4AC = B2, which is consistent with F (x, y) = 0 being a degenerate parabola, though we cannot obtain this figure as the intersection of a cone and a plane if the vertex of the cone is in R3. Note also that AE2 BDE + CD2 = 0 9 Exercise. If the cross product of Ai + Bj + Ck and D2i + 2DEj + E2k is 0 show that AE2 BDE + CD2 = 0. Case iib): F (x, y) = ( x + y)(x + y + ), which will be a pair of intersecting lines provided = . Expanding the product this time we find 6 4AC B2 = ( )2 < 0 AE2 BDE + CD2 = 0 In particular, a pair of intersecting lines is a degenerate hyperbola. Conversely, these two conditions imply that the conic is a pair of intersecting lines since they allow us to factor F (x, y). Apparently, the condition AE2 BDE + CD2 = 0 is equivalent to degeneracy for the conic F (x, y) = 0. This suggests that this cubic form in the coeffi cients has an invariant representation. Let 2ABD Q = B 2CE 0 DE 0 1 @ A Then F (x, y) = 0 can be written x x y 1 Q y = 0 0 1 1 @ A We say that the symmetric matrix Q represents the affi ne conic. Further, since det Q = 2 (AE2 BDE + CD2) we have: Theorem. The conic F (x, y) = 0 is degenerate if and only if det Q = 0.