## Lecture Notes 19: Projective Conics

Throughout this section we will be working in the Desarguesian projective plane PG(2,F) with F a field.

After studying points and lines in a plane, the next type of structure to look at are those of a quadratic nature (circles, ellipses, parabolas, etc.). The term quadratic refers to the fact that the coordinates of the points in these structures satisfy quadratic equations.

A quadratic form Q on a vector space V defined over the field F is a map Q: V -> F which satisfies Q(kv) = k2Q(v), for all vectors v in V and scalars k in F and Q(v+w) = Q(v) + Q(w) + b(v,w), where b(v,w) is a bilinear form. In terms of coordinates, if v = (x,y,z) then the most general quadratic form is:

Q(v) = Ax2 + Bxy + Cy2 + Dxz + Eyz + Fz2.
A quadric in PG(2,F) is the set of points whose coordinates satisfy the equation Q(v) = 0 for some quadratic form Q.

Examples: Consider PG(2,5), and the quadratic form Q(x,y,z) = x2 + y2 + z2. The quadric determined by this quadratic form consists of the points with coordinates (1,2,0), (1,3,0), (2,0,1), (0,2,1), (3,0,1), and (0,3,1).
In the same plane, the quadratic form Q(x,y,z) = x2 - z2 gives the quadric with points (0,1,0), (1,0,1),(1,1,1),(1,2,1),(1,3,1),(1,4,1),(4,0,1),(4,1,1),(4,2,1),(4,3,1), and (4,4,1). Note that these points lie on the two lines with homogeneous equations x + z = 0 and x - z = 0 (or with non-homogeneous equations x = 1 and x = 4).
Again in this plane, the quadratic form Q(x,y,z) = x2 - 3z2 gives the quadric whose only point is (0,1,0).

There are 5 possible types of point sets for a quadric in a projective plane. The first four types are known as degenerate quadrics. A degenerate quadric may be a pair of lines, a single line (repeated twice), a point or the empty set (but this last only in infinite planes). A non-degenerate quadric is called a conic. In the above examples, the first is a conic and the others are degenerate quadrics. [The term degenerate conic used by the text will not be tolerated in my classes!]

It may be shown that no three of the points of a conic lie on a line and through each point of the conic there passes exactly one tangent line (a line which meets the conic in only one point). In general, these geometric conditions can be satisfied by sets of points which are not conics as well. Also, if the projective plane is finite, i.e., PG(2,q), then we have that a conic consists of q+1 points.

We will now examine what happens to a conic when we pass to the Desarguesian affine plane AG(2,F) obtained by removing the points of the line z = 0. First, consider the intersection of the line z = 0 with the general quadric Ax2 + Bxy + Cy2 + Dxz + Eyz + Fz2 = 0. We must determine the solutions of

Ax2 + Bxy + Cy2 = 0.
Note that the point (0,1,0) will satisfy this if and only if C = 0. All other points on z = 0 have the form (1,m,0) and upon substitution we have the quadratic equation
Cm2 + Bm + A = 0.
If C = 0 and B is not zero, then there are two points (0,1,0) and (1,-A/B,0) in the intersection. If C = 0 and B = 0, then only the point (0,1,0) is in the intersection (and A must be 0). If C is not 0 and the field does not have characteristic 2, we may use the familiar quadratic formula to find the solutions of this quadratic equation. By considering the discriminant, B2 - 4AC, we can determine the number of solutions. If B2 - 4AC is not a square, then the equation has no solutions in the field, so the intersection is empty and the conic is contained entirely in the affine plane. In this case we call the conic an ellipse. If B2-4AC = 0 then there is one solution and the points of the conic in the affine plane form a parabola. Finally, if B2-4AC is a square, there will be two solutions and the points of the conic in the affine plane form a hyperbola. When the field has characteristic 2 we can not use the quadratic formula, but alternate means of solving the quadratic equation exist and we get results that are the same as those in odd characteristic fields.

Example: In our example of a conic in PG(2,5) we see that there are two points on the line z = 0, so the remaining 4 points form a hyperbola in AG(2,5). This can also be seen without determining all the points of the conic. From the quadratic form in that example we see that B = 0, A = C = 1. Thus, the discriminant is 0 -4(1)(1) = -4 = 1 in Z5 which is a square.

It is easy to see what the degenerate quadrics will be in an affine plane. If in the projective plane the degenerate quadric is a point, then in the affine plane we either get a point or the empty set (when the point is on the line z = 0). If the degenerate quadric is a repeated line, we get either a repeated line or the empty set (if the repeated line is z = 0). Finally, if the degenerate quadric is a pair of lines we get either a pair of lines, a pair of parallel lines (when the point of intersection is on z = 0) or a single line (when one of the pair is the line z = 0).