This is a sequel to ProjectiveCoordinates.html handling the description of conics in a homogeneous or projective coordinate system. Projective coordinate systems are used to represent algebraic curves. In particular lines, defined by homogeneous equations of degree 1: ax1 + bx2 + cx3 = 0, and conics, defined by homogeneous equations of degree two: (0) ax12 + bx22 + cx32 + 2dx1x2 + 2ex2x3 + 2fx3x1 = 0.
If the conic passes through the basis points of the system, then {(1,0,0),(0,1,0),(0,0,1)} must satisfy the equation. This implies a = b = c = 0, hence the equation obtains the form: (1) dx1x2 + ex2x3 + fx3x1 = 0. If in addition the conic passes through the fourth (unit) point D calibrating the coordinate system, then the coefficients must satisfy: (2) d+e+f = 0. Replacing f = -(d+e) into the equation, we find that the conic is represented by an equation of the form: (3) d x1(x2 - x3) + e x3(x2 - x1) = 0. This means that every conic passing through the four points is given as a linear combination of the two reducible conics (i.e. conics that can be split into a product of lines): (4) x1(x2 - x3) = 0, and x3(x2 - x1) = 0. The first represents the degenerate conic which is the product of two lines: (5) BC (x1 = 0) and AD (x2 - x3 = 0). The second represents similarly the conic which is the product of: (6) AB(x3 = 0) and CD(x2 - x1 = 0). Notice that (d,e) are defined modulo a non zero multiplicative constant, so that they determine a one parameter pencil or family of conics. This implies that a fifth point E completely determines the conic passing through all five points {A,B,C,D,E}.
Another interesting case is that of the conic passing through two points B, C and having there tangents intersecting at a point A. Selecting a coordinate system as shown we have: (i) B, C lying on the conic ==> b = c = 0. (ii) AC resp. AB tangent at C resp. B ==> d = 0 resp. e = 0, see the remark in last section. Thus the conic reduces to the form: (7) ax12 + fx2x3 = 0. Thus it belongs to the so-called bitangent pencil or family of conics, represented by linear combinations of two degenerate conics. The first conic consists of the product of lines: (8) x2x3 = 0 (representing lines AB and AC). The second conic consists of the double line: (9) x12 = 0 (representing BC in product with itself). In particular, if D belongs to the conic, then a + f = 0, and the conic is represented by: (10) x12 = x2x3. This shows that all non-degenerate conics of the projective plane are equivalent. This because for every such conic we can select a system of coordinates representing the conic through an equation like the one in (10). The transformation of projective coordinates which matches two such coordinate systems, matches also the corresponding conics expressed in the form of (10).
Another interesting case is that of the conic referred to a self-polar triangle ABC. This means that each vertex is the pole of the opposite side with respect to that conic. Taking a projective basis {A,B,C,D} as shown (see ProjectiveBasis.html ), the polar of A being BC, implies d = f = 0. Similarly the polar of B being AC implies d = e = 0. Thus the conic has the form: (11) ax12 + bx22 + cx32 = 0.
Remark The argument applied here is that the conic being given by an equation of the (matrix) form:
The polar of a point (u1,u2,u3) is given by the equation:
Thus, if BC (x1 = 0) is the polar of A (1,0,0) which is ax1+dx2+fx3=0, we must have d = f = 0. Analogous arguments have been applied to the case of bitangent conic, to reduce it to the form x12 = x2x3. The representation of the polar in this way is proved for the case of cartesian coordinates in the file Conic_Equation.html . The proof for the case of homogeneous coordinates is essentially the same.