In this chapter we consider real functions from R x
R x R to R. The image of a triplet (x,y,z) is a real
number. Example: f(x,y,z) = 3 x y - x z + 6 y -4
If
F(x,y,z) is a homogeneous
polynomial in x, y, and z then the corresponding function F(x,y,z) is a
homogeneous polynomial function and F(x,y,z) = 0 is a homogeneous polynomial
equation. Example:
F(x,y,z) = x z + y z - x2
If
(xo,yo,zo) is a solution (different from
(0,0,0)) of a homogeneous polynomial equation F(x,y,z) = 0, then
(r.xo,r.yo,r.zo) is a solution too. With all
this solutions corresponds exactly one point. The set of all points
P(xo,yo,zo) such that
F(xo,yo,zo) = 0, is called the graph of the
equation F(x,y,z) = 0.
Such graph is an algebraic curve. F(x,y,z) = 0 is the equation of the
algebraic curve. F(x,y,1) = 0 is the cartesian equation of the algebraic
curve.
We say a curve c is degenerated if and only if F(x,y,z) =
f(x,y,z) . g(x,y,z) with f(x,y,z) and g(x,y,z) homogeneous polynomial functions
in x,y and z. If c1 is the graph of f(x,y,z)=0 and c2 is the graph of
g(x,y,z)=0, then c is the union of c1 and c2. c1 and c2 are called the
components of the degenerated curve c.
Example:
F(x,y,z) = (x2 + y2 - 9) (x + y - 1)
The curve of F = 0 is a degenerated curve. The components are
a circle with equation (x2 + y2 - 9) = 0 and
a line with equation (x + y - 1) = 0
An
algebraic curve with an equation of the form
a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2= 0
is called a conic section.
In all following pages, we write this equation as F(x,y,z) = 0. The
coefficients are real numbers. The cartesian equation of the conic section is :
a x2 + 2 b" x y + a' y2 + 2 b' x + 2 b y + a" = 0
A conic section is affine if and only if the ideal line
is not a component of the conic section.
The three partial derivatives of the equation of a
conic section are:
Fx'(x,y,z) = 2 a x + 2 b" y + 2 b' z = 2 ( a x + b" y + b' z )
Fy'(x,y,z) = 2 b" x + 2 a' y + 2 b z = 2 ( b" x + a' y + b z )
Fz'(x,y,z) = 2 b' x + 2 b y + 2 a" z = 2 ( b' x + b y + a" z )
The matrix formed by half the coefficients of x, y and z is
[ a b" b']
C = [ b" a' b ]
[ b' b a"]
It is the symmetric cubic matrix of the conic section. The determinant of
this matrix is usually written as the Greek capital delta. Here we denote this
determinant as 'DELTA'.
| a b" b'|
DELTA = | b" a' b |
| b' b a"|
The matrix
[ a b"]
[ b" a']
is the quadratic matrix of the conic section. The determinant of this
matrix is usually written as the Greek delta. Here we denote this determinant as
'delta'.
| a b"|
delta = | b" a'| = a a' - b"2
The cofactors of the elements of the cubic matrix are denoted as
A, A', A", B, B', B"
In the following sections we denote :
[x] [x1] [x2]
P = [y] P1 = [y1] P2 = [y2]
[z] [z1] [z2]
T [ a b" b'] [x]
P C P = [x y z].[ b" a' b ].[y]
[ b' b a"] [z]
[ a x + b" y + b' z ]
= [x y z].[ b" x + a' y + b z ]
[ b' x + b y + a" z ]
= x(a x + b" y + b' z)+y(b" x + a' y + b z)+z(b' x + b y + a" z)
= a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2
So, the equation of a conic section is
PT C P = 0
It is
easy to prove that
F(x1,y1,z1) = P1T C P1
F(kx1,ky1,kz1) = (kP1)T C (kP1) = k (P1T C P1)
F(x1 + x2, y1 + y2, z1 + z2) = (P1 + P2)T C (P1 + P2)
[ a b" b'] [x1] [Fx'(x1,y1,z1)]
C.P1 = [ b" a' b ]. [y1] = (1/2).[Fy'(x1,y1,z1)]
[ b' b a"] [z1] [Fz'(x1,y1,z1)]
Theorem:
If F(x,y,z) = a x2 + 2 b" x y + a' y 2 + 2 b' x z + 2 b y z + a" z2
then
x1.Fx'(x2,y2,z2) + y1.Fy'(x2,y2,z2) + z1.Fz'(x2,y2,z2)
= x2.Fx'(x1,y1,z1) + y2.Fy'(x1,y1,z1) + z2.Fz'(x1,y1,z1)
proof:
x1.Fx'(x2,y2,z2) + y1.Fy'(x2,y2,z2) + z1.Fz'(x2,y2,z2)
[Fx'(x2,y2,z2)]
= [x1 y1 z1] [Fy'(x2,y2,z2)] = 2 P1T C P2
[Fz'(x2,y2,z2)]
and
x2.Fx'(x1,y1,z1) + y2.Fy'(x1,y1,z1) + z2.Fz'(x1,y1,z1)
[Fx'(x1,y1,z1)]
= [x2 y2 z2] [Fy'(x1,y1,z1)] = 2 P2T C P1
[Fz'(x1,y1,z1)]
But P1T C P2 is a number; and the transpose of a number is that number itself.
So,
P1T C P2 = (P1T C P2)T = P2T C P1
If F(x,y,z) = a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2
then
x1.Fx'(x1,y1,z1) + y1.Fy'(x1,y1,z1) + z1.Fz'(x1,y1,z1) = 2 F(x1,y1,z1)
Proof:
x1.Fx'(x1,y1,z1) + y1.Fy'(x1,y1,z1) + z1.Fz'(x1,y1,z1)
[Fx'(x1,y1,z1)]
= [x1 y1 z1] [Fy'(x1,y1,z1)]
[Fz'(x1,y1,z1)]
= [x1 y1 z1] .2 C P1 = 2 P1T C P1 = 2 F(x1,y1,z1)
If F(x,y,z) = a x + 2 b" x y + a' y + 2 b' x z + 2 b y z + a" z
then
F(kx1 + lx2, ky1 + ly2, kz1 + lz2)
= k2 F(x1,y1,z1)
+ kl(x1.Fx'(x2,y2,z2) + y1.Fy'(x2,y2,z2) + z1.Fz'(x2,y2,z2))
+ l2 F(x2,y2,z2)
Proof:
F(kx1 + lx2, ky1 + ly2, kz1 + lz2)
= (k P1 + l P2)T C (k P1 + l P2)
= (k P1T + l P2 ).(k C P1 + l C P2)
= k2 P1T C P1 + kl(P1T C P2 + P2T C P1) + l2 P2T C P2
= k2 F(x1,y1,z1)
+ kl(x1.Fx'(x2,y2,z2) + y1.Fy'(x2,y2,z2) + z1.Fz'(x2,y2,z2))
+ l2 F(x2,y2,z2)
We know that the transformation formulas are
[x']
P = M P' with M = the transformation matrix and P' = [y']
[z']
Then
F(x,y,z) = 0 (condition for old x,y,z)
<=>
PT C P = 0
<=> (P = M P')
P'T MT C M P' = 0 (condition for new x',y',z')
<=>
P'T C1 P' = 0 (New equation of a conic section)
So the connection between the old C and the new C1 is
C1 = MT C M
C1 = MT C M
We take the determinant of both sides
DELTA1 = determinant(MT ) DELTA determinant(M)
<=>
DELTA1 = DELTA . (determinant(M))2
The sign of
DELTA is invariant by a projective coordinate transformation.
It can be
proved that
delta1 = delta . (determinant(M))2
The sign of
delta is invariant by a affine coordinate transformation.
It can be proved that a + a' is
invariant for a metric transformation.
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