- Components

- Lines through (0,0,1)

- Lines through (0,1,0)

- Lines through (1,0,0)

- Direction of two lines through (0,0,1)

- Nature of the components of two lines through (0,0,1)

- Double points and simple points of a degenerated conic section

- Theorem

- Inverse theorem

- Formula to calculate the double points of a conic section

- Criterion for degenerated conic section

- Classification of the degenerated affine conic sections

Since F(x,y,z) is homogeneous with a degree = 2 , f(x,y,z) and g(x,y,z) are homogeneous with a degree = 1. From this we see that a degenerated conic section has 2 lines as components.

This equation has the form

a xConversely, every equation of the last form can be written as (ux + vy)(u'x + v'y) = 0 and is a degenerated conic section through the origin (0,0,1). The lines can be imaginary lines.^{2}+ 2 b" x y + a' y^{2}= 0

Example :

9 x^{2}+ 7 y^{2}- 24 x y = 0 <=> (3 x - y) (3 x - 7 y) = 0

The conic section is degenerated in two lines through (0,1,0) if and only if The conic section has equation a x^{2}+ 2 b' x z + a"z^{2}= 0

The conic section is degenerated in two lines through (1,0,0) if and only if The conic section has equation a'y^{2}+ 2 b y z + a"z^{2}= 0

a xWith this formula, you can calculate the slopes of the lines.^{2}+ 2 b" x y + a' y^{2}= 0 m is the slope of a component <=> (1,m,0) is on the conic section <=> a + 2 b" m + a' m^{2}= 0

Remark:

If a' = 0, the equation of the conic section is

a x^{2}+ 2 b" x y = 0 Then the two components are x = 0 and a x + 2 b" y = 0 If a' = b" = 0, the equation of the conic section is a x^{2}= 0 Then the two components are x = 0 and x = 0.

- a' is not 0

The slopes of the lines are the roots ofa + 2 b" m + a' m

The nature of these roots depend on the sign of the discriminant D.^{2}= 0D = 4 b"

^{2}- 4 a a' = 4(b"^{2}- a a') = -4(a a' - b"^{2})= -4. delta- delta > 0

Two conjugate imaginary lines and therefore two conjugate imaginary ideal points. - delta = 0

Two coinciding regular lines and therefore two coinciding regular ideal points. - delta < 0

Two different regular lines and therefore two different regular ideal points.

This gives the condition a + a' = 0. - delta > 0
- a' = 0

The lines have the equationa x

^{2}+ 2 b" x y = 0 <=> x(a x + 2 b" y) = 0 and delta = - b"^{2}- delta > 0 can not occur
- delta = 0

Two coinciding regular lines and therefore two coinciding regular ideal points. - delta < 0

Two different regular lines and therefore two different regular ideal points.

Properties:

- A conic section is degenerated <=> a conic section has a double point.
- If the components are different, the double point is unique.
- If the components are parallel, the double point is ideal.
- if the components are coinciding, all points are double points.

FProof:_{x}'(x_{o},y_{o},z_{o}) = 0 and F_{y}'(x_{o},y_{o},z_{o}) = 0 and F_{z}'(x_{o},y_{o},z_{o}) = 0

If D(x

F(x,y,z) = (ux + vy + wz)(u'x + v'y + w'z) = 0So, u x

FSimilarly for F_{x}'(x_{o},y_{o},z_{o}) = u(u'x_{o}+ v'y_{o}+ w'z_{o}) + u'( u x_{o}+ v y_{o}+ w z_{o}) = u.0 + u'.0 = 0

If Fthen D(x_{x}'(x_{o},y_{o},z_{o}) = 0 and F_{y}'(x_{o},y_{o},z_{o}) = 0 and F_{z}'(x_{o},y_{o},z_{o}) = 0

Proof:

From Euler's formula we have

2 F(xThus, D(x_{o},y_{o},z_{o}) = x_{o}.F_{x}'(x_{o},y_{o},z_{o}) + y_{o}.F_{y}'(x_{o},y_{o},z_{o}) + z_{o}.F_{z}'(x_{o},y_{o},z_{o}) = x_{o}.0 + y_{o}.0 + z_{o}.0 = 0

Now, choose another point P(x

A variable point of the line DP is ( kx

This variable point is permanently on the conic section because

F( kxThe line DP is a component of the conic section and thus the conic section is degenerated._{o}+ lx_{1}, ky_{o}+ ly_{1}, kz_{o}+ lz_{1}) = k^{2}F(x_{o},y_{o},z_{o}) + kl(x_{o}.F_{x}'(x_{1},y_{1},z_{1}) + y_{o}.F_{y}'(x_{1},y_{1},z_{1}) + z_{o}.F_{z}'(x_{1},y_{1},z_{1})) + l^{2}F(x_{1},y_{1},z_{1}) = k^{2}F(x_{o},y_{o},z_{o}) + kl(x_{1}.F_{x}'(x_{o},y_{o},z_{o}) + y_{1}.F_{y}'(x_{o},y_{o},z_{o}) + z_{1}.F_{z}'(x_{o},y_{o},z_{o})) + l^{2}F(x_{1},y_{1},z_{1}) = k^{2}.0 + kl.0 + l^{2}.0 = 0

If the conic section contains another point P' not on DP then DP' is a component of the conic section and then D is double point.

If P' does not exist, the conic section is degenerated in two coinciding lines and D is double point.

D(x_{o},y_{o},z_{o}) is double point of a conic section if and only if x_{o},y_{o},z_{o}is a non-trivial solution of the system / F_{x}'(x,y,z) = 0 | | F_{y}'(x,y,z) = 0 | \ F_{z}'(x,y,z) = 0

Proof:

The conic section F(x,y,z) = 0 is degenerated <=> The conic section has a double point <=> The system / F_{x}'(x,y,z) = 0 | | F_{y}'(x,y,z) = 0 | \ F_{z}'(x,y,z) = 0 has a solution different from (0,0,0) <=> The system / 2 ( a x + b" y + b' z ) = 0 | | 2 ( b" x + a' y + b z ) = 0 | \ 2 ( b' x + b y + a" z ) = 0 has a solution different from (0,0,0) <=> DELTA = 0

The ideal points are the solutions of / |a xFrom above we know that^{2}+ 2 b" x y + a' y^{2}+ 2 b' x z + 2 b y z + a" z^{2}= 0 | \ z = 0 <=> / | a x^{2}+ 2 b" x y + a' y^{2}= 0 | \ z = 0

- delta > 0

Two conjugate imaginary ideal points.

Two conjugate imaginary components.

We say that the conic section is a degenerated ellipse. - delta = 0

Two coinciding ideal points.

Two coinciding components.

We say that the conic section is a degenerated parabola. - delta < 0

Two different simple ideal points.

The components are two different real lines.

We say that the conic section is a degenerated hyperbola.

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