Degenerated conic sections and classification

• 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

Components

We know that a conic section 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.
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.

Lines through (0,0,1)

Two lines trough (0,0,1) have an equation of the form ux + vy = 0 and u'x + v'y = 0. The conic section who is degenerated in these lines has the equation (ux + vy)(u'x + v'y) = 0.
This equation has the form

 
 a x2 + 2 b" x y + a' y2 = 0

Conversely, 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).

Lines through (0,1,0)

In the same way as above, you'll find:

 
The conic section is degenerated in two lines through (0,1,0)
 
 if and only if
 
The conic section has equation a x2 + 2 b' x z + a"z2 = 0

Lines through (1,0,0)

In the same way as above, you'll find:

 
The conic section is degenerated in two lines through (1,0,0)
 
 if and only if
 
The conic section has equation a'y2 + 2 b y z + a"z2 = 0

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

The lines have an equation of the form

 
 a x2 + 2 b" x y + a' y2 = 0
 
 
 m is the slope of a component
 
<=>
 (1,m,0) is on the conic section
<=>
 a + 2 b" m + a' m2 = 0

With this formula, you can calculate the slopes of the lines.
Remark:
If a' = 0, the equation of the conic section is

 
 a x2 + 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 x2 = 0
 
Then the two components are x = 0 and x = 0.

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

• a' is not 0
  The slopes of the lines are the roots of

· ·  
· ·  a + 2 b" m + a' m2 = 0

The nature of these roots depend on the sign of the discriminant D.

 
 D = 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.

The lines are orthogonal if and only if the product of the slopes = -1.
This gives the condition a + a' = 0.

• a' = 0
  The lines have the equation

· ·  
· ·  a x2 + 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.

The lines are orthogonal if and only if a = 0 <=> a + a' = 0.

Double points and simple points of a degenerated conic section

Each common point of the two components of a conic section is a double point. All the other points of the conic section are simple 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.

Theorem

If D(x_o,y_o,z_o) is a double point of a conic section, then

 
  Fx'(xo,yo,zo) = 0 and Fy'(xo,yo,zo) = 0 and Fz'(xo,yo,zo) = 0

Proof:
If D(x_o,y_o,z_o) is a double point, it is on both components of

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

So, u x_o + v y_o + w z_o = 0 and u'x_o + v'y_o + w'z_o = 0 and then

 
 Fx'(xo,yo,zo) = u(u'xo + v'yo + w'zo ) + u'( u xo + v yo + w zo )
 = u.0 + u'.0 = 0

Similarly for F_y'(x_o,y_o,z_o) = 0 and F_z'(x_o,y_o,z_o) = 0

Inverse theorem

 
If Fx'(xo,yo,zo) = 0 and Fy'(xo,yo,zo) = 0 and Fz'(xo,yo,zo) = 0

then D(x_o,y_o,z_o) is a double point of a conic section.

Proof:
From Euler's formula we have

 
 2 F(xo,yo,zo)
 
 = xo.Fx'(xo,yo,zo) + yo.Fy'(xo,yo,zo) + zo.Fz'(xo,yo,zo)
 
 = xo.0 + yo.0 + zo.0 = 0

Thus, D(x_o,y_o,z_o) is on the conic section.
Now, choose another point P(x₁,y₁,z₁) of the conic section.
A variable point of the line DP is ( kx_o + lx₁, ky_o + ly₁, kz_o + lz₁).
This variable point is permanently on the conic section because

 
F( kxo + lx1, kyo + ly1, kzo + lz1)
 
  = k2 F(xo,yo,zo)
 
 + kl(xo.Fx'(x1,y1,z1) + yo.Fy'(x1,y1,z1) + zo.Fz'(x1,y1,z1))
 
 + l2 F(x1,y1,z1)
 
 = k2 F(xo,yo,zo)
 
 + kl(x1.Fx'(xo,yo,zo) + y1.Fy'(xo,yo,zo) + z1.Fz'(xo,yo,zo))
 
 + l2 F(x1,y1,z1)
 
 = k2 .0 + kl.0 + l2 .0 = 0

The line DP is a component of the conic section and thus the conic section is degenerated.
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.

Formula to calculate the double points of a conic section

From previous theorems we conclude:

 
 D(xo,yo,zo) is double point of a conic section
 
  if and only if
 
 xo,yo,zo is a non-trivial solution of the system
 / Fx'(x,y,z) = 0
 |
 | Fy'(x,y,z) = 0
 |
 \ Fz'(x,y,z) = 0

Criterion for degenerated conic section

A conic section is degenerated if and only if DELTA = 0

Proof:

 
 The conic section F(x,y,z) = 0 is degenerated
 
<=>
 
 The conic section has a double point
 
<=>
 The system
 / Fx'(x,y,z) = 0
 |
  | Fy'(x,y,z) = 0
 |
 \ Fz'(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
 

Classification of the degenerated affine conic sections

 
 The ideal points are the solutions of
 
 
 /
 |a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2 = 0
 |
 \ z = 0
 
<=>
 /
 | a x2 + 2 b" x y + a' y2 = 0
 |
 \ z = 0

From above we know that

• 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.