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

 
        F_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:
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

 
        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

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

Inverse theorem

 
If F_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

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(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

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( kx_o + lx₁, ky_o + ly₁, kz_o + lz₁)

        = k² F(x_o,y_o,z_o)

          + kl(x_o.F_x'(x₁,y₁,z₁) + y_o.F_y'(x₁,y₁,z₁) + z_o.F_z'(x₁,y₁,z₁))

          + l² F(x₁,y₁,z₁)

        = k² F(x_o,y_o,z_o)

          + kl(x₁.F_x'(x_o,y_o,z_o) + y₁.F_y'(x_o,y_o,z_o) + z₁.F_z'(x_o,y_o,z_o))

          + 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(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

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
                /  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

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.

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