Pole and polar line




Polar line of a point A relative to a conic section

  1. Point A is a simple point.
    The polar line of A(x1,y1,z1) relative to a conic section with equation F(x,y,z) = 0 is the line
     
            x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z) = 0
    <=>
            x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1) = 0
    
    with matrix notation:
                            [Fx' (x1,y1,z1)]
    <=>        [x   y   z ] [Fy' (x1,y1,z1)] = 0
                            [Fz' (x1,y1,z1)]
    
    <=>
                    PT C P1 = 0
    
    <=>
                    P1T C P = 0
    
    
  2. Point A is a double point.
    Each line is a polar line of A.

Corollaries

Polar line and harmonic conjugate points

Point A(x1,y1,z1) is not on conic section F(x,y,z) = 0.
Theorem:
The polar line of point A is the set of all points B, such that A and B are harmonic conjugate points relative to the intersection points of the conic section and a variable line through A.

Proof:
A(x1,y1,z1) is not on conic section F(x,y,z) = 0.
Say B(x,y,z) is an arbitrary point.
A variable point P of the line AB has coordinates

 
        (x + h x1, y + h y1, z + h z1)

        Point P is on the conic section
<=>
        F(x + h x1, y + h y1, z + h z1) = 0
<=>
        F(x,y,z)

        + h (x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1))

        + h2  F(x1,y1,z1)  = 0
This is a quadratic equation in h. The roots h1 and h2 correspond with the intersection points of the conic section and AB.

Points A and B are harmonic conjugate points relative to the intersection points P1 and P2 of the conic section and the variable line AB

 
<=>
        (P1,P2,A,B) = -1
<=>
        (A,B,P1,P2) = -1
<=>
        h1 = - h2
<=>
        h1 + h2 = 0
<=>
        x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1) = 0
<=>
        B is element of the stated set
We see that the equation of the polar line and the equation of the stated set are the same.

If a point is on its polar line then it is on the conic section

Proof:
If that point is a double point then it is trivial.

If that point A(x1,y1,z1) is a simple point then:

 
        A(x1,y1,z1) is on its polar line
<=>
        x1.Fx' (x1,y1,z1) + y1.Fy' (x1,y1,z1) + z1.Fz' (x1,y1,z1) = 0
<=>
        2.F(x,y,z) = 0
<=>
        point A(x1,y1,z1) is on the conic section

Polar line and a degenerated conic section

The polar line of a point A, different from a double point, contains each double point.

Proof:
Take a point A(x1,y1,z1), different from a double point.
The polar line of A is x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z) = 0 Say D(xo,yo,zo) is a double point.
We investigate if D is on the polar line.

 
        x1.Fx' (xo,yo,zo) + y1.Fy' (xo,yo,zo) + z1.Fz' (xo,yo,zo)

      = x1 . 0 + y1 . 0 + z1 . 0

      = 0
Thus, D is on the polar line.

If A is on a polar line of B, then B is on a polar line of A.

proof:
If A or B is a double point, then it is trivial.
Now, suppose that A and B are not a double point.
 
        A(x1,y1,z1) is on the polar line of B(x2,y2,z2)
<=>
        x1.Fx' (x2,y2,z2) + y1.Fy' (x2,y2,z2) + z1.Fz' (x2,y2,z2) = 0
<=>
        x2.Fx' (x1,y1,z1) + y2.Fy' (x1,y1,z1) + z2.Fz' (x1,y1,z1) = 0
<=>
        B(x2,y2,z2) is on the polar line of A(x1,y1,z1)

Pole of a line relative to a conic section

 
        Point A is a pole of line a relative to a conic section
<=>
        Line a is a polar line of A.

Each line has exactly one pole relative to a non-degenerated conic section

Proof:
We use matrix notation.
 
The conic section has equation  PT C P = 0

The line a has equation
        u x + v y + w z = 0
<=>
                  [x]
        [u  v  w] [y] = 0
                  [z]

<=>
        U.P = 0         with U = [u  v  w]



         [x1]
Say P1 = [y1]  are the coordinates of  a possible pole of line a.
         [z1]

The polar line  is      P1T C P = 0

This polar line is also the line U.P = 0

From this we have :     k U =   P1T C

<=>                     P1T = k U C-1

From that last result, we see that there is exactly one pole of the line a.
The last formula gives a method to calculate the pole.

Calculation of a pole

 
Take the conic section  y2  - 2 x = 0 and the line x + y + 1 = 0
We'll show three methods to calculate the pole of this line.

Conjugated points

 
        The points A and B are conjugated relative to a conic section
<=>
        A is on a polar line of point B
<=>
        B is on a polar line of point A.
Remark :
A double point is conjugated to any point.

Criterion for conjugated points

  1. Points A and B aren't double points
     
            A(x1,y1,z1) and B(x2,y2,z2) are conjugated points
    <=>
            A is on a polar line of point B
    <=>
            x1.Fx' (x2,y2,z2) + y1.Fy' (x2,y2,z2) + z1.Fz' (x2,y2,z2) = 0
    
  2. A is a double point
     
            A(x1,y1,z1) and B(x2,y2,z2) are conjugated points
    <=>
            B is an arbitrary point
    <=>
            x2.0 + y2.0 + z2.0 = 0
    <=>
            x2.Fx' (x1,y1,z1) + y2.Fy' (x1,y1,z1) + z2.Fz' (x1,y1,z1) = 0
    <=>
            x1.Fx' (x2,y2,z2) + y1.Fy' (x2,y2,z2) + z1.Fz' (x2,y2,z2) = 0
    
  3. A is a double point
    Analogous as in the previous case.
Conclusion :
 
        A(x1,y1,z1) and B(x2,y2,z2) are conjugated points
<=>
        x2.Fx' (x1,y1,z1) + y2.Fy' (x1,y1,z1) + z2.Fz' (x1,y1,z1) = 0
<=>
        x1.Fx' (x2,y2,z2) + y1.Fy' (x2,y2,z2) + z1.Fz' (x2,y2,z2) = 0

Conjugated lines

 
        The lines a and b are conjugated relative to a conic section
<=>
        a contains each pole of line b AND
        b contains each pole of line a
Remark :
If the conic section is not degenerated, we can write
 
        The lines a and b are conjugated relative to a conic section
<=>
        a contains the  pole of b
<=>
        b contains the  pole of a

Polar transformation relative to a non-degenerated conic section

Say V is the set of all points and all lines.
Now, we take the transformation of V, who sends each point to its polar line and each line to its pole.
This is a polar transformation relative to the non-degenerated conic section.

Remark :
Each ordered quartet of concurrent lines a,b,c,d is transformed in an ordered quartet of collinear points A,B,C,D.
Each ordered quartet of concurrent points A,B,C,D is transformed in an ordered quartet of collinear lines a,b,c,d .

A polar transformation preserves the cross ratio

Take an ordered quartet of collinear points A,B,C,D.
 
        A(x1,y1,z1) B(x2,y2,z2)

        C(x1 + h x2, y1 + h y2, z1 + h z2)

        D(x1 + h'x2, y1 + h'y2, z1 + h'z2)
Then the cross ratio (A,B,C,D) = h/h'.
The polar transformation transforms the four points in their polar lines. These lines a,b,c,d have line coordinates:
 
        a( Fx' (x1,y1,z1) , Fy' (x1,y1,z1) , Fz' (x1,y1,z1) )
        b( Fx' (x2,y2,z2) , Fy' (x2,y2,z2) , Fz' (x2,y2,z2) )
        c( Fx' (x1,y1,z1) + h Fx' (x2,y2,z2) , ..., ...)
        d( Fx' (x1,y1,z1) + h' Fx' (x2,y2,z2) , ..., ...)
The cross ratio (a,b,c,d) is h/h' .

Polar triangle

A triangle is a polar triangle of a conic section if and only if each side is a polar line of the opposite vertex.

We say that the polar triangle is conjugated to the conic section or that the conic section is conjugated to the polar triangle.

Equation of a conic section conjugated to a triangle

Theorem 1

The equations of the sides of a triangle ABC are for short: A=0 ; B=0 and C=0.
Each non-degenerated conic section conjugated to this triangle has an equation
 
        k A2  + l B2  + m C2  = 0
k,l and m are homogeneous parameters (not all 0).

Proof:

 
        B = 0 is polar line of point B
=>
        (B,C,S1,S2) = -1
=>
        (AB,AC,AS1,AS2) = -1
=>
        There is a value of h such that
        Line AS1 has equation B + h C = 0  and
        Line AS2 has equation B - h C = 0
Now we take the system of conic sections with basic conic sections: The four basic points are S1, S1, S2, S2. The given non-degenerated conic section is an element of the system and therefore it has an equation of the form
 
        (B + h C).(B - h C) + k A.A = 0
<=>
        B2  - h2  C2  + k A2  = 0

Theorem 2

Each non-degenerated conic section with equation
 
        k A2  + l B2  + m C2  = 0
is conjugated to a triangle with A=0 ; B=0 and C=0 as equations of the sides.

Proof:
Since the conic section is non-degenerated, the parameter l is not 0.
Dividing the equation by l, the conic section has equation

 
        B2 + (m/l) C2  + (k/l) A2 = 0
                                    2
                We denote m/l as - h
<=>
        B2 - h2 C2  + (k/l) A2 = 0
<=>
        (B - h C) (B + h C) + (k/l) A2 = 0
This is an element of the system of conic sections with basic conic sections: This means that A=0 is the tangent chord of the intersection point A of the lines (B + h C) = 0 and (B - h C) = 0.
So, A=0 is the polar line of the point A.
Similarly, you can show the same property for B = 0 and C = 0.

Theorem 3

The equations of the sides of a triangle ABC are for short: A=0 ; B=0 and C=0.
Each degenerated conic section conjugated to this triangle has an equation
 
        k A2  + l B2  + m C2  = 0
k,l and m are homogeneous parameters (not all 0).

Proof:.. Say the conic section is degenerated in the lines d1 and d2.

  1. First case: d1 is different from d2

     
            C = 0 is polar line of point C.
    
    =>      (B,C,S1,S2) = -1
    
    =>      (AB,AC,AS1,AS2) = -1
    
    =>      There is a value of h such that
            Line AS1 has equation B + h C = 0  and
            Line AS2 has equation B - h C = 0
    
    =>      The degenerated conic section has equation
    
            B2  - h2  C2  = 0
    
  2. Second case: d1 coinciding with d2
    The equation of the conic section is
     
            A2 = 0  or  B2 = 0  or C2 = 0
    

Theorem 4

Each degenerated conic section with equation
 
        k A2  + l B2  + m C2  = 0
is conjugated to a triangle with A=0 ; B=0 and C=0 as equations of the sides.

Proof:

  1. First case: 2 parameters are zero ; take l = m = 0
    The equation of the conic section is A.A = 0 and the triangle is conjugated with the conic section.
  2. Second case : 1 parameter is zero ; take k = 0
    The equation of the conic section can be written as
     
            B2  - h2  C2  = 0  <=> (B - h C) (B + h C) = 0
    
    Then the conic section is conjugated to the triangle.
  3. Third case: no parameter is zero. It can be proved that this case cannot occur.



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