- Polar line of a point A relative to a conic section

- Polar line and harmonic conjugate points

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

- Polar line and a degenerated conic section

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

- Pole of a line relative to a conic section

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

- Calculation of a pole

- Conjugated points

- Criterion for conjugated points

- Conjugated lines

- Polar transformation relative to a non-degenerated conic section

- A polar transformation preserves the cross ratio

- Polar triangle

- Equation of a conic section conjugated to a triangle

- Point A is a simple point.

The polar line of A(x_{1},y_{1},z_{1}) relative to a conic section with equation F(x,y,z) = 0 is the linex

_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0 <=> x.F_{x}' (x_{1},y_{1},z_{1}) + y.F_{y}' (x_{1},y_{1},z_{1}) + z.F_{z}' (x_{1},y_{1},z_{1}) = 0 with matrix notation: [F_{x}' (x_{1},y_{1},z_{1})] <=> [x y z ] [F_{y}' (x_{1},y_{1},z_{1})] = 0 [F_{z}' (x_{1},y_{1},z_{1})] <=> P^{T}C P_{1}= 0 <=> P_{1}^{T}C P = 0 - Point A is a double point.

Each line is a polar line of A.

- If point A is on the conic section, the polar line is the tangent line in A.
- If point A is not on a non-degenerated conic section the polar line is the tangent chord of point A.

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(x_{1},y_{1},z_{1}) 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 xThis is a quadratic equation in h. The roots h_{1}, y + h y_{1}, z + h z_{1}) Point P is on the conic section <=> F(x + h x_{1}, y + h y_{1}, z + h z_{1}) = 0 <=> F(x,y,z) + h (x.F_{x}' (x_{1},y_{1},z_{1}) + y.F_{y}' (x_{1},y_{1},z_{1}) + z.F_{z}' (x_{1},y_{1},z_{1})) + h^{2}F(x_{1},y_{1},z_{1}) = 0

Points A and B are harmonic conjugate points relative to the
intersection points P_{1} and P_{2} of the conic section and the variable line AB

<=> (PWe see that the equation of the polar line and the equation of the stated set are the same._{1},P_{2},A,B) = -1 <=> (A,B,P_{1},P_{2}) = -1 <=> h_{1}= - h_{2}<=> h_{1}+ h_{2}= 0 <=> x.F_{x}' (x_{1},y_{1},z_{1}) + y.F_{y}' (x_{1},y_{1},z_{1}) + z.F_{z}' (x_{1},y_{1},z_{1}) = 0 <=> B is element of the stated set

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

If that point A(x_{1},y_{1},z_{1}) is a simple point then:

A(x_{1},y_{1},z_{1}) is on its polar line <=> x_{1}.F_{x}' (x_{1},y_{1},z_{1}) + y_{1}.F_{y}' (x_{1},y_{1},z_{1}) + z_{1}.F_{z}' (x_{1},y_{1},z_{1}) = 0 <=> 2.F(x,y,z) = 0 <=> point A(x_{1},y_{1},z_{1}) is on the conic section

Proof:

Take a point A(x_{1},y_{1},z_{1}), different from a double point.

The polar line of A is x_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0
Say D(xo,yo,zo) is a double point.

We investigate if D is on the polar line.

xThus, D is on the polar line._{1}.F_{x}' (xo,yo,zo) + y_{1}.F_{y}' (xo,yo,zo) + z_{1}.F_{z}' (xo,yo,zo) = x_{1}. 0 + y_{1}. 0 + z_{1}. 0 = 0

If A or B is a double point, then it is trivial.

Now, suppose that A and B are not a double point.

A(x_{1},y_{1},z_{1}) is on the polar line of B(x_{2},y_{2},z_{2}) <=> x_{1}.F_{x}' (x_{2},y_{2},z_{2}) + y_{1}.F_{y}' (x_{2},y_{2},z_{2}) + z_{1}.F_{z}' (x_{2},y_{2},z_{2}) = 0 <=> x_{2}.F_{x}' (x_{1},y_{1},z_{1}) + y_{2}.F_{y}' (x_{1},y_{1},z_{1}) + z_{2}.F_{z}' (x_{1},y_{1},z_{1}) = 0 <=> B(x_{2},y_{2},z_{2}) is on the polar line of A(x_{1},y_{1},z_{1})

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

We use matrix notation.

The conic section has equation PThe last formula gives a method to calculate the pole.^{T}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] [x_{1}] Say P1 = [y_{1}] are the coordinates of a possible pole of line a. [z_{1}] The polar line is P_{1}^{T}C P = 0 This polar line is also the line U.P = 0 From this we have : k U = P_{1}^{T}C <=> P_{1}^{T}= k U C^{-1}From that last result, we see that there is exactly one pole of the line a.

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

- First method : w use the formula from previous theorem.
[ 0, 0, -1 ] C = [ 0, 1, 0 ] U = [1 1 1] [ -1, 0, 0 ] -1 [ 0, 0, -1 ] C = [ 0, 1, 0 ] [ -1, 0, 0 ] T [ 0, 0, -1 ] P1 = k [1 1 1]. [ 0, 1, 0 ] = k [-1 1 -1] [ -1, 0, 0 ] The pole is point (-1,1,-1) or (1,-1,1)

- Second method
If P(x
_{1},y_{1},z_{1}) is the pole, then the polar line isx

This line has to be the line x + y + z = 0. So the pole is point (1,-1,1)._{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0 <=> x_{1}(-2z) + y_{1}(2y) + z_{1}(-2x) = 0 <=> z_{1}x - y_{1}y + x_{1}z = 0 - Third method
We choose two simple points P
_{1}(0,1,-1) and P_{2}(1,0,-1) on line x + y + 1 = 0. Say P is the pole of that line.

The polar line of P_{1}x + y = 0 contains point P. The polar line of P_{2}z - x = 0 contains point P. Thus, P is the intersection point of these polar lines. So the pole is point (1,-1,1).

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.

- Points A and B aren't double points
A(x

_{1},y_{1},z_{1}) and B(x_{2},y_{2},z_{2}) are conjugated points <=> A is on a polar line of point B <=> x_{1}.F_{x}' (x_{2},y_{2},z_{2}) + y_{1}.F_{y}' (x_{2},y_{2},z_{2}) + z_{1}.F_{z}' (x_{2},y_{2},z_{2}) = 0 - A is a double point
A(x

_{1},y_{1},z_{1}) and B(x_{2},y_{2},z_{2}) are conjugated points <=> B is an arbitrary point <=> x_{2}.0 + y_{2}.0 + z_{2}.0 = 0 <=> x_{2}.F_{x}' (x_{1},y_{1},z_{1}) + y_{2}.F_{y}' (x_{1},y_{1},z_{1}) + z_{2}.F_{z}' (x_{1},y_{1},z_{1}) = 0 <=> x_{1}.F_{x}' (x_{2},y_{2},z_{2}) + y_{1}.F_{y}' (x_{2},y_{2},z_{2}) + z_{1}.F_{z}' (x_{2},y_{2},z_{2}) = 0 - A is a double point

Analogous as in the previous case.

A(x_{1},y_{1},z_{1}) and B(x_{2},y_{2},z_{2}) are conjugated points <=> x_{2}.F_{x}' (x_{1},y_{1},z_{1}) + y_{2}.F_{y}' (x_{1},y_{1},z_{1}) + z_{2}.F_{z}' (x_{1},y_{1},z_{1}) = 0 <=> x_{1}.F_{x}' (x_{2},y_{2},z_{2}) + y_{1}.F_{y}' (x_{2},y_{2},z_{2}) + z_{1}.F_{z}' (x_{2},y_{2},z_{2}) = 0

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 aRemark :

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

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(xThen the cross ratio (A,B,C,D) = h/h'._{1},y_{1},z_{1}) B(x_{2},y_{2},z_{2}) C(x_{1}+ h x_{2}, y_{1}+ h y_{2}, z_{1}+ h z_{2}) D(x_{1}+ h'x_{2}, y_{1}+ h'y_{2}, z_{1}+ h'z_{2})

The polar transformation transforms the four points in their polar lines. These lines a,b,c,d have line coordinates:

a( FThe cross ratio (a,b,c,d) is h/h' ._{x}' (x_{1},y_{1},z_{1}) , F_{y}' (x_{1},y_{1},z_{1}) , F_{z}' (x_{1},y_{1},z_{1}) ) b( F_{x}' (x_{2},y_{2},z_{2}) , F_{y}' (x_{2},y_{2},z_{2}) , F_{z}' (x_{2},y_{2},z_{2}) ) c( F_{x}' (x_{1},y_{1},z_{1}) + h F_{x}' (x_{2},y_{2},z_{2}) , ..., ...) d( F_{x}' (x_{1},y_{1},z_{1}) + h' F_{x}' (x_{2},y_{2},z_{2}) , ..., ...)

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

Each non-degenerated conic section conjugated to this triangle has an equation

k Ak,l and m are homogeneous parameters (not all 0).^{2}+ l B^{2}+ m C^{2}= 0

Proof:

B = 0 is polar line of point B => (B,C,SNow we take the system of conic sections with basic conic sections:_{1},S_{2}) = -1 => (AB,AC,AS_{1},AS_{2}) = -1 => There is a value of h such that Line AS_{1}has equation B + h C = 0 and Line AS_{2}has equation B - h C = 0

- The degenerated conic section with equation (B + h C).(B - h C) = 0
- The degenerated conic section with equation A.A = 0

(B + h C).(B - h C) + k A.A = 0 <=> B^{2}- h^{2}C^{2}+ k A^{2}= 0

k Ais conjugated to a triangle with A=0 ; B=0 and C=0 as equations of the sides.^{2}+ l B^{2}+ m C^{2}= 0

Proof:

Since the conic section is non-degenerated, the parameter l is not 0.

Dividing the equation by l, the conic section has equation

BThis is an element of the system of conic sections with basic conic sections:^{2}+ (m/l) C^{2}+ (k/l) A^{2}= 0 2 We denote m/l as - h <=> B^{2}- h^{2}C^{2}+ (k/l) A^{2}= 0 <=> (B - h C) (B + h C) + (k/l) A^{2}= 0

- The degenerated conic section with equation (B + h C).(B - h C) = 0
- The degenerated conic section with equation A.A = 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.

Each degenerated conic section conjugated to this triangle has an equation

k Ak,l and m are homogeneous parameters (not all 0).^{2}+ l B^{2}+ m C^{2}= 0

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

- First case: d1 is different from d2

C = 0 is polar line of point C. => (B,C,S

_{1},S_{2}) = -1 => (AB,AC,AS_{1},AS_{2}) = -1 => There is a value of h such that Line AS_{1}has equation B + h C = 0 and Line AS_{2}has equation B - h C = 0 => The degenerated conic section has equation B^{2}- h^{2}C^{2}= 0 - Second case: d1 coinciding with d2

The equation of the conic section isA

^{2}= 0 or B^{2}= 0 or C^{2}= 0

k Ais conjugated to a triangle with A=0 ; B=0 and C=0 as equations of the sides.^{2}+ l B^{2}+ m C^{2}= 0

Proof:

- 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. - Second case : 1 parameter is zero ; take k = 0

The equation of the conic section can be written asB

Then the conic section is conjugated to the triangle.^{2}- h^{2}C^{2}= 0 <=> (B - h C) (B + h C) = 0 - Third case: no parameter is zero. It can be proved that this case cannot occur.

The tutorial address is http://home.scarlet.be/math/

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