Tangent lines




Intersection points

Theorem:
If a line is not a component of a conic section, then that line intersects the conic section in two points. (projective property)
Proof:
The conic section is F(x,y,z) = 0. The line d is defined by the points A(x1,y1,z1) and B(x2,y2,z2); Choose B not on the conic section.
A variable point of line d is D( x1 + h x2, y1+ h y2, z1 + h z2).
 
        D is on conic section
<=>
        F(x1 + h x2, y1+ h y2, z1 + h z2) = 0
<=>
        F(x1,y1,z1)

           + h (x1.Fx' (x2,y2,z2) + y1.Fy' (x2,y2,z2) + z1.Fz' (x2,y2,z2))

           + h2 F(x2,y2,z2) = 0
Since F(x2,y2,z2) is not 0, the previous equation is a quadratic equation in h. The roots are h1 and h2. With these h-values correspond two intersection points with the conic section.
These points can be coinciding or imaginary or ideal.

Remark :

Tangent line in a point of a non-degenerated conic section.

A tangent line in a point D of a non-degenerated conic section is a line through D, such that the intersection points are coinciding.

Equation of a tangent line in a point D of a non-degenerated conic section

Take point D(x1,y1,y1) of the conic section.
We search for all points P(x,y,z) such that DP is a tangent line.
A variable point of the line DP has coordinates (x1 + h x, y1 + h y, z1 + h z). The intersection points of DP and the conic section are defined by the values of h such that
 
        F(x1 + h x, y1+ h y, z1 + h z) = 0
<=>
                F(x1,y1,z1)

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

           + h2 F(x,y,z) = 0
<=>
        h (x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z))

           + h2F(x,y,z) = 0

Thus,

        The line DP is a tangent line

<=>
        The previous equation has two coinciding roots for h.
        Since one root = 0. The other has to be 0.

<=>
        x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z) = 0
The last condition is the necessary and sufficient condition for the coordinates of P such that DP is a tangent line. It is the equation of the tangent line through D.

Remark:
Appealing on the switching property, the equation of the tangent line is also

 
        x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1) = 0

Example :

We'll calculate the tangent line in D(1,1,1) of conic section
 
        3 x2 + 4 xy + 2 xz - 9 z2 = 0


        Fx' (1,1,1) = 12 ; Fy' (1,1,1) = 4 ; Fz' (1,1,1) = -16

The tangent line in D(1,1,1) is

        12 x + 4 y -16 = 0 <=> 3 x + y - 4 = 0

Tangent line in point D of a degenerated conic section.

  1. The point D is a simple point. A tangent line in a simple point D of a degenerated conic section is the component through that point D(x1,y1,z1).

    Equation:
    The line

     
            x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1) = 0
    
    is the same line as
     
            x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z) = 0
    
    From the first equation we see that the line contains point D(x1,y1,z1) because
     
    x1.Fx' (x1,y1,z1) + y1.Fy' (x1,y1,z1) + z1.Fz' (x1,y1,z1) = 2 F(x1,y1,z1) =0
    
    From the second equation we see that the line contains the double point.
    Thus, the tangent line in a simple point D of a degenerated conic section is
     
            x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1) = 0
    or
            x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z) = 0
    
  2. The point D is a double point. By definition, we say that each line through a double point is a tangent line.

Tangent lines through a point D not on a conic section

Take D(x1,y1,z1) not on the conic section F(x,y,z) = 0.
 
         P(xo,yo,zo) is a point of tangency.
<=>
        P(xo,yo,zo) is on the conic section
        D(x1,y1,z1) is on the tangent line through P
<=>
        F(xo,yo,zo) = 0
        xo.Fx' (x1,y1,z1) + yo.Fy' (x1,y1,z1) + zo.Fz' (x1,y1,z1)  = 0
<=>
        (xo,yo,zo) is a solution of the system
        / F(x,y,z) = 0
        \ x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1)  = 0
From these system, we see that the points of tangency are the intersection points of the conic section and the line with equation
 
         x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1)  = 0
Therefore, we call this line the tangent chord of point D.
This equation is equivalent with the equation
 
        x1.Fx' (x,y,z) + y1.Fy' (x,y,z) + z1.Fz' (x,y,z) = 0
and it is the same formula as the formula of the tangent line in a point of the conic section.

We can calculate the tangent lines through a point D not on a conic section in three steps.

Example

We'll calculate the tangent lines through a point D(1,0,1) at conic section
 
        x2  + 2 x y - y2  + 4 x z - 6 z2  = 0
The tangent chord has equation 3 x + y - 4 z = 0
The intersection points P1 and P2 of the tangent chord and the conic section are the solutions of the system
 
        / x2  + 2 x y - y2  + 4 x z - 6 z2  = 0
        \ 3 x + y - 4 z = 0
These solutions are P1(1,1,1) and P2(11,-5,7).
The tangent line DP1 is x - z = 0
The tangent line DP2 is 5 x + 4 y - 5 z = 0

Quadratic equation of the tangent lines through a point

Say P(x1,y1,z1) is a point not on a non-degenerated conic section.
Q(x,y,z) is a point different from P.
A variable point D of the line PQ is
 
        (x + h x1, y+ h y1, z + h z1)

        Point D is on conic section F(x,y,z) = 0
<=>
        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.

Well,
        PQ is a tangent line
<=>

        The quadratic equation in h has two equal roots
<=>
        (x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1))2
                - 4 F(x,y,z).F(x1,y1,z1) = 0
<=>
        point Q is on a tangent line through P(x1,y1,z1)
Thus, the quadratic equation of the tangent lines through point P is:
 
        (x.Fx' (x1,y1,z1) + y.Fy' (x1,y1,z1) + z.Fz' (x1,y1,z1))2
                - 4 F(x,y,z).F(x1,y1,z1) = 0

Example:

We'll calculate the tangent lines through a point D(1,0,1) at conic section
 
        x2 + 2 x y - y2  + 4 x z - 6 z2  = 0

Fx' (x1,y1,z1) = 2 x1 + 2 y1 + 4 z1 = 6
Fy' (x1,y1,z1) = 2 x1 - 2 y1 = 2
Fz' (x1,y1,z1) = 4 x - 12 z = -8
F(x1,y1,z1) = -1

The quadratic equation of the tangent lines through point P is:

        (6 x + 2 y - 8 z)2  + 4 (x2  + 2 x y - y2  + 4 x z - 6 z2 ) = 0
<=>
        40 x2  - 32 y z + 32 x y - 80 x z + 40 z2 = 0
<=>
        5 x2  - 4 y z + 4 x y - 10 x z + 5 z2 =  0
<=>
        (5 x + 4 y - 5 z) (x - z) = 0

Special pair of lines through the origin

F(x,y,z) = 0 is the equation of a conic section and ux + vy + wz = 0 is the equation of a line d.
The intersection points of d and the conic section are the solutions of the system.
 
        / ux + vy + wz = 0
        \ F(x,y,z) = 0
This system is equivalent with
 
        /     -(u x + v y)
        | z = ------------
        |          w
        |
        |          -(u x + v y)
        | F(x, y , ------------ ) = 0
        |             w
        \
The last equation of that system is a quadratic and homogeneous equation in x and y. Thus, it is the equation of a pair of lines through the origin. These lines through the origin go through the intersection points of d and the conic section.
Conclusion:
If we eliminate z between the line d and the equation of the conic section, the resulting equation is the equation of the lines through the origin and through the intersection points of d with the conic section.



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