Tangent lines

• Intersection points
• Tangent line in a point of a non-degenerated conic section.
• Equation of a tangent line in a point D of a non-degenerated conic section
• Example :
• Tangent line in point D of a degenerated conic section.
• Tangent lines through a point D not on a conic section
• Example
• Quadratic equation of the tangent lines through a point
• Example:
• Special pair of lines through the origin

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 :

• Previous theorem holds even if the line d is imaginary.
• If a line and a conic section have more than two common points, the conic section is degenerated and the line is a component.

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. 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. 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.

• calculate the equation of the tangent chord of point D .
• calculate the intersection points P1 and P2 of the tangent chord and the conic section
• calculate the tangent lines DP1 and DP2

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.