- 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

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

- Tangent lines through a point D not on a conic section

- Special pair of lines through the origin

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(x

A variable point of line d is D( x

D is on conic section <=> F(xSince F(x_{1}+ h x_{2}, y_{1}+ h y_{2}, z_{1}+ h z_{2}) = 0 <=> F(x_{1},y_{1},z_{1}) + h (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})) + h^{2}F(x_{2},y_{2},z_{2}) = 0

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.

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 (x

F(xThe 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._{1}+ h x, y_{1}+ h y, z_{1}+ h z) = 0 <=> F(x_{1},y_{1},z_{1}) + h (x_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z)) + h^{2}F(x,y,z) = 0 <=> h (x_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z)) + h^{2}F(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. <=> x_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0

Remark:

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

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

3 x^{2}+ 4 xy + 2 xz - 9 z^{2}= 0 F_{x}' (1,1,1) = 12 ; F_{y}' (1,1,1) = 4 ; F_{z}' (1,1,1) = -16 The tangent line in D(1,1,1) is 12 x + 4 y -16 = 0 <=> 3 x + y - 4 = 0

- 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(x
_{1},y_{1},z_{1}).Equation:

The linex.F

is the same line as_{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}) = 0x

From the first equation we see that the line contains point D(x_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0_{1},y_{1},z_{1}) becausex

From the second equation we see that the line contains the double point._{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}) = 2 F(x_{1},y_{1},z_{1}) =0

Thus, the tangent line in a simple point D of a degenerated conic section isx.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 or x_{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0 - The point D is a double point. By definition, we say that each line through a double point is a tangent line.

P(xFrom these system, we see that the points of tangency are the intersection points of the conic section and the line with equation_{o},y_{o},z_{o}) is a point of tangency. <=> P(x_{o},y_{o},z_{o}) is on the conic section D(x_{1},y_{1},z_{1}) is on the tangent line through P <=> F(x_{o},y_{o},z_{o}) = 0 x_{o}.F_{x}' (x_{1},y_{1},z_{1}) + y_{o}.F_{y}' (x_{1},y_{1},z_{1}) + z_{o}.F_{z}' (x_{1},y_{1},z_{1}) = 0 <=> (x_{o},y_{o},z_{o}) is a solution of the system / F(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

x.FTherefore, we call this line the tangent chord of point D._{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

This equation is equivalent with the equation

xand it is the same formula as the formula of the tangent line in a point of the conic section._{1}.F_{x}' (x,y,z) + y_{1}.F_{y}' (x,y,z) + z_{1}.F_{z}' (x,y,z) = 0

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

xThe tangent chord has equation 3 x + y - 4 z = 0^{2}+ 2 x y - y^{2}+ 4 x z - 6 z^{2}= 0

The intersection points P1 and P2 of the tangent chord and the conic section are the solutions of the system

/ xThese solutions are P1(1,1,1) and P2(11,-5,7).^{2}+ 2 x y - y^{2}+ 4 x z - 6 z^{2}= 0 \ 3 x + y - 4 z = 0

The tangent line DP1 is x - z = 0

The tangent line DP2 is 5 x + 4 y - 5 z = 0

Q(x,y,z) is a point different from P.

A variable point D of the line PQ is

(x + h xThus, the quadratic equation of the tangent lines through point P is:_{1}, y+ h y_{1}, z + h z_{1}) Point D is on conic section F(x,y,z) = 0 <=> 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 This is a quadratic equation in h. Well, PQ is a tangent line <=> The quadratic equation in h has two equal roots <=> (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}))^{2}- 4 F(x,y,z).F(x_{1},y_{1},z_{1}) = 0 <=> point Q is on a tangent line through P(x_{1},y_{1},z_{1})

(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}))^{2}- 4 F(x,y,z).F(x_{1},y_{1},z_{1}) = 0

x^{2}+ 2 x y - y^{2}+ 4 x z - 6 z^{2}= 0 F_{x}' (x_{1},y_{1},z_{1}) = 2 x_{1}+ 2 y_{1}+ 4 z_{1}= 6 F_{y}' (x_{1},y_{1},z_{1}) = 2 x_{1}- 2 y_{1}= 2 F_{z}' (x_{1},y_{1},z_{1}) = 4 x - 12 z = -8 F(x_{1},y_{1},z_{1}) = -1 The quadratic equation of the tangent lines through point P is: (6 x + 2 y - 8 z)^{2}+ 4 (x^{2}+ 2 x y - y^{2}+ 4 x z - 6 z^{2}) = 0 <=> 40 x^{2}- 32 y z + 32 x y - 80 x z + 40 z^{2}= 0 <=> 5 x^{2}- 4 y z + 4 x y - 10 x z + 5 z^{2}= 0 <=> (5 x + 4 y - 5 z) (x - z) = 0

The intersection points of d and the conic section are the solutions of the system.

/ ux + vy + wz = 0 \ F(x,y,z) = 0This 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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