** suivant:** Orthogonal polynomials
** monter:** Arithmetic and polynomials
** précédent:** Sylvester matrix of two
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##

Resultant of two polynomials : `resultant`

`resultant` takes as argument two polynomials and
returns the resultant of the two polynomials.

The resultant of two polynomials is the determinant of their
Sylvester matrix *S*.
The Sylvester matrix *S* of two polynomials
*A*(*x*) = *a*_{i}*x*^{i}
and
*B*(*x*) = *b*_{i}*x*^{i}
is a square matrix with *m* + *n* rows and columns; its first *m* rows
are made from the coefficients of *A*(*X*):

and the following *n* rows are made in the same way from the
coefficients of *B*(*x*) :

If *A* and *B* have integer coefficients with non-zero resultant *r*,
then the polynomials equation

*AU* + *BV* = *r*

has a unique solution *U*, *V* such that degree(*U*) <degree(*B*) and
degree(*V*) <degree(*A*), and this solution has integer coefficients.
Input :

`resultant(x``^`

3-p*x+q,3*x`^`

2-p,x)

Output :
`-4*p``^`

3-27*q`^`

2

**Remark**

discriminant(P)=resultant(P,P').
**An example using resultant**

Let, *F*1 and *F*2 be 2 fixed points in the plan and
*A*, a variable point on the circle of center *F*1 and radius 2*a*.
Find the cartesian equation of the set of points *M*, intersection of
the line *F*1*A* and of the segment bisector of *F*2*A*.

Geometric answer :

*MF*1 + *MF*2 = *MF*1 + *MA* = *F*1*A* = 2*a*

hence *M* is on an ellipse with focus *F*1, *F*2 and major axis 2*a*.
Analytic answer :
In the Cartesian coordinate system of center *F*1
and *x*-axis having the same
direction than the vector *F*1*F*2, the coordinates of *A* are :

*A* = (2

*a* cos(

);2

*a* sin(

))

where is the (*Ox*, *OA*) angle.
Now choose
*t* = tan(/2) as parameter, such that the coordinates
of *A* are rational functions with respect to *t*.
More precisely :

*A* = (

*ax*;

*ay*) = (2

*a*;2

*a*)

If *F*1*F*2 = 2*c* and if *I* is the midel point of *AF*2,
since the coordinates of *F*2 are *F*2 = (2*c*, 0), the coordinates
of *I*

*I* = (

*c* +

*ax*/2;

*ay*/2) = (

*c* +

*a*;

*a*)

*IM* is orthogonal to *AF*2, hence *M* = (*x*;*y*) verify the equation
*eq*1 = 0 where

*eq*1 : = (*x* - *ix*)*(*ax* - 2**c*) + (*y* - *iy*)**ay*

But *M* = (*x*;*y*) is also on *F*1*A*, hence *M* verify the equation *eq*2 = 0

*eq*2 : = *y*/*x* - *ay*/*ax*

The resultant of both equations with respect to *t*
`resultant(eq1,eq2,t)` is a polynomial *eq*3 depending on the
variables *x*, *y*, independant of *t* which is the cartesian equation
of the set of points *M* when *t* varies.
Input :

`ax:=2*a*(1-t``^`

2)/(1+t`^`

2);ay:=2*a*2*t/(1+t`^`

2);

`ix:=(ax+2*c)/2; iy:=(ay/2)`

`eq1:=(x-ix)*(ax-2*c)+(y-iy)*ay`

`eq2:=y/x-ay/ax`

`factor(resultant(eq1,eq2,t))`

Output gives as resultant :

The factor
is always different from zero,
hence the locus equation of *M* :

If the frame origin is *O*, the middle point of *F*1*F*2,
we find the cartesian equation of an ellipse.
To make the change of origin
, input :

Output :

or if
*b*^{2} = *a*^{2} - *c*^{2}, input :

Output :

that is to say, after division by *a*^{2}**b*^{2}, *M* verifies the equation :

+

= 1

**Another example using resultant**

Let *F*1 and *F*2 be fixed points and *A* a variable point on the
circle of center *F*1 and radius 2*a*.
Find the cartesian equation of the hull of *D*, the segment bisector
of *F*2*A*.

The segment bisector of *F*2*A* is tangent to the ellipse of focus
*F*1, *F*2 and major axis 2*a*.

In the Cartesian coordinate system of center *F*1 and *x*-axis having the same
direction than the vector *F*1*F*2, the coordinates of *A* are :

*A* = (2

*a* cos(

);2

*a* sin(

))

where is the (*Ox*, *OA*) angle.
Choose
*t* = tan(/2) as parameter, such that the coordinates of *A* are
rational functions with respect to *t*.
More precisely :

*A* = (

*ax*;

*ay*) = (2

*a*;2

*a*)

If *F*1*F*2 = 2*c* and if *I* is the middle point of *AF*2:

*F*2 = (2

*c*, 0),

*I* = (

*c* +

*ax*/2;

*ay*/2) = (

*c* +

*a*;

*a*)

Since *D* is orthogonal to *AF*2, the equation of *D* is
*eq*1 = 0 where

*eq*1 : = (*x* - *ix*)*(*ax* - 2**c*) + (*y* - *iy*)**ay*

So, the hull of *D* is the locus of *M*, the intersection point of *D*
and *D'* where *D'* has equation
*eq*2 : = *diff* (*eq*1, *t*) = 0.
Input :

`ax:=2*a*(1-t``^`

2)/(1+t`^`

2);ay:=2*a*2*t/(1+t`^`

2);

`ix:=(ax+2*c)/2; iy:=(ay/2)`

`eq1:=normal((x-ix)*(ax-2*c)+(y-iy)*ay)`

`eq2:=normal(diff(eq1,t))`

`factor(resultant(eq1,eq2,t))`

Output gives as resultant :

The factor
is always different from zero,
therefore the locus equation is :

If *O*, the middle point of *F*1*F*2, is choosen as origin,
we find again the cartesian equation of the ellipse :

+

= 1

** suivant:** Orthogonal polynomials
** monter:** Arithmetic and polynomials
** précédent:** Sylvester matrix of two
** Table des matières**
** Index**
giac documentation written by Renée De Graeve and Bernard Parisse