Resultant of two polynomials : resultant

### 5.30.19 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)=∑i=0i=n aixi and B(x)=∑i=0i=m bixi is a square matrix with m+n rows and columns; its first m rows are made from the coefficients of A(X):

⎛ ⎜ ⎜ ⎜ ⎝ |
s11=an | s12=an−1 | ⋯ | s1(n+1)=a0 | 0 | ⋯ | 0
s21=0 | s22=an | ⋯ | s2(n+1)=a1 | s2(n+2)=a0 | ⋯ | 0
⋮ | ⋮ | ⋮ | ⋱ | ⋮ | ⋱ | ⋮
sm1=0 | sm2=0 | ⋯ | sm(n+1)=am−1 | sm(n+2)=am−2 | ⋯ | a0
| ⎞ ⎟ ⎟ ⎟ ⎠

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

⎛ ⎜ ⎜ ⎝ |
s(m+1)1=bm | s(m+1)2=bm−1 | ⋯ | s(m+1)(m+1)=b0 | 0 | ⋯ | 0
⋮ | ⋮ | ⋮ | ⋱ | ⋮ | ⋱ | ⋮
s(m+n)1=0 | s(m+n)2=0 | ⋯ | s(m+n)(m+1)=bn−1 | bn−2 | ⋯ | b0
| ⎞ ⎟ ⎟ ⎠

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 the resultant
Let, F1 and F2 be 2 fixed points in the plane and A, a variable point on the circle of center F1 and radius 2a. Find the cartesian equation of the set of points M, intersection of the line F1A and of the perpendicular bisector of F2A.

Geometric answer :

MF1+MF2=MF1+MA=F1A=2a

hence M is on an ellipse with focus F1,F2 and major axis 2a.

Analytic answer : In the Cartesian coordinate system with center F1 and x-axis having the same direction as the vector F1F2, the coordinates of A are :

A= (2acos(θ),2asin(θ))

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

A=(ax,ay)=(2a |
1−t2
1+t2
| ,2a |
2t
1+t2
| )

If F1F2=2c and if I is the midpoint of AF2, since the coordinates of F2 are F2=(2c,0), the coordinates of I

I=(c+ax/2;ay/2)=(c+a |
1−t2
1+t2
| ;a |
2t
1+t2
| )

IM is orthogonal to AF2, hence M=(x;y) satisfies the equation eq1=0 where

eq1:=(x−ix)*(ax−2*c)+(y−iy)*ay

But M=(x,y) is also on F1A, hence M satisfies the equation eq2=0

eq2:=y/x−ay/ax

The resultant of both equations with respect to t resultant(eq1,eq2,t) is a polynomial eq3 depending on the variables x,y, independent 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 :

-(64·(x2+y2)·(x2· a2-x2· c2+-2· x· a2· c+2· x· c3-a4+2· a2· c2+
a2· y2-c4))

The factor -64· (x2+y2) is always different from zero, hence the locus equation of M :

x2a2-x2c2+-2xa2c+2xc3-a4+2a2c2+a2y2-c4=0

If the frame origin is O, the middle point of F1F2, we find the cartesian equation of an ellipse. To make the change of origin F1M=F1O+OM, input :

normal(subst(x2· a2-x2· c2+-2· x· a2· c+2· x· c3-a4+2· a2· c2+

a2· y2-c4,[x,y]=[c+X,Y]))

Output :

-c2*X2+c2*a2+X2*a2-a4+a2*Y2

or if b2=a2−c2, input :

normal(subst(-c2*X2+c2*a2+X2*a2-a4+a2*Y2,c2=a2-b2))

Output :

-a2*b2+a2*Y2+b2*X2

that is to say, after division by a2*b2, M verifies the equation :

X2
a2
| + |
Y2
b2
| =1

Another example using the resultant
Let F1 and F2 be fixed points and A a variable point on the circle of center F1 and radius 2a. Find the cartesian equation of the hull of D, the segment bisector of F2A.

The segment bisector of F2A is tangent to the ellipse of focus F1,F2 and major axis 2a.

In the Cartesian coordinate system of center F1 and x-axis having the same direction than the vector F1F2, the coordinates of A are :

A= (2acos(θ);2asin(θ))

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

A=(ax;ay)=(2a |
1−t2
1+t2
| ;2a |
2t
1+t2
| )

If F1F2=2c and if I is the middle point of AF2:

F2=(2c,0), I=(c+ax/2;ay/2)=(c+a |
1−t2
1+t2
| ;a |
2t
1+t2
| )

Since D is orthogonal to AF2, the equation of D is eq1=0 where

eq1:=(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 eq2:=diff(eq1,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 :

(-(64· a`^`2))·(x`^`2+y`^`2)·(x`^`2· a`^`2-x`^`2· c`^`2+-2· x· a`^`2· c+
2· x· c`^`3-a`^`4+2· a`^`2· c`^`2+a`^`2· y`^`2-c`^`4)

The factor -64· (x2+y2) is always different from zero, therefore the locus equation is :

x2a2-x2c2+-2xa2c+2xc3-a4+2a2c2+a2y2-c4=0

If O, the middle point of F1F2, is chosen as origin, we find again the cartesian equation of the ellipse :

X2
a2
| + |
Y2
b2
| =1