Given two polynomials A(x)=∑i=0i=n aixi and B(x)=∑i=0i=mbixi, their Sylvester matrix is a square matrix of size m+n where m=degree(B(x)) and n=degree(A(x)). The m first lines are made with the A(x) coefficients, so that:
⎛ ⎜ ⎜ ⎜ ⎝ |
| ⎞ ⎟ ⎟ ⎟ ⎠ |
and the n further lines are made with the B(x) coefficients, so that:
⎛ ⎜ ⎜ ⎝ |
| ⎞ ⎟ ⎟ ⎠ |
The determinant of a Sylvester polynomial is the resultant of the two polynomials. 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.
Remark.
The discriminant of a polynomial is the resultant of the polynomial
and its derivative.
The sylvester command computes Sylvester matrices.
The resultant command computes the resultant of two polynomials.
Example.
Input:
Output:
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ |
| ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ |
Input:
Output:
−4 p3+27 q2 |
Input:
Output:
−4 p3+27 q2 |
Examples using the resultant.
MF1+MF2=MF1+MA=F1A=2a |
A= (2acos(θ),2asin(θ)) |
A=(ax,ay)=(2a |
| ,2a |
| ) |
I=(c+ax/2;ay/2)=(c+a |
| ;a |
| ) |
eq1:=(x−ix)*(ax−2*c)+(y−iy)*ay |
eq2:=y/x−ay/ax |
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)) |
−(64·(x2+y2)·(x2· a2−x2· c2+−2· x· a2· c+2· x· c3−a4+2· a2· c2+a2· y2−c4)) |
x2a2−x2c2+−2xa2c+2xc3−a4+2a2c2+a2y2−c4=0 |
normal(subst(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,[x,y]=[c+X,Y])) |
X2 a2−X2 c2+Y2 a2−a4+a2 c2 |
X2 b2+Y2 a2−a2 b2 |
| + |
| =1 |
The segment bisector of F2A is tangent to the ellipse of focus F1,F2 and major axis 2a.
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. 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 |
| ;2a |
| ) |
If F1F2=2c and I is the midpoint of AF2:
F2=(2c,0), I=(c+ax/2;ay/2)=(c+a |
| ;a |
| ) |
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:
Output gives as resultant:
(−(64a2))(x2+y2)(x2 a2−x2 c2+−2x a2c+ 2x c3−a4+2a2c2+a2y2−c4) |
The factor −64· a2·(x2+y2) is always different from zero, therefore the locus equation is:
x2a2−x2c2+−2xa2c+2xc3−a4+2a2c2+a2y2−c4=0 |
If O, the midpoint of F1F2, is chosen as origin, you find again the cartesian equation of the ellipse:
| + |
| =1 |