..Elimination of
homogeneous linear parameters
..Elimination of
non-homogeneous linear parameters
..Elimination of
quadratic parameter
..Elimination and
trigonometry
Elimination of homogeneous
linear parameters
Associated curves - Example
Take in a plane two lines a and b with resp. equations
/ l x - m y = 0
\ m x + l y = 5l (1)
The real numbers l and m are real parameters, not both zero. If we
substitute l and m with r.l and r.m then the lines don't change.We say : l and m are homogeneous parameters. For each choice of l and m, the lines are fixed. Therefore we say that the lines a and b are associated.
If l and m are changing, the associated lines are moving and the intersection point plots a curve c.
The curve c is the locus of all intersection points of the associated lines.
Equation of the locus c
P(x,y) is a point of the curve c
<=> P(x,y) is an intersection points of the associated lines
<=> There is an value of l and m (not both zero) such that
l x - m y = 0
m x + l y = 5l
<=> There is an value of l and m (not both zero) such that
x l - y m = 0
(y - 5) l + x m = 0
(from theory about systems)
<=> | x -y |
| | = 0
|y - 5 x |
<=> x2 + y2 - 5 y = 0 (2)
The last equation (2) is the condition in order that the system (1) has a
solution for l and m, different from the obvious solution.The equation (2) contains neither l nor m.
We have eliminated l and m from the system (1).
The equation (2) is the equation of a circle. It is the locus of all intersection points of the associated lines.
Homogeneous parameters
To eliminate homogeneous parameters from a system :- consider the parameters as unknowns
- calculate the condition in order that the system has a solution different from the obvious one.
Example 1
Eliminate l and m from
2lx + (l+m)y = 0
4mx + (l-m)y = 0
We consider the parameters as unknowns
(2x+y) l + y m = 0
y l + (4x-y) m = 0
the condition in order that the system has a solution different from the
obvious one is
| 2x+y y |
| | = 0
| y 4x-y|
<=> ...
<=> 4 x2 - y2 + xy = 0
Elimination of non-homogeneous linear parameters
Associated Curves - Example
Take in a plane two curves a and b with resp. equations
/ y = x2 + mx
\ y = 2x + m (1)
The number m is a non-homogeneous parameter.For each choice of m, the curves are fixed. Therefore we say that the curves a and b are associated.
If m is changing, the associated curves are moving and the intersection point plots a curve c.
The curve c is the locus of all intersection points of the associated curves.
Equation of locus c
P(x,y) is a point of the curve c
<=> P(x,y) is an intersection points of the associated curves
<=> There is an value of m such that
2
/ y = x + mx
\ y = 2x + m
<=> There is an value of m such that
/ mx = y - x2
\ m = y - 2x
(from theory about systems)
<=> | 1 y-2x |
| | = 0
| x y - x2|
<=> x2 - xy + y = 0 (2)
The last equation (2) is the condition in order that the system (1) has a
solution for m.The equation (2) does not contain the parameter m.
We have eliminated m from the system (1).
The equation (2) is the equation of the locus of all intersection points of the associated curves.
Non-homogeneous parameters
To eliminate non-homogeneous parameters from a system :- consider the parameters as unknowns
- calculate the condition in order that the system has a solution.
Example 2
Eliminate x from
3 x = a
5b x = 2
Consider the parameter x as unknown. Calculate the condition in order that
the system has a solution
| 3 a |
|5b 2 | = 6 - 5ab = 0
Elimination of quadratic parameter
Example 3
Take the following equations with parameter m.
(x - 3)2 + y2 = (5 - m)2
(x + 3)2 + y2 = (5 + m)2
For each value of m we have two fixed circles, the associated curves. When
m is a variable, the intersection points of the two circles, form a locus
c.If we consider m as unknown in previous system, the system takes the form
a m2 + b m + c = 0
a'm2 + b'm + c'= 0
The coefficients are functions of x and y.P(x,y) is a point of the locus if and only if last system has a solution for m. In other words if and only if the two quadratic equations have a common root.
In the following theorem, we search for the condition for (a, b, c, a', b', c') such that the two quadratic equations have a common root.
Common root of two quadratic equations.
Theorem:Consider the two quadratic equations
a m2 + b m + c = 0
a'm2 + b'm + c'= 0
with a and a' not zero, together with 3 determinants
|b c| |a c| |a b|
A = | | B = - | | C = | |
|b' c'| |a' c'| |a' b'|
The equations have a common root if and only ifA.C = B2.
The expression A.C = B2 is called the eliminant of Sylvester.
Proof:
We multiply the first equation with a', the second one with a, and
we calculate the difference of these two results.
(ba' - a'b) m + (a'c - ac')
= 0.
From this, the system of the two quadratic equations is equivalent with
/ (ba' - a'b) m + (a'c - ac') = 0.
\ a m2 + b m + c = 0
These equations have a common root if and only if the root of the first one,
is a root of the second equation. The root of the first equation is
(a'c - a
c') / (b'a - a b') = B/C .
The condition then is:
a (B/C)2 + b (B/C) + c = 0
<=>
a B2 + b B C + c C2 = 0
<=>
a B2 = -C ( b B + c C)
<=>
...
<=>
a B2 = a ( C A) and since a is not 0
<=>
A.C = B2
Quadratic elimination
If, in a system with two equations, the parameter m is quadratic, then the system can be written in the form
a m2 + b m + c = 0
a'm2 + b'm + c'= 0
The condition in order that the system has a solution for m, is A.C = B.B
(see previous theorem).We say that m is eliminated from the system.
Example 4
Eliminate m from the system
(x - 3)2 + y2 = (5 - m)2
(*)
(x + 3)2 + y2 = (5 + m)2
Considering m as unknown, we write the system as
- m2 + 10 m + (x2 - 6 x + y2 - 16) = 0
- m2 - 10 m + (x2 + 6 x + y2 - 16) = 0
Then,
|10 x2 - 6 x + y2 - 16 |
A = | | = 20 x2 + 20 y2 - 320
|-10 x2 + 6 x + y2 - 16 |
B = ... = 12 x
C = ... = 20
The eliminant of Sylvester is
A.C = B.B
<=> 400 x2 + 400 y2 - 6400 = 144 x2
<=> 16 x2 + 25 y2 - 400 = 0
It is the condition such that the given system (*) has a common root for
m.It is the locus of all intersection points of the circles, given by the system (*).
An extensive example
Given :f(x) = x3 + p x2 + q x + r (1)a) What is the connection between p, q and r if you know that a root of f(x)=0 is the product of the two other roots.
b) Apply on a) to find the roots of x3 - x2 - 35 x - 49 = 0
A solution:
a)
Denote the roots of f(x)=0 with a,b and a.b .
Then f(x) = (x-a)(x-b)(x-ab)
= x3 - (a + b + a b) x2 + (a2 b + a b2 + ab) x - a2 b2
So we have:
p = - (a + b + a b) (2)
q = a b (a + b + 1) (3)
r = - a2 b2 (4)
The connection between p, q and r can be found by eliminating a and b from
the equations (2) (3) (4) .
Since these equations only contain (a+b) and a.b, we denote (a+b) = s and a.b
= t.
Then we have to eliminate s and t from
p = - (s+t) (5)
q = t (s+1) (6)
r = - t.t (7)
We calculate s from (5) and put the result in (6). Then we have to
eliminate t from
/ q = t (1 - p -t )
\ r = - t.t
<=>
/
| t2 + (p - 1) t + q = 0
|
| t2 + r = 0
\
Now we have two quadratic equations in t.We can eliminate t with the help of Sylvester's eliminant.
A = (p-1)r
B = q-r
C = -(p-1)
A.C = B.B
<=>
- (p - 1)2 r = (q - r)2
<=>
(q - r)2 + (p - 1)2 r = 0 (10)
b)In x3 - x2 - 35 x - 49 = 0
p = -1; q = -35; r = -49. For these values (10) holds.
So,
- a2 b2 = -49 => a.b = 7 or a.b = -7
But -7 is not a root. Thus x = 7 is a root.
Dividing by (x - 7) we have
x3 - x2 - 35 x - 49 = (x - 7)(x2 + 6 x + 7)
Now it is easy to find the three roots.
Elimination and trigonometry
Formula
Given: a system of the form
/ a = cos(t)
\ b = sin(t)
To eliminate t we search for the condition in order that the given system
has a solution for t.
- If the system has a solution for t, then
a2 + b2 = 1 If a2 + b2 = 1 ,
then point P(a,b) is a point of the trigonometric circle. Then, there is an angle with value t such that/ a = cos(t) \ b = sin(t)
/ a = cos(t)
\ b = sin(t)
we write
a2 + b2 = 1
A simple example
Eliminate t from
/ x cos(t) + 2 sin(t) = y
\ x cos(t) + sin(t) = 1
First we calculate cos(t) and sin(t). We find
2 - y
cos(t) = ----- and sin(t) = y - 1
x
Now we write the condition
(2 - y)2
(y - 1)2+ -------- = 1
x2
The parameter t is eliminated.
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