Question

How to homogenising a equation?

Answer 1

Homogenization is the process of making homogenous. That is, it should make the degree of every term the same. Here’s an example:

Given the non-homogeneous quadratic x2+y2+5x+2y=14 and the line 3x+2y=7.

Write the equation of the line as 3x+2y7=1.

Our goal is to make the degree of each term of the quadratic into a 2. The first two terms are fine. If we multiply the 5x by 3x+2y7, it becomes 15x2+10xy7, which is of degree 2. Remember that we really just multiplied by 1. Do the same with 2y to get 6xy+4y27. Finally, to deal with the 14, our multiplication trick does not quite work because it will only result in a term oaf degree 1. So multiply it by (3x+2y7)2 and enjoy the new term, 18x2+24xy+8y27.

The quadratic is now homogeneous and looks, after collecting terms like:

4x2−8xy+3y2=0.

This equation has a powerful feature. It can be written (2x−y)(2x−3y)=0.

It can ALWAYS be factored into linear factors.

Answer 2

Answer:

Homogenization is the process of making homogenous. That is, it should make the degree of every term the same.

Example:

Let the curve equation be  $ax^2+by^2+cxy+dx+ey+f=0$ ----1

Now, let the equation  line intersecting it be $px+qy+r=0$-------2

Now the equation needs to be analyzed.

$ax^2+by^2+cxy+dx+ey+f=0$

2        2          2     1        1    0           degree

now let us multiply 1 in 1-degree equation and $1^2$ in 0 degree equation.

$ax^2+by^2+cxy+1*dx+1*ey+1^2*f=0$ --------3

Now in the line equation,

$px+qy+r=0$

$px+qy=-r$

$\frac{-p}{r} x+\frac{-q}{r} y=1$ -----4

Substitute 4 to 3 and get a homogenized equation.