How does the homogenization of a curve using a given line work?

Question

I am given a curve $$C_1:2x^2 +3y^2 =5$$ and a line $$L_1: 3x-4y=5$$ and I needed to find curve joining the origin and the points of intersection of $C_1$ and $L_1$ so I was told to "homogenize" the line with the curve . They basically said that the required curve would be $$ 2x^2 +3y^2 -5\left(\frac{3x-4y}{5}\right)^2=0$$ What is this ? And how does this give required curve??

Answer 1

A homogeneous equation, ie, an equation with all terms of same degree, will always represent a set of straight lines passing through the origin. With this fact in mind, consider the following example:

I am given a curve of the form $ax^2+by^2+cxy+dy+ex+f=0$ and the line $px+qy=r$ which intersects it at points $A$ and $B$. I need to find the joint equation of the lines $OA$ and $OB$.

To achieve this, I homogenize the equation of the curve by multiplying any term of less than second degree with a factor that doesn't change anything except the degree, ie , unity. Since for any point on the given line, $\dfrac{px+qy}{r}=1,$ I homogenize the curve in this manner: $$ax^2+by^2+cxy+dy\left(\frac{px+qy}{r}\right)+ex\left(\frac{px+qy}{r}\right)+f\left(\frac{px+qy}{r}\right)^2=0\tag{i}$$ To justify that (i) is, in fact, the joint equation of $OA$ and $OB$, I give two reasons:

1. This is a homogeneous equation, so it is for sure a pair of straight lines through the origin.
2. It passes through the points $A$ and $B$ as they are the points of intersection of the given curve and line, so both the relations $\dfrac{px+qy}{r}=1$ and $ax^2+by^2+cxy+dy+ex+f=0$ are satisfied.

With a similar argument we can say that this 'trick' of homogenization can be applied to any curve of any degree but I think a rigorous proof would require knowledge of projective geometry,scale invariance of homogeneous curves etc.