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

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??

• What are the properties of wanted curve? – science Mar 5 '15 at 6:51
• I've been told this is applicable to all curves – Tesla Mar 5 '15 at 7:00

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:
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.
• “Always” is a bit strong, I think. The equation $x^2+y^2=0$, for example, only has a single real point as its solution set. – amd Oct 31 '19 at 9:01