# Homogeneous second degree equation problem

We are given a curve having the equation:

$ax^2+2hxy+by^2+2gx+2fy+c=0$

and a line having the equation:

$lx+my=n$

$\frac{lx+my}{n} = 1$

While making the equation of the curve "homogeneous", we multiply all the terms having degree one with 1 and the terms having degree 0 with $1^2$. Then we substitute the value of 1 as shown in the above equation. And get the result

$ax^2+2hxy+by^2+2gx(\frac{lx+my}{n})+2fy(\frac{lx+my}{n})+c(\frac{lx+my}{n})^2 = 0$

But why do we do this? Is there any concrete reason or theorem supporting this?

• Could you provide some clarification? Are you asking why this is a valid operation? Are you asking why this is an interesting step in your problem of interest? – Michael Burr Oct 2 '18 at 17:59
• @MichaelBurr I am asking why this is a valid operation. – S.Nep Oct 2 '18 at 18:07

Working backwards, it seems that the question must have something to do with trying to find the intersection between the given conic and line.

Therefore, at an intersection point, the $$(x,y)$$ will satisfy $$\frac{lx+my}{n}=1$$ as well as the formula for the conic. Therefore, at an intersection point, you can make the substitution $$\frac{lx+my}{n}=1$$ to construct a homogeneous equation in $$x$$ and $$y$$.

The derived homogeneous equation will have more solutions than the original intersection, but if you intersect the solutions to the homogeneous equation with solutions to the line, you get the original set of solutions.