Homogenization of a Conic with a Line

Joint equation of lines joining the origin to the points of intersection of the conic ax²+2hxy+by²+2gx+2fy+c=0 and the line lx+my+n=0. The result is always a homogeneous second degree equation.

Problem: Given conic $S: ax^2+2hxy+by^2+2gx+2fy+c=0$ and line $L: lx+my=n$, find the joint equation of lines $OP_1$ and $OP_2$, where $O$ is the origin and $P_1, P_2$ are the intersections of $L$ with $S$.

The technique

From $L$: $\dfrac{lx+my}{n} = 1$. This equals $1$ at every point on $L$.

Replace degree-1 terms in $S$ using this factor, and the constant using its square:

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

Why it works

The result is homogeneous of degree 2 in $x$ and $y$, so it represents a pair of lines through the origin. At any point $P$ on both $S$ and $L$, the substitution $\frac{lx+my}{n}=1$ reduces the equation back to $S(P)=0$. So the equation is satisfied by $O$ (trivially) and by $P_1$, $P_2$ — hence it is the joint equation of $OP_1$ and $OP_2$.

Standard application

The chord $P_1P_2$ subtends a right angle at $O$ if and only if $OP_1 \perp OP_2$, i.e. the homogenized equation represents a perpendicular pair. The condition:

$$\text{(coeff. of } x^2\text{)} + \text{(coeff. of } y^2\text{)} = 0$$

This gives a direct condition relating the parameters of $S$ and $L$ without finding $P_1$ and $P_2$ explicitly.

Why homogenization cannot be avoided

There is no other method that produces the joint equation of $OP_1$ and $OP_2$ without first solving the conic-line system. Homogenization bypasses this computation entirely.