Homogenization of a Conic with a Line

Combined equation of the pair of lines joining the origin to the intersection points of conic S = ax²+2hxy+by²+2gx+2fy+c = 0 and line lx+my+n = 0. Obtained by substituting 1 = −(lx+my)/n into S. The result is a homogeneous second-degree equation — always a pair of lines through the origin.

Setup: Conic $S \equiv ax^2+2hxy+by^2+2gx+2fy+c=0$ meets line $L \equiv lx+my+n=0$ at two points $A$ and $B$. Find the combined equation of lines $OA$ and $OB$ (where $O$ is the origin).

Key idea: Make $S$ homogeneous using the line equation.

From $L = 0$: $lx+my = -n$, so $\dfrac{lx+my}{-n} = 1$.

Any point on $S=0$ that also lies on $L=0$ satisfies both. To write $S$ in a form where every term has degree 2, replace the implicit "1" in constant and linear terms:

• Replace $c$ with $c \cdot 1^2 = c \cdot\left(\dfrac{lx+my}{-n}\right)^2$
• Replace $2gx+2fy$ with $(2gx+2fy) \cdot 1 = (2gx+2fy)\cdot\dfrac{lx+my}{-n}$

Substituting:

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

Multiplying through by $n^2$:

$$n^2(ax^2+2hxy+by^2) - n(2gx+2fy)(lx+my) + c(lx+my)^2 = 0$$

This is homogeneous of degree 2 — it represents a pair of lines through the origin, namely $OA$ and $OB$.

Why this works: The homogenized equation vanishes at $O(0,0)$ (every term has degree $\geq 2$, and at the origin all are zero). It also vanishes at $A$ and $B$ (since both satisfy $S=0$ and $L=0$). So it contains the lines $OA$ and $OB$.

Applications:

1. Angle between $OA$ and $OB$ — use the angle formula on the homogenized equation's coefficients
2. $OA \perp OB$ — apply the perpendicularity condition $a+b=0$ to the homogenized coefficients
3. Finding the locus of a point such that $OA \perp OB$ for all positions — equate the perpendicularity condition to get the locus

Standard JEE result: The chord $AB$ of conic $S=0$ subtends a right angle at the origin iff the homogenized equation satisfies coefficient of $x^2$ + coefficient of $y^2$ $= 0$.