Homogeneous Second-Degree Equation

Represents a pair of straight lines through the origin. Every second-degree homogeneous equation (no constant term, no linear terms) factors into two linear factors through the origin, real or imaginary.

Consider $ax^2 + 2hxy + by^2 = 0$.

Case 1: $b \neq 0$. Divide by $x^2$ (for $x \neq 0$):

$$a + 2h\frac{y}{x} + b\left(\frac{y}{x}\right)^2 = 0$$

This is a quadratic in $m = y/x$ with roots $m_1, m_2$:

$$bm^2 + 2hm + a = 0$$

The roots give two lines $y = m_1 x$ and $y = m_2 x$, both passing through the origin.

The original equation factors as:

$$b(y - m_1x)(y - m_2x) = 0$$

which expands to $b(y^2 - (m_1+m_2)xy + m_1m_2x^2) = 0$, consistent with $ax^2+2hxy+by^2=0$ by Vieta's.

Case 2: $b = 0$. The equation becomes $ax^2+2hxy = x(ax+2hy) = 0$, which represents $x = 0$ and $ax+2hy = 0$ — two lines through the origin.

Why both lines pass through the origin: A homogeneous equation $f(x,y) = 0$ satisfies $f(0,0) = 0$ trivially, so the origin lies on both lines.

Converse: Any pair of lines through the origin $y = m_1x$ and $y = m_2x$ has joint equation $(y-m_1x)(y-m_2x) = 0$, which expands to $y^2-(m_1+m_2)xy+m_1m_2x^2 = 0$ — a homogeneous second-degree equation.