Condition for Real and Distinct Lines ($h^2>ab$)

Given a homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$, we must ask: does this always represent two real straight lines through the origin? To answer this, treat the equation as a quadratic in the ratio $x/y$ (assuming $b \neq 0$; the case $b=0$ is handled separately below). Dividing throughout by $y^2$,

This is an ordinary quadratic equation in the single unknown $t = x/y$, so by the quadratic formula,

Each root gives one value of the slope-like ratio $x/y$, and hence one line through the origin (since a fixed ratio $x/y = k$ describes the line $x = ky$). So the homogeneous equation always yields two such ratios algebraically — the question is whether they are real and distinct, real and equal, or not real at all, and this is governed entirely by the discriminant $h^2 - ab$ appearing under the square root:

• If $h^2 > ab$, the discriminant is positive, so $\sqrt{h^2-ab}$ is real and non-zero, giving two distinct real values of $x/y$ — two real, distinct lines through the origin.
• If $h^2 = ab$, the discriminant is zero, so both roots of $x/y$ coincide — the two "lines" are actually the same line counted twice (a coincident pair).
• If $h^2 < ab$, the discriminant is negative, so $x/y$ has no real value — the equation has no real linear factors, i.e. it represents no real lines through the origin (only the origin itself satisfies it as a real point; the two lines are a complex-conjugate pair).

When $b = 0$ the same conclusion is reached directly: the equation becomes $ax^2 + 2hxy = 0$, i.e. $x(ax + 2hy) = 0$, always giving the real line $x=0$ together with $ax+2hy=0$, which are distinct precisely when $h \neq 0$ (and coincide only in the degenerate case $a=h=0$) — consistent with $h^2 > ab = 0$ reading as $h \neq 0$. …