Show that every homogeneous equation of degree two in \(x\) and \(y\) i.e. \(ax^2 + 2hxy + by^2 = 0\), represents a pair of lines passing through the origin, if \(h^2 - ab \ge 0\)

Answer: Proved by factorising the equation into two linear factors through the origin.

Consider \(ax^2 + 2hxy + by^2 = 0\) ... (1)

Case 1: \(b \ne 0\). Multiply (1) by \(b\): \(b^2y^2 + 2hbxy + abx^2 = 0\).

\((by)^2 + 2(by)(hx) + h^2x^2 - h^2x^2 + abx^2 = 0\)

\((by + hx)^2 - (h^2 - ab)x^2 = 0\)

Since \(h^2 - ab \ge 0\), \(\sqrt{h^2 - ab}\) is real, so

\(\left(by + hx + \sqrt{h^2 - ab}\,x\right)\left(by + hx - \sqrt{h^2 - ab}\,x\right) = 0\)

i.e. \(\left(h + \sqrt{h^2 - ab}\right)x + by = 0\) or \(\left(h - \sqrt{h^2 - ab}\right)x + by = 0\).

Each is a first degree equation with no constant term, so each represents a line through the origin. Hence (1) represents a pair of lines through the origin.

Case 2: \(b = 0\). Then (1) becomes \(ax^2 + 2hxy = 0\), i.e. \(x(ax + 2hy) = 0\), so \(x = 0\) or \(ax + 2hy = 0\). Both are lines through the origin (if \(h = 0\) as well, \(a \ne 0\) and we get \(x^2 = 0\), the coincident lines \(x = 0\)).

Hence in every case, \(ax^2 + 2hxy + by^2 = 0\) represents a pair of lines through the origin when \(h^2 - ab \ge 0\).