Answer: Proof below; for \(x^2 + y^2 = 0\), \(h^2 - ab = -1 \lt 0\), so it is not a pair of lines.
Case 1: \(b \ne 0\). Multiply \(ax^2 + 2hxy + by^2 = 0\) by \(b\):
\(b^2y^2 + 2hbxy + abx^2 = 0\)
\(\Rightarrow (b^2y^2 + 2hbxy + h^2x^2) - h^2x^2 + abx^2 = 0\)
\(\Rightarrow (by + hx)^2 - (h^2 - ab)x^2 = 0\)
\(\Rightarrow (by + hx)^2 - \left(\sqrt{h^2 - ab}\,x\right)^2 = 0\) (real since \(h^2 - ab \ge 0\))
\(\Rightarrow \left(by + hx + \sqrt{h^2 - ab}\,x\right)\left(by + hx - \sqrt{h^2 - ab}\,x\right) = 0\).
So the equation is equivalent to \(\left(h + \sqrt{h^2 - ab}\right)x + by = 0\) or \(\left(h - \sqrt{h^2 - ab}\right)x + by = 0\). Each is a first degree equation with no constant term, i.e. a line through the origin.
Case 2: \(b = 0\). The equation becomes \(ax^2 + 2hxy = 0\), i.e. \(x(ax + 2hy) = 0\), which represents the lines \(x = 0\) and \(ax + 2hy = 0\), both through the origin. (Here \(h^2 - ab = h^2 \ge 0\) automatically.)
Hence \(ax^2 + 2hxy + by^2 = 0\) represents a pair of lines through the origin if \(h^2 - ab \ge 0\).
Hence part: For \(x^2 + y^2 = 0\): \(a = 1\), \(h = 0\), \(b = 1\), so \(h^2 - ab = 0 - 1 = -1 \lt 0\). The condition fails; indeed \(x^2 + y^2 = 0\) is satisfied only by the single point \((0, 0)\). So it does not represent a pair of lines.
