Q.28 Prove that the homogeneous equation of degree two in \(x\) and \(y\), \(ax^2 + 2hxy + by^2 = 0\) represents a pair of lines passing through the origin if \(h^2 - ab \ge 0\).
Best answer
Answer: Proof (by splitting the equation into two linear factors).
Consider \(ax^2 + 2hxy + by^2 = 0\) ... (1).
Case 1: \(b \ne 0\). Multiply (1) by \(b\): \(b^2y^2 + 2hbxy + abx^2 = 0\). Completing the square, \((by + hx)^2 - h^2x^2 + abx^2 = 0\), i.e. \((by + hx)^2 = (h^2 - ab)x^2\).
Since \(h^2 - ab \ge 0\), \(\sqrt{h^2 - ab}\) is real, and
\(by + hx = \pm\sqrt{h^2 - ab}\,x\), 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 linear equation with no constant term, so each represents a line through the origin. Hence (1) represents a pair of lines through the origin (coincident lines if \(h^2 = ab\)).
Case 2: \(b = 0\). Then (1) is \(ax^2 + 2hxy = 0\), i.e. \(x(ax + 2hy) = 0\), which gives the lines \(x = 0\) and \(ax + 2hy = 0\), both passing through the origin. (Here \(h^2 - ab = h^2 \ge 0\) automatically.)
Thus \(ax^2 + 2hxy + by^2 = 0\) represents a pair of lines through the origin if \(h^2 - ab \ge 0\).
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