Show that the acute angle \(\theta\) between the lines represented by \(ax^2 + 2hxy + by^2 = 0\) is given by, \(\tan\theta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right|\)

Answer: Proof using the sum and product of slopes.

Let \(b \ne 0\) and let the lines be \(y = m_1x\) and \(y = m_2x\). Then their combined equation is \((y - m_1x)(y - m_2x) = 0\), i.e. \(m_1m_2x^2 - (m_1 + m_2)xy + y^2 = 0\). Dividing \(ax^2 + 2hxy + by^2 = 0\) by \(b\) and comparing:

\(m_1 + m_2 = -\frac{2h}{b}\), \(m_1m_2 = \frac{a}{b}\).

\((m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4(h^2 - ab)}{b^2}\), so \(m_1 - m_2 = \pm\frac{2\sqrt{h^2 - ab}}{b}\).

Also \(1 + m_1m_2 = 1 + \frac{a}{b} = \frac{a + b}{b}\).

The acute angle \(\theta\) between the lines is given by \(\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{2\sqrt{h^2 - ab}/b}{(a + b)/b}\right| = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right|\).

(If \(b = 0\), one line is the Y-axis and the same result follows by taking slopes of \(x\) in terms of \(y\).)