Question

If \( \theta \) is the acute angle between the lines represented by \( ax^2 + 2hxy + by^2 = 0 \), then prove that \( \tan \theta = \frac{2 \sqrt{h^2 - ab}}{a + b} \).

Hint:

For angle between lines, use the slope difference formula; ensure \( h^2 \geq ab \) for real lines.

Answer 1

The equation \( ax^2 + 2hxy + by^2 = 0 \) represents two lines through the origin. Let their slopes be \( m_1, m_2 \). The quadratic in \( m = \frac{y}{x} \) is: \[ bm^2 + 2hm + a = 0. \] Sum and product of slopes: \[ m_1 + m_2 = -\frac{2h}{b}, \quad m_1 m_2 = \frac{a}{b}. \] The angle \( \theta \) between lines satisfies: \[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|. \] Compute \( m_1 - m_2 \): \[ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4 m_1 m_2 = \frac{4h^2}{b^2} - 4 \cdot \frac{a}{b} = \frac{4(h^2 - ab)}{b^2}. \] \[ m_1 - m_2 = \frac{2 \sqrt{h^2 - ab}}{|b|}. \] \[ 1 + m_1 m_2 = 1 + \frac{a}{b} = \frac{a + b}{b}. \] \[ \tan \theta = \left| \frac{\frac{2 \sqrt{h^2 - ab}}{|b|}}{\frac{a + b}{b}} \right| = \frac{2 \sqrt{h^2 - ab}}{|a + b|}. \] For acute \( \theta \), assume \( a + b>0 \), so: \[ \tan \theta = \frac{2 \sqrt{h^2 - ab}}{a + b}. \]