Q.If $\theta$ is the measure of the acute angle between the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$, then prove that $\tan\theta = \left|\dfrac{2\sqrt{h^2 - ab}}{a+b}\right|$ where $a + b \ne 0$ and $b \ne 0$. Find the condition for coincident lines.

Write the pair as $y=m_1x, y=m_2x$, use $m_1+m_2, m_1m_2$ from the quadratic, then the angle-between-lines formula.

The equation $ax^2+2hxy+by^2=0$ represents two lines through the origin $y=m_1x$ and $y=m_2x$, where $m_1,m_2$ are roots of $bm^2+2hm+a=0$ (dividing by $x^2$ and setting $m=y/x$).

So $m_1+m_2 = -\dfrac{2h}{b}$, $\ m_1m_2 = \dfrac{a}{b}$.

If $\theta$ is the acute angle between the lines:

$\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

$(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2 = \dfrac{4h^2}{b^2} - \dfrac{4a}{b} = \dfrac{4(h^2-ab)}{b^2}$

$\Rightarrow m_1-m_2 = \pm\dfrac{2\sqrt{h^2-ab}}{b}$

$1+m_1m_2 = 1+\dfrac{a}{b} = \dfrac{a+b}{b}$

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