If the slope of one of the straight lines $ax^2 + 2hxy + by^2 = 0$ is three times the other, then show that $3h^2 = 4ab$.
Let $m_1$ and $m_2$ be the slopes of the two lines.
Given: $\hspace{1em}$ $m_1 = 3m_2$ $\;\;\; \cdots \; (1)$
For a pair of lines,
$m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (2b)$
$\begin{aligned}
\text{Now, } \left(m_1 - m_2\right)^2 & = \left(m_1 + m_2\right)^2 - 4 m_1 m_2 \\\\
& = \dfrac{4h^2}{b^2} - \dfrac{4a}{b} \hspace{1em} \left[\text{from equations }(2a) \text{ and } (2b)\right] \\\\
& = \dfrac{4 \left(h^2 - ab\right)}{b^2}
\end{aligned}$
i.e. $m_1 - m_2 = \pm \dfrac{2 \sqrt{h^2 - ab}}{b}$
Consider the case $\hspace{1em}$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (2c)$
Adding equations $(2a)$ and $(2c)$ gives
$2 m_1 = \dfrac{-2h + 2 \sqrt{h^2 - ab}}{b}$ $\implies$ $m_1 = \dfrac{-h + \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3a)$
Substituting the value of $m_1$ from equation $(3a)$ in equation $(2a)$ gives
$m_2 = \dfrac{-2h}{b} + \dfrac{h - \sqrt{h^2 -ab}}{b}$ $\implies$ $m_2 = \dfrac{-h - \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3b)$
$\therefore$ $\;$ In view of equations $(1)$, $(3a)$ and $(3b)$ we have
$\dfrac{-h + \sqrt{h^2 - ab}}{b} = - 3 \times \left(\dfrac{h + \sqrt{h^2 - ab}}{b}\right)$
i.e. $- h + \sqrt{h^2 - ab} = - 3h - 3 \sqrt{h^2 - ab}$
i.e. $4 \sqrt{h^2 - ab} = - 2h$ $\implies$ $2\sqrt{h^2 - ab} = - h$ $\;\;\; \cdots \; (4)$
Squaring both sides of equation $(4)$ we have,
$4 \left(h^2 - ab\right) = h^2$
i.e. $4 h^2 - 4ab = h^2$ $\implies$ $3h^2 = 4ab$ $\;\;$ Hence proved.
Note:
The same result can be obtained by taking $\;$ $m_1 - m_2 = \dfrac{-2 \sqrt{h^2 - ab}}{b}$ $\;$ in equation $(2c)$.
The values of $m_1$ and $m_2$ will be interchanged.