Prove that one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the coordinate axes if $\left(a + b\right)^2 = 4h^2$
Let $m_1$ and $m_2$ be the slopes of the lines given by the equation $\;\;$ $ax^2 + 2hxy + by^2 = 0$ $\;\;\; \cdots \; (1)$
The two lines given by equation $(1)$ are
$y - m_1 x = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - m_2 x = 0$ $\;\;\; \cdots \; (2b)$
Now, $\;$ $m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;\;$ $m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (3b)$
Let the line given by equation $(2a)$ bisect the coordinate axes.
Then the angle between the line and the coordinate axis (say, X axis) is $\dfrac{\pi}{4}$
$\therefore$ $\;$ Slope of line given by equation $(2a)$ is $m_1 = \tan \left(\dfrac{\pi}{4}\right) = 1$
Substituting $m_1 = 1$ in equation $(3a)$ gives
$1 + m_2 = \dfrac{-2h}{b}$ $\implies$ $m_2 = \dfrac{-2h}{b} - 1$ $\;\;\; \cdots \; (4a)$
Substituting $m_1 = 1$ in equation $(3b)$ gives
$m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (4b)$
$\therefore$ $\;$ We have from equations $(4a)$ and $(4b)$,
$\dfrac{-2h}{b} - 1 = \dfrac{a}{b}$
i.e. $- 2h - b = a$ $\implies$ $a + b = - 2h$ $\implies$ $\left(a + b\right)^2 = 4 h^2$
Hence, one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the coordinate axes if $\left(a + b\right)^2 = 4h^2$.