Prove that the equation of the pair of lines through the origin perpendicular to the pair whose equation is $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ is $\;$ $bx^2 - 2hxy + ay^2 = 0$.
Equation of given pair of lines is: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $m_1$ $\;$ and $\;$ $m_2$ $\;$ be the slopes of the two lines given by equation $(1)$.
Then, $\;$ $m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (2a)$
$m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (2b)$
and $\;$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{\left|b\right|}$ $\;\;\; \cdots \; (2c)$
From equations $(2a)$ and $(2c)$,
$2m_1 = \dfrac{-2h}{b} + \dfrac{2 \sqrt{h^2 - ab}}{b}$
$\implies$ $m_1 = \dfrac{-h + \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3a)$
and $\;$ $2m_2 = \dfrac{-2h}{b} - \dfrac{2 \sqrt{h^2 - ab}}{b}$
$\implies$ $m_2 = \dfrac{-h - \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3b)$
Let $\;$ $m_1'$ $\;$ $m_2'$ $\;$ be the slopes of the lines perpendicular to those given by equation $(1)$.
Then we have,
$m_1' = \dfrac{-1}{m_1} = \dfrac{-b}{-h + \sqrt{h^2 - ab}}$ $\;\;\; \cdots \; (4a)$ $\;\;$ [in view of equation $(3a)$]
and $\;$ $m_2' = \dfrac{-1}{m_2} = \dfrac{-b}{-h - \sqrt{h^2 - ab}} = \dfrac{b}{h + \sqrt{h^2 - ab}}$ $\;\;\; \cdots \; (4b)$ $\;\;$ [in view of equation $(3b)$]
We have from equations $(4a)$ and $(4b)$,
$m_1' + m_2' = \dfrac{-b}{-h + \sqrt{h^2 - ab}} + \dfrac{b}{h + \sqrt{h^2 - ab}}$
i.e. $\;$ $m_1' + m_2' = -b \left[\dfrac{h + \sqrt{h^2 - ab} - \left(-h + \sqrt{h^2 - ab}\right)}{\left(-h + \sqrt{h^2 - ab}\right) \left(h + \sqrt{h^2 - ab}\right)}\right]$
i.e. $\;$ $m_1' + m_2' = - b \left[\dfrac{2h}{h^2 - ab - h^2}\right] = \dfrac{2bh}{ab}$
i.e. $\;$ $m_1' + m_2' = \dfrac{2h}{a}$ $\;\;\; \cdots \; (5a)$
and $\;$ $m_1' \times m_2' = \left(\dfrac{-b}{-h + \sqrt{h^2 - ab}}\right) \left(\dfrac{b}{h + \sqrt{h^2 - ab}}\right)$
i.e. $\;$ $m_1' m_2' = \dfrac{-b^2}{h^2 - ab - h^2} = \dfrac{-b^2}{-ab}$
i.e. $\;$ $m_1' m_2' = \dfrac{b}{a}$ $\;\;\; \cdots \; (5b)$
In view of equations $(5a)$ and $(5b)$, the combined equation of pair of lines with slopes $m_1'$ and $m_2'$ is
$bx^2 - 2hxy + ay^2 = 0$ $\;\;\; \cdots \; (6)$
i.e. $\;$ The equation of pair of lines through the origin perpendicular to the pair given by equation $(1)$ are given by equation $(6)$.