If the pair of straight lines $x^2 - 2 kxy - y^2 = 0$ bisect the angle between the pair of straight lines $x^2 - 2 \ell x y - y^2$, show that the later pair also bisects the angle between the former.
Comparing the equation $\;\;$ $x^2 - 2 \ell x y - y^2$ $\;\;\; \cdots \; (1)$
with the standard equation $\;\;$ $ax^2 = 2hxy + by^2 = 0$ $\;\;\; \cdots \; (2)$ $\;$
gives $\;\;$ $a = 1$, $\;\;\;$ $h = - \ell$, $\;\;\;$ $b = -1$
Equation of angle bisector of pair of lines given by equation $(2)$ is
$\dfrac{x^2 - y^2}{ab} = \dfrac{xy}{h}$ $\;\;\; \cdots \; (3)$
Substituting the values of $a$, $b$ and $h$ in equation $(3)$, the equation of angle bisector of the pair of lines given by equation $(1)$ is
$\dfrac{x^2 - y^2}{1 \times \left(-1\right)} = \dfrac{xy}{- \ell}$
i.e. $\ell x^2 - \ell y^2 = xy$ $\implies$ $\ell x^2 - xy - \ell y^2 = 0$ $\;\;\; \cdots \; (4)$
Given: The pair of straight lines $x^2 - 2kxy - y^2 = 0$ $\;\;\; \cdots \; (5)$
is the angle bisector the pair of lines given by equation $(1)$.
Since equations $(4)$ and $(5)$ represent the same angle bisector, comparing the like terms in both equations gives
$\dfrac{\ell}{1} = \dfrac{1}{2k} = \dfrac{\ell}{1}$ $\implies$ $k = \dfrac{1}{2 \ell}$ $\;\;\; \cdots \; (6)$
Comparing equation $(5)$ with the standard equation $\;\;$ $Ax^2 + 2 Hxy + By^2 = 0$
gives $\;\;$ $A = 1$, $\;\;\;$ $H = -k$, $\;\;\;$ $B = -1$
Equation of angle bisector of pair of lines given by equation $(5)$ is
$\dfrac{x^2 - y^2}{1 \times \left(-1\right)} = \dfrac{xy}{-k}$
i.e. $kx^2 - ky^2 = xy$ $\implies$ $k x^2 - xy - ky^2 = 0$ $\;\;\; \cdots \; (7)$
Substituting the value of $k$ from equation $(6)$ in equation $(7)$ gives
$\dfrac{x^2}{2 \ell} - xy - \dfrac{y^2}{2 \ell} = 0$
i.e. $x^2 - 2 \ell xy - y^2 = 0$ $\;\;\;$ which are pair of lines given by equation $(1)$.
$\therefore$ $\;$ The pair of equations given by $(1)$ bisect the angle between the pair of lines given by equation $(5)$.
Hence proved.