The slope of angular bisectors of pair of lines $(ax+by{)}^2=c(bx-ay{)}^2,(c>0)$ are

The combined equation of pair of lines is $(ax+by{)}^2=c(bx-ay{)}^2,(c>0)$

$$\begin{gathered} a^2{x}^2+b^2{y}^2+2abxy=c{b}^2{x}^2+c{a}^2{y}^2-2abcxy \\ {x}^2\left(a^2-b^{2}c\right)+2abxy\left(1+c\right)+y^2\left(b^2-c{a}^2\right)=0 \end{gathered}$$

The combined equation of angular bisector of pair of lines is $\begin{array}{rcl}ab\left(1+c\right)\left(x^2-y^2\right)& =& xy\left(a^2-b^{2}c-b^2+c{a}^2\right)\\ ab\left(x^2-y^2\right)& =& xy\left(a^2-b^2\right)\end{array}$

$ab{x}^2-xy\left(a^2-b^2\right)-y^2\left(2ab\right)=0$

Product of the slope is $-1$ and the sum of the slopes is $\frac{a^2-b^2}{ab}=\frac{a}{b}-\frac{b}{a}$

Hence, the slopes are $\frac{a}{b},-\frac{b}{a}$