If the line $y=mx$ is one of bisector of $x^2+4xy-y^2=0$ then the value of 2m=

Angular bisector eq $\cancel{2}(x^2-y^2)=\cancel{2}xy$
$x^2-xy-y^2=0$
Put y=mx
$\begin{array}{l}{x}^2-x\left(mx\right)-m^2{x}^2=0\\ 1-m-m^2\simeq 0\Rightarrow m^2+m-1\simeq 0\\ \Rightarrow m=\frac{-1\pm \sqrt{5}}{2}\Rightarrow 2m=1\pm \sqrt{5}\\ 2m=-1+\sqrt{5}\end{array}$