If the two pair of lines $2x^2+6xy+y^2=0and4x^2+18xy+b{y}^2=0$ are equally inclined to each other then b =

Given Equation 

$$\begin{gathered} 2x^2+6xy+y^2=0---\left(i\right) \\ 4x^2+18xy+b{y}^2=0---\left(ii\right) \end{gathered}$$

Angle bisector of (i) 

$$\begin{gathered} 3(x^2-y^2)=(2-1)xy \\ \Rightarrow 3(x^2-y^2)=xy---\left(iii\right) \end{gathered}$$

Angle bisector of (ii)

$9(x^2-y^2)=(4-b)xy---\left(iv\right)$

(iii) and (iv) represent same lines

$\therefore \frac{9}{3}=\frac{4-b}{1}\Rightarrow 3=4-b\Rightarrow b=4-3\Rightarrow b=1$