Question

The lines represented by $x^2+2\lambda xy+2y^2=0$ and $(1+\lambda ){\lambda }^2-8xy+y^2=0$ are equally inclined then find $\lambda$.

Answer 1

The lines $x^2{+2\lambda xy+2y}^2=0.....\left(1\right)(1+\lambda )x^2-8xy+y^2=0....\left(2\right)$

are eqyally inclined if bisector of the angles are equal

equation of bisector of angles of pair (1)

$\frac{x^2{-y}^2}{a-b}=\frac{xy}{h}\Rightarrow \frac{x^2{-y}^2}{1-2}=\frac{xy}{\lambda }\Rightarrow \lambda (x^2-y^2)+xy=0....\left(3\right)$

equation of bisector of angles of pair (2)

$\frac{x^2{-y}^2}{a-b}=\frac{xy}{h}\Rightarrow \frac{x^2{-y}^2}{1+\lambda -1}=\frac{xy}{-4}\Rightarrow -4(x^2-y^2)-\lambda xy=0\Rightarrow 4(x^2-y^2)+\lambda xy=0.....\left(4\right)$

from (3) and (4)

$\frac{\lambda }{4}=\frac{1}{\lambda }\Rightarrow {\lambda }^2=4\Rightarrow \lambda =\pm 2$