If a and b positive numbers $(a<b)$, then the range of values of k for which a real  $\lambda$ be found such that equation $a{x}^2+2\lambda xy+b{y}^2+2k(x+y+1)=0$ represents a pair of straight lines is

$a=a,2h=2\lambda ,b=b,2g=2k,2f=2k,c=2k$

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$\Rightarrow ab\left(2k\right)+2\left(k\right)\left(k\right)\lambda -a{k}^2-b{k}^2-2k{\lambda }^2=0$

$\Rightarrow 2{\lambda }^2-2k\lambda +(ak+bk-2ab)=0$

it is quadratic equation $\lambda$

$\lambda \in R\Rightarrow$roots are real

$\Rightarrow {(-2k)}^2-4\left(2\right)(ak+bk-2ab)\ge 0$

$\Rightarrow (k-2a)(k-2b)\ge 0$

$\Rightarrow k\le 2a$ or $k\ge 2b$