If $\theta$ is the acute angle between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the straight line $x+y+2=0$then $\cos \theta =$

The equation to the pair of lines joining the origin to the points of intersection of the curve

 $x^2+xy+y^2+x+3y+1=0$ and x+y+2=o is $x^2+2y+y^2+ \left(x+3y\right) \left(\frac{{\displaystyle x+y}}{{\displaystyle -2}}\right)+{\left(\frac{{\displaystyle x+y}}{{\displaystyle -2}}\right)}^2=0$

$\Rightarrow$$4\left(x^2+xy+y^2\right)-2\left(x^2+4xy+3y^2\right)+\left(x^2+2xy+y^2\right)=0\Rightarrow 3x^2-2xy-y^2=0$.

$\theta$ is the acute angle between the lines then

$\cos \theta =$$\frac{1}{\sqrt{5}}$