Proving the equation of pair of lines is $x^2+2xy\sec(2\alpha)+y^2$ Prove that the equation of the pair of lines having an angle $\alpha$ with $x+y=0$ and passing through origin is $x^2+2xy\sec(2\alpha)+y^2=0$  I think we have to use rotation matrix but can't think of any concrete step to solve it.

The equation of any straight line passing through $(0,0)$ is $y=mx$
Now $$\tan\alpha=\pm\dfrac{m-(-1)}{1+m(-1)}=\pm\dfrac{1+m}{1-m}$$
Squaring we get  $$\tan^2\alpha=\left(\dfrac{1+m}{1-m}\right)^2$$
Rearrange to form a Quadratic Equation in $m$
If the roots are $m_1,m_2;$ our required equation will be $$0=(y-m_1x)(y-m_2x)=y^2-xy(m_1+m_2)+m_1m_2x^2$$
