Solved Problems on Existence of Second Degree Cones

Let the general equation of a cone with vertex at the origin be $$ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0.$$ If the cone contains the coordinate axes, then $$a=b=c=0,$$ and the equation reduces to $$\begin{equation} fyz+gzx+hxy=0. \end{equation}$$ Let $(l_1,m_1,n_1)$, $(l_2,m_2,n_2)$ and $(l_3,m_3,n_3)$ be the direction cosines of three mutually perpendicular lines through the origin. These are taken as a new set of rectangular axes. Then, $$\begin{aligned} l_1m_1+l_2m_2+l_3m_3&=0,\\ m_1n_1+m_2n_2+m_3n_3&=0,\\ n_1l_1+n_2l_2+n_3l_3&=0.\quad (2) \end{aligned}$$ If the cone (1) contains the lines with direction cosines $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$, then $$fm_1n_1+gn_1l_1+hl_1m_1=0,$$ $$fm_2n_2+gn_2l_2+hl_2m_2=0.$$ Adding and using relations (2), it follows that $$fm_3n_3+gn_3l_3+hl_3m_3=0.$$ Thus, the third axis also lies entirely on the cone. Hence, a second-degree cone can be drawn through any two sets of rectangular axes through the origin. $\blacksquare$