Analytic Geometry

If the intersect curve of $Ax + By + Cz + D = 0$ and $x^2 + y^2 - z^2 = 1$ are two lines

show that: $A^2 + B^2 = C^2 + D^2$

My idea is: joint the two equations: $\left\{\begin{matrix} Ax + By + Cz + D = 0 \\<br /> x^2 + y^2 - z^2 = 1 \end{matrix}\right.$

then elimination one parameter: because $Ax + By + Cz + D = 0$ is a plane, so $A^2 + B^2 + C^2 \not = 0$

so let's Assume $A\not = 0$ , then $x = - \frac {1}{A}\left(By + Cz + D\right)$ , take into $x^2 + y^2 - z^2 = 1$ in order to eliminate $x$ , we

get: $\left(1 + \frac {B^2}{A^2}\right)y^2 + \left(\frac {C^2}{A^2} - 1\right)z^2 + \frac {2BD}{A^2}y + \frac {2CD}{A^2}z + \frac {D^2}{A^2} - 1 = 0$

we know: it is a plane curve, and we know ,it replaces two plane lines if and only if : $I_2 \leq 0, I_3 = 0$

where: $I_3 = \begin{vmatrix} 1 + \frac {B^2}{A^2} & 0 & \frac {BD}{A^2} \\<br /> 0 & \frac {C^2}{A^2} - 1 & \frac {CD}{A^2} \\<br /> \frac {BD}{A^2} & \frac {CD}{A^2} & \frac {D^2}{A^2} - 1 \end{vmatrix}$ $I_2 = \begin{vmatrix} 1 + \frac {B^2}{A^2} & 0 \\<br /> 0 & \frac {C^2}{A^2} - 1 \end{vmatrix}$
and I want to get the conclusion from : $I_2 \leq 0, I_3 = 0$ , But I failed, Help me :o