1. ## 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: $\left\{\begin{matrix} Ax + By + Cz + D = 0 \\
x^2 + y^2 - z^2 = 1 \end{matrix}\right.$

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: $I_3 = \begin{vmatrix} 1 + \frac {B^2}{A^2} & 0 & \frac {BD}{A^2} \\
0 & \frac {C^2}{A^2} - 1 & \frac {CD}{A^2} \\
\frac {BD}{A^2} & \frac {CD}{A^2} & \frac {D^2}{A^2} - 1 \end{vmatrix}$
$I_2 = \begin{vmatrix} 1 + \frac {B^2}{A^2} & 0 \\
0 & \frac {C^2}{A^2} - 1 \end{vmatrix}$

and I want to get the conclusion from : $I_2 \leq 0, I_3 = 0$, But I failed, Help me

2. You have made a mistake here:
$\left(1 + \frac {B^2}{A^2}\right)y^2 + \left(\frac {C^2}{A^2} - 1\right)z^2 + \frac {2BD}{A^2}y + \frac {2CD}{A^2}z + \frac {D^2}{A^2} - 1 = 0$

It should be
$\left(1 + \frac {B^2}{A^2}\right)y^2 + \left(\frac {C^2}{A^2} - 1\right)z^2 + \frac{2BC}{A^2}yz + \frac {2BD}{A^2}y + \frac {2CD}{A^2}z + \frac {D^2}{A^2} - 1 = 0$
since the intersection are two lines, consider the above expression as a equation with y as unknown variable, simplify the equation by multiplying $A^2$, its discriminant
$\triangle_{y}=(A^4+B^2A^2-C^2A^2)z^2-2A^2CDz+(A^4+B^2A^2-D^2A^2)$
must be a perfect square expression containning z as variable. that is to say,
$(A^2+B^2-C^2)z^2-2CDz+(A^2+B^2-D^2)$
is a perfect square expression with respect to z. thus its discriminant
$\triangle_{z}=4C^2D^2-4(A^2+B^2-C^2)(A^2+B^2-D^2)=0$
which gives $A^2+B^2=C^2+D^2$.