How prove this equation $A^2+B^2=C^2+D^2$ define:
plane $W:Ax+By+Cz+D=0$ and the 
 hyperboloid of one sheet $U:x^2+y^2-z^2=1$
if $$W\bigcap U=l_{1},W\bigcap U=l_{2}$$
where $l_{1},l_{2}$ be two  straight lines
show that :$$A^2+B^2=C^2+D^2$$

My try: since
  $$\begin{cases}
Ax+By+Cz+D=0\\
x^2+y^2-z^2=1
\end{cases}$$
  then
  $$(Ax+By+D)^2=C^2(x^2+y^2-1)$$
  then
  $$(A^2-C^2)x^2+(B^2-C^2)y^2+2ABxy+2BDy+2ADx+D^2+C^2=0$$

Follow is user44197 idea
$$(A^2-C^2)x^2+(2ABy+2AD)x+(B^2-C^2)y^2+D^2+C^2=0$$
$$\Delta (y)=(2ABy+2AD)^2-4(A^2-C^2)[(B^2-C^2)y^2+D^2+C^2]$$
$$\Delta=4A^2B^2y^2+8A^2BDy+4A^2D^2-4(A^2B^2-A^2C^2-B^2C^2+C^4)y^2-4(A^2D^2+A^2C^2-C^2D^2-C^4)$$
so
$$\Delta(y)=4[C^2(A^2+B^2-C^2)y^2+2A^2BDy+C^2(C^2+D^2-A^2)]$$
then I can't.Thank you very much!
 A: You need to make use of the fact that the intersection is a pair of straight lines.
Look at the last equation as a quadratic in $x$, i.e of the form $a x^2+b x +c=0$
Its discriminant is $b^2-4ac$ is given by
$$
\Delta(y) = 4\,{C}^{2}\,\left( {D}^{2}+2\,y\,B\,D-{y}^{2}\,{C}^{2}+{C}^{2}+{y}^{2}\,{B}^{2}+{y}^{2}\,{A}^{2}-{A}^{2}\right) $$
which is a quadratic in $y$. Now taking its discriminant (as a quadratic in $y$) gives
$$
\Delta =
64\,{C}^{4}\,\left( C-A\right) \,\left( C+A\right) \,\left( {D}^{2}+{C}^{2}-{B}^{2}-{A}^{2}\right) $$
This has to be zero to get the intersection to be straight lines.
Ruling $C=A$ and $C=-A$ and $C=0$ you get the desired answer.
A: Here's an answer with a bit of a geometric flavor.
Clearly, the hyperboloid contains no lines that are parallel to any of the coordinate axes; it also contains no points $(x,y,z)$ such that $x^2 + y^2 < 1$. Consequently, an embedded line $\ell$ must pass through a point $P$ on the $xy$-plane, which must in fact lie on the unit circle; we can write $P = (\cos\theta, \sin\theta, 0)$ for some $\theta$. Moreover, the projection of $\ell$ into the $xy$-plane must be tangent to the unit circle, so its direction vector, $v$, has the form $(\sin\theta,-\cos\theta,c)$ for some $c$. That is, the line has this parameterization:
$$\ell : (\cos\theta, \sin\theta, 0 ) + t (\sin\theta,-\cos\theta, c)$$
and we have
$$0 = x^2 + y^2 - z^2 - 1 = (\cos\theta+t\sin\theta)^2+(\sin\theta-t\cos\theta)^2+(ct)^2-1 =t^2(c^2-1)$$
which must hold for all $t$, so that $c=\pm 1$.
Since $P$ is on the plane:
$$A \cos\theta + B \sin\theta + D = 0 \qquad \implies \qquad D = -\left(A \cos\theta+ B \sin\theta\right)$$ 
Since $v \perp (A,B,C)$:
$$v\cdot(A,B,C) = A\sin\theta - B\cos\theta \pm C = 0 \qquad \implies \qquad C = \mp \left( A \sin\theta - B \cos\theta \right)$$
Therefore,
$$C^2 + D^2 = A^2 + B^2 \qquad (\star)$$

Note: We have shown that $(\star)$ holds whenever the plane meets the hyperboloid in even a single line. This is consistent with the fact that $(\star)$ removes only a single degree of freedom in the plane equation.
