Find the sets of lines on a quadric First of all, let me apologize for my English: I'll be making up all the terms of which I don't know the translation.
This is my issue:

In the real projective space $\mathbb{P}^3$, consider the quadric surface $Q$ defined by the following equation:
  $$ 8x_1x_4+2x_2x_3-4x_2x_4-4x_3x_4=0 $$
  Recognize $Q$, determine if it contains lines, and if it does, describe through homogeneous cartesian equations the two ranks of lines it contains.

The associated matrix is:
$$\begin{bmatrix} 0 & 0 & 0 & 4 \\ 0 & 0 & 1 & -2 \\ 0 & 1 & 0 & -2 \\ 4 & -2 & -2 & 0\end{bmatrix}$$
It has obviously rank 4, and I have been able to determine that it has two positive eigenvalues and two negative ones.
Therefore it should be a real, non degenerate quadric, and contain lines, but I have no idea how to find them.
I know that the quadric is similar (through a projectivity) to that of equation $x_1^2+x_2^2-x_3^2-x_4^2=0$, and I know how to find the lines: through the equations
$$ \begin{cases} h(x_1-x_3)=k(x_4-x_2) \\
k(x_1+x_3)=h(x_2+x_4)
\end{cases} $$
and the other, similar system of equation.
But I cannot make it work in the general case: am I supposed to determine the projectivity that maps $Q$ in its standard form, and then use its inverse to determine the lines?
 A: If $x_4 \neq 0$, then
$$x_1 = -\frac{x_2 x_3}{4 x_4} + \frac{1}{2} x_2 + \frac{1}{2} x_3.$$
This shows that, however we choose $x_2, x_3, x_4 \in \mathbb{R}$ with $x_4 \neq 0$, we obtain a point of $Q$ as $[x_1, x_2, x_3, x_4]$ with $x_1$ given by the relation above.
Using this, we want to define an injective map $S\colon \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ such that the image of $S$ is contained in $Q$. In this way, if we choose $[h, k] \in \mathbb{P}^1$, we obtain two lines contained in $Q$ as the images of $\{[h, k]\} \times \mathbb{P}^1$ and of $\mathbb{P}^1 \times \{[h, k]\}$ under $S$.
We can construct $S$ as follows. Fix $[v_0, v_1], [w_0, w_1] \in \mathbb{P}^1$ and write $[x_1, x_2, x_3, x_4]$ for $S([v_0, v_1], [w_0, w_1])$. First of all, we let $x_4 = v_0 w_0$, since it is arbitrary and must be a homogeneous polynomial in both $v_0, v_1$ and $w_0, w_1$. Then, we need $x_2 x_3$ to be divisible by $4x_4$, so we choose $x_2 = 2 v_0 w_1$ and $x_3 = 2 v_1 w_0$. Finally, in order for $[x_1, x_2, x_3, x_4]$ to belong to $Q$, we substitute $x_2$, $x_3$ and $x_4$ in the relation above, so we must have $x_1 = -v_1w_1 + v_0 w_1 + v_1 w_0$.
To sum up,
$$S([v_0, v_1], [w_0, w_1]) = [-v_1 w_1 + v_0 w_1 + v_1 w_0, 2 v_0 w_1, 2 v_1 w_0, v_0 w_0].$$
Notice that the map induces a bijection between $\mathbb{P}^1 \times \mathbb{P}^1$ and the quadric $Q$.
As $[h, k] \in \mathbb{P}^1$ varies, let us describe the images of $\{[h, k]\} \times \mathbb{P}^1$ and $\mathbb{P}^1 \times \{[h, k]\}$ under $S$, which provide two families of lines contained in $Q$.
The first line associated to $[h, k]$ is $S([h, k], [1, 0]) \vee S([h, k], [0, 1])$, i.e.
$$[k, 0, 2k, h] \vee [h-k, 2h, 0, 0]$$
which has equations
$$\begin{cases} hx_3 = 2k x_4 \\ 2h x_1 = (h - k) x_2 + 2k x_4. \end{cases}$$
The second line associated to $[h, k]$ is $S([1, 0], [h, k]) \vee S([0, 1], [h, k])$, i.e.
$$[k, 2k, 0, h] \vee [h-k, 0, 2h, 0]$$
which has equations
$$\begin{cases} h x_2 = 2k x_4 \\ 2h x_1 = (h - k) x_3 + 2k x_4. \end{cases}$$
A: I think that the general idea in these cases is to find a smart way to manipulate the equations so that the two families of lines stand out clearly. In this specific case I would do the following:
$$ 8x_1x_4-4x_2x_4=4x_3x_4-2x_2x_3 $$
$$ 4x_4(2x_1-x_2)= 2x_3(2x_4-x_2) $$
and then you easily get the families of lines:
$$\mathcal{F}_1:\begin{cases}
\mu4x_4=\lambda(2x_4-x_2)\\
\lambda(2x_1-x_2)=\mu2x_3
\end{cases}$$
$$\mathcal{F}_2:\begin{cases}
\mu'4x_4=\lambda'2x_3\\
\lambda'(2x_1-x_2)=\mu'(2x_4-x_2)
\end{cases} $$
Now for each $(\mu,\lambda),(\mu',\lambda')\in\mathbb{P}^1$ you get a line from one of each family (as the intersection of two planes). Note that they are already expressed through cartesian equations.
