Why $xy+yz+xz=1$ is two-sheeted hyperboloid? I can't see why $xy+yz+xz=1$ is two-sheeted hyperboloid.
I know that the equation for two-sheeted hyperboloid is: $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1$. 
 A: You need to perform Gauß' reduction to see it:
\begin{align*}
xy+yz+zx&=(x+z)(y+z)-z^2=\frac14\Bigl((x+z+y+z)^2-(x+z-y-z)^2\Bigr)-z^2\\
&=\Bigl(\frac{x+y+2z}2\Bigr)^2-\Bigl(\frac{x-y}2\Bigr)^2-z^2.
\end{align*}
Hence the equation of the quadric can be written as 
$$\Bigl(\frac{x-y}2\Bigr)^2+z^2-\Bigl(\frac{x+y+2z}2\Bigr)^2=-1.$$
A: Change coordinates! Set $x_1 = \frac{1}{2}(x+y), \ y_1 = \frac{1}{2}(x-y),\ z_1 = z$. This gets you $x = x_1+y_1$ and $y = x_1 - y_1$. Since $yz + xz = x_1z_1$,
$$x_1^2 -y_1^2 + x_1z_1 = 1$$
Either stop here if you like this, or set $y_2 = y_1,\ z_2=z_1$ and complete the square,
$$ x_1 = x_2 - \frac{1}{2}z_2   $$
so that $x_1^2 + x_1z_1 = x_2^2 - \frac{1}{4}z_2^2$. 
This gives us the required form after setting $x_3=x_2,y_3=y_2,z_3 = \frac{z_2}{2}$,
$$x_3^2 -  z_3^2 - y_3^2  = 1$$
Or 
$$z_3^2 +  y_3^2 - x_3^2  = -1$$
We did not use orthogonal change of coordinates so we have 'warped' the picture but this does not change the classification of the quadric.
This is indeed(!) a two-sheet hyperboloid,

