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I can't see why $xy+yz+xz=1$ is two-sheeted hyperboloid.

I know that the equation for two-sheeted hyperboloid is: $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1$.

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  • $\begingroup$ Hint: rotated axes. $\endgroup$ – hardmath Mar 26 '16 at 13:57
  • $\begingroup$ Do you know about eigenvalues and eigenvectors? $\endgroup$ – Andrew D. Hwang Mar 26 '16 at 13:57
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    $\begingroup$ The equation has the form $Q(x, y, z) = c$ for some homogeneous quadratic polynomial $Q$ and constant $c$, so it defines a quadric, and glancing at the coefficients shows that it is nondegenerate. We can see that $\pm \frac{1}{\sqrt{3}}(1, 1, 1)$ are both on the quadric and some easy algebra shows that the plane $x + y + z = 0$ dividing them does not intersect the quadric, so the quadric has at least two components, but the only type of nondegenerate quadric that is not connected is the hyperboloid of two sheets. $\endgroup$ – Travis Willse Mar 26 '16 at 14:05
  • $\begingroup$ @AndrewD.Hwang yes I Do. $\endgroup$ – F1sargyan Mar 26 '16 at 14:26
  • $\begingroup$ @F1sargyan: Travis has provided a good sketch. Alternatively, orthogonally diagonalize the coefficient matrix of your quadratic form,$$\tfrac{1}{2}\left[\begin{array}{@{}ccc@{}}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end{array}\right].$$ $\endgroup$ – Andrew D. Hwang Mar 26 '16 at 14:40
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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 $$\Bigl(\frac{x-y}2\Bigr)^2+z^2-\Bigl(\frac{x+y+2z}2\Bigr)^2=-1.$$

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  • $\begingroup$ Oops! You're right. I've corrected the computation. Thank you for pointing it! $\endgroup$ – Bernard Mar 28 '16 at 17:52
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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,

enter image description here

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  • $\begingroup$ I still can't see how the last equation gives us the required form. there must be something that I'm missing here. $\endgroup$ – F1sargyan Mar 26 '16 at 14:54
  • $\begingroup$ @F1sargyan I finished the argument. FWIW i don't understand why you have the same coefficient in front of $x^2$ and $y^2$ in your equation of a two-sheeted hyperboloid; in particular if you already commit the mild sin of not using orthogonal change of coordinates, then WLOG you can replace all coefficients $a_i$ by $\text{sgn} (a_i)$. $\endgroup$ – Calvin Khor Mar 26 '16 at 15:11
  • $\begingroup$ That's the definition of a hyperboloid with $a=b=c=1$ $\endgroup$ – Stella Biderman Mar 26 '16 at 15:11
  • $\begingroup$ shouldn't be -1 on the right side instead of 1 ? $\endgroup$ – F1sargyan Mar 26 '16 at 15:14
  • $\begingroup$ @F1sargyan Oh no, I made a mistake. Let me fix this. $\endgroup$ – Calvin Khor Mar 26 '16 at 15:19

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