Reduced equation of quadratic forms

[grimTesseract]

Homework Statement

Given the following quadric surfaces:

1. Classify the quadric surface.
2. Find its reduced equation.
3. Find the equation of the axes on which it takes its reduced form.

Homework Equations

The quadric surfaces are:

(1) $3x^2 + 3y^2 + 3z^2 - 2xz + 2\sqrt{2}(x+z)-2 = 0$

(2) $2x^2 + 2y^2 -z^2 -2xy -2xz -4yz + 2x + 2y +10z -7 = 0$

(3) $x^2 + y^2 -2 z^2 + 2xy -3 \sqrt{2}x +3 \sqrt{2}y = 0$

The Attempt at a Solution

I was able to classify (1) and (3) as an ellipsoid and a hyperbolic paraboloid, respectively. However, the characteristic polynomial of (2) has decimal roots and I wasn't able to continue past that... Wolfram Alpha tells me it should be a two-sheeted hyperboloid.

However, I do not know how to compute the 2nd or 3rd parts of my problem. I'm sure it will emerge from the work I've already done, so I include my methodology:

I found the eigenvalues of the matrix associated to the quadric form. In the case of (1) these were $2, 3, 4$ with the associated eigenvectors $(1,0,1),(0,1,0),(-1,0,1)$ respectively.

I normalized these, and my basis became $(\frac{1}{\sqrt2}, 0, \frac{1}{\sqrt2}), (0, 1, 0), (\frac{-1}{\sqrt2}, 0, \frac{1}{\sqrt2})$. Not sure how to proceed from here. Knowing that the determinant of the original matrix was nonzero, I was able to deduce this was an ellipsoid from the signs of the eigenvalues and of the lower order terms.

In the case of (3), the eigenvalues are $-2, 2, 0$ and the normalized eigenvectors are $(0, 0, 1), (\frac{1}{\sqrt2}, \frac{1}{\sqrt2}, 0), (\frac{-1}{\sqrt2}, \frac{1}{\sqrt2}, 0)$. These are the columns of my invertible matrix $Q$ and I calculated $Q^TB$, where $B$ is the matrix of the coefficients of my lower order terms. I was again able to deduce from a list of properties that this is a hyperbolic paraboloid.

And yet now I'm stuck. All help appreciated.

 

[SammyS]

(1) $3x^2 + 3y^2 + 3z^2 - 2xz + 2\sqrt{2}(x+z)-2 = 0$
...

However, I do not know how to compute the 2nd or 3rd parts of my problem. I'm sure it will emerge from the work I've already done, so I include my methodology:

I found the eigenvalues of the matrix associated to the quadric form. In the case of (1) these were $2, 3, 4$ with the associated eigenvectors $(1,0,1),(0,1,0),(-1,0,1)$ respectively.

I normalized these, and my basis became $(\frac{1}{\sqrt2}, 0, \frac{1}{\sqrt2}), (0, 1, 0), (\frac{-1}{\sqrt2}, 0, \frac{1}{\sqrt2})$. Not sure how to proceed from here. Knowing that the determinant of the original matrix was nonzero, I was able to deduce this was an ellipsoid from the signs of the eigenvalues and of the lower order terms.
.

This gives a matrix corresponding to a rotation about the $\ y\$ axis of ±45 ° .

$\left [ \begin{matrix} \displaystyle \frac 1 {\sqrt{2}} & 0 & \displaystyle \frac 1{\sqrt{2}} \\ 0 & 1 & 0 \\ \displaystyle \frac {-1} {\sqrt{2}} & 0 & \displaystyle \frac 1 {\sqrt{2}} \end{matrix} \right ]$

Use it to find x and z in terms of x' and z' . Plug them in. Do some algebra.