Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag

Mathematics-Online course: Linear Algebra - Analytic Geometry - Quadrics

Euclidean Normal Form of three-dimensional Quadrics


[previous page]   [table of contents][page overview]

There exist appropriate Cartesian coordinate systems with respect to which the equations defining quadrics have the following normal forms.


conical quadrics


normal form name
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\frac{x_3^2}{a_3^2}=0$ point
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}-\frac{x_3^2}{a_3^2}=0$ (double) cone
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}=0$ line
$ \frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}=0$ intersecting planes
$ \frac{x_1^2}{a_1^2}=0$ coincident planes


central quadrics


normal form name
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\frac{x_3^2}{a_3^2}+1=0$ (empty set)
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}-\frac{x_3^2}{a_3^2}+1=0$ hyperboloid of 2 sheets
$ \frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}-\frac{x_3^2}{a_3^2}+1=0$ hyperboloid of 1 sheet
$ -\frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}-\frac{x_3^2}{a_3^2}+1=0$ ellipsoid
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+1=0$ (empty set)
$ \frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}+1=0$ hyperbolic cylinder
$ -\frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}+1=0$ elliptic cylinder
$ \frac{x_1^2}{a_1^2}+1=0$ (empty set)
$ -\frac{x_1^2}{a_1^2}+1=0$ parallel planes


parabolic quadrics


normal form name
$ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+2x_3=0$ elliptic paraboloid
$ \frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}+2x_3=0$ hyperbolic paraboloid
$ \frac{x_1^2}{a_1^2}+2x_2=0$ parabolic cylinder


The normal forms are uniquely determined up to permutation of subscripts and in the case of conical quadrics up to multiplication by a constant $ c\ne 0$.

The values $ a_i$ are set to be positive and are called lengths of the principal axes of the quadric.

(double) cone intersecting planes
   
\includegraphics[width=.4\moimagesize]{quadriken_kegel} \includegraphics[width=.4\moimagesize]{quadriken_schneidende_ebenen}

hyperboloid of 2 sheets hyperboloid of 1 sheet
   
\includegraphics[width=.4\moimagesize]{quadriken_zweischaliges_hyperboloid} \includegraphics[width=.4\moimagesize]{quadriken_einschaliges_hyperboloid}

ellipsoid hyperbolic cylinder
   
\includegraphics[width=.4\moimagesize]{quadriken_ellipsoid} \includegraphics[width=.4\moimagesize]{quadriken_hyperbolischer_zylinder}

elliptic cylinder elliptic paraboloid
   
\includegraphics[width=.4\moimagesize]{quadriken_elliptischer_zylinder} \includegraphics[width=.4\moimagesize]{quadriken_paraboloid}

hyperbolic paraboloid parabolic cylinder
   
\includegraphics[width=.4\moimagesize]{quadriken_hyperbolisches_paraboloid} \includegraphics[width=.4\moimagesize]{quadriken_parabolischer_zylinder}


The quadric

$\displaystyle Q: \frac{1}{2}
x^{\operatorname t}\left(\begin{array}{rrr}5 & 4 & 0 \\ 4 & 3 & 4 \\ 0 & 4 & 1\end{array}\right)x+
\left(-2,1,2\right)x+1=0
$

is to put into normal form.

The characteristical polynomial of the corresponding matrix,

$\displaystyle p(\lambda)=\lambda^3-9\lambda^2-9\lambda+81\,,
$

has the zeros

$\displaystyle \lambda_1=9\,,\quad \lambda_2=3\,,\quad \lambda_3=-3\,.
$

A basis of eigenvectors is

$\displaystyle v_9=\left(\begin{array}{r}2\\ 2\\ 1\end{array}\right)\,,\quad
v_3...
...y}\right)\,,\quad
v_{-3}=\left(\begin{array}{r}1\\ -2\\ 2\end{array}\right)\,.
$

We obtain the following orthogonal transformation

$\displaystyle U=\frac{1}{3}\left(\begin{array}{rrr} 2 & -2 & 1\\ 2 & 1 &-2\\ 1&2&2\end{array}\right)\,.
$

Transforming the equation by $ x=Uy$ yields

$\displaystyle \frac{1}{2}
y^{\operatorname t}\left(\begin{array}{rrr}9 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -3\end{array}\right)y+
\left(0,3,0\right)y+1=0\,.
$

Completing squares gives

0 $\displaystyle =$ $\displaystyle 9y_1^2+3y_2^2-3y_3^2+6y_2+2$  
  $\displaystyle =$ $\displaystyle 9y_1^2+3(y_2+1)^2-6y_2-3-3y_3^2+6y_2+2$  
  $\displaystyle =$ $\displaystyle 9z_1^2+3z_2^2-3z_3^2-1$  

or

$\displaystyle \frac{z_3^2}{\frac{1}{3}}-\frac{z_1^2}{\frac{1}{9}}-\frac{z_2^2}{\frac{1}{3}}+1=0\,.
$

Thus, the equation describes a hyperboloid of 1 sheet.


  automatically generated 4/21/2005