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

Euclidean Normal Form of two-dimensional Quadrics

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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}=0$ point
$ \frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}=0$ intersecting pair of lines
$ \frac{x_1^2}{a_1^2}=0$ coincident lines


central quadrics


normal form name
$ \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$ hyperbola
$ -\frac{x_1^2}{a_1^2}-\frac{x_2^2}{a_2^2}+1=0$ ellipse
$ \frac{x_1^2}{a_1^2}+1=0$ (empty set)
$ -\frac{x_1^2}{a_1^2}+1=0$ parallel pair of lines


parabolic quadrics


normal form name
$ \frac{x_1^2}{a_1^2}+2x_2=0$ parabola


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.

intersecting pair of lines coincident lines
\includegraphics[width=.4\moimagesize]{a_normalform_quadrik_2d_5} \includegraphics[width=.4\moimagesize]{a_normalform_quadrik_2d_6}

hyperbola ellipse
\includegraphics[width=.4\moimagesize]{a_normalform_quadrik_2d_3} \includegraphics[width=.4\moimagesize]{a_normalform_quadrik_2d_1}

parallel pair of lines parabola
\includegraphics[width=.4\moimagesize]{a_normalform_quadrik_2d_4} \includegraphics[width=.4\moimagesize]{a_normalform_quadrik_2d_2}

The equation

$\displaystyle 3x_1^2+3x_2^2+10x_1x_2-12\sqrt{2}x_1-4\sqrt{2}x_2-8=0 $

can be written as quadratic form

$\displaystyle x^{\operatorname t}A x+2b^{\operatorname t}x+c= x^{\operatorname ... ...array}{rr}3 & 5 \\ 5 & 3\end{array}\right)x+ 2\sqrt{2}\left(-6,-2\right)x-8\,. $

In order to put this quadratic form into normal form, first of all, find the eigenvalues and eigenvectors of matrix $ A$. The characteristic polynomial

$\displaystyle (3-\lambda)^2-25=9-6\lambda +\lambda^2-25 =\lambda^2-6\lambda-16 $

has the zeros

$\displaystyle \lambda_{1,2}=\frac{6\pm\sqrt{36+64}}{2}=3\pm5\,. $

For $ \lambda_1=-2$ we obtain the linear system of equations

$\displaystyle \left(\begin{array}{rr}5 & 5\\ 5 &5\end{array}\right)x=0 $

and find the eigenvector $ v_1=(-1,1)^{\operatorname t}$. Since the second eigenvector must be orthogonal to the first one, we obtain $ v_2=(1,1)^{\operatorname t}$. This yields the following orthogonal transformation matrix

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

Transforming the equation by $ x=Uy$ leads to
0 $\displaystyle =$ $\displaystyle y^{\operatorname t}\tilde{A}y+2\tilde{b}^{\operatorname t}y+c$  
  $\displaystyle =$ $\displaystyle y^{\operatorname t}U^{\operatorname t} AU y+2\left(b^{\operatorname t}U\right)y + c$  
  $\displaystyle =$ $\displaystyle -2y_1^2+8y_2^2+8y_1-16y_2-8\,.$  

Completing squares yields

0 $\displaystyle =$ $\displaystyle -2y_1^2+8y_2^2+8y_1-16y_2-8$  
  $\displaystyle =$ $\displaystyle -2(y_1+2)^2-8y_1+8+8(y_2-1)^2+16y_2-8+8y_1-16y_2-8$  
  $\displaystyle =$ $\displaystyle -2z_1^2+8z_2^2-8$  

or

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

Thus the conic section is a hyperbola.

  automatically generated 4/21/2005