# A quadric depending on a parameter

The picture shows the quadric with equation $$2x_1^2 + ax_1x_2 + x_2^2 = 1$$ (in red), with the principal axes (in blue).

You may adjust the parameter $$a$$ by dragging it with your mouse.

The quadric will be an ellipse as long as $$|a|<\sqrt{8}$$.
If $$|a|$$ reaches $$\sqrt{8}$$ (that's the point marked in dark violet) then the quadric degenerates to a pair of lines.
If $$|a|$$ exceeds $$\sqrt{8}$$ then the quadric becomes a hyperbola; the asymptotes will be shown in yellow.

Using the symmetric matrix $$A = \left[\begin{matrix} 2 & a/2 \\ a/2 & 1 \end{matrix}\right]$$ and $${\bf x} = \left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right]$$ we can write the equation as $${\bf x}^T A \, {\bf x} = 1$$.

The matrix $$A$$ has characteristic polynomial $$\lambda^2 - 3\lambda +2 -a^2/4$$; this gives the eigenvalues $$\lambda_1 = (3+\sqrt{1+a^2})/2$$ and $$\lambda_2 = (3-\sqrt{1+a^2})/2$$. We abbreviate $$w = \sqrt{1+a^2}$$ (this value is also shown on the slider that you use to adjust $$a$$ in the picture).

If $$a=0$$ then $$A$$ is a diagonal matrix, and the equation is easy to understand (actually, our applet does not work well for $$a=0$$ because a division by $$a$$ is involved in the geometric construction behind the screen).

The eigenspaces are $$E_{\lambda_1} = \text{span}\left(\left[\begin{matrix} a \\ w-1 \end{matrix}\right]\right)$$ and $$E_{\lambda_2} = \text{span}\left(\left[\begin{matrix} 1-w \\ a \end{matrix}\right]\right)$$ (orthogonal to $$E_{\lambda_1}$$). Using $$N = \sqrt{a^2+(w-1)^2} = \sqrt{2w(w-1)}$$ we find the orthonormal basis $${\cal{B}} = \left\{ \frac1N \left[\begin{matrix} a \\ w-1 \end{matrix}\right], \frac1N \left[\begin{matrix} 1-w \\ a \end{matrix}\right] \right\}$$ and the orthogonal matrix $$Q = \frac1N \left[\begin{matrix} a & 1-w \\ w-1 & a \end{matrix}\right]$$ which in fact describes a rotation.

In coordinates $${\bf y} = \left[\begin{matrix} y_1 \\ y_2 \end{matrix}\right]$$ with respect to our new basis (i.e., for $${\bf x} = Q{\bf y}$$), the equation of the quadric becomes easy: we have $$\lambda_1y_1^2+\lambda_2y_2^2 = 1$$ because $${\bf x}^T A\, {\bf x} = {\bf y}^T (Q^T A\, Q)\, {\bf y}$$ and $$Q^T A\, Q = \left[\begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix}\right]$$ is a diagonal matrix.

Whether this quadric is an ellipse or a hyperbola is decided by the signs of the eigenvalues: while $$\lambda_1$$ is clearly positive for any choice of $$a$$, the second eigenvalue $$\lambda_2$$ may be negative. This happens if, and only if, we have $$|a|>\sqrt{8}$$. The boundary case $$a=\sqrt{8}$$ leads to the equation $$3y_1^2 = 1$$ which describes two lines parallel to the $$y_2$$-axis, at distance $$1/\sqrt3$$ from that axis.

You can see it all in the picture, go ahead and play with the value of $$a$$.

(Just in case you wondered whether $$N=0$$ is possible: this only occurs if $$a=0$$. In that case, the matrix $$A$$ is diagonal, anyway; the eigenspaces are the standard axes. In our computation of the eigenspaces above, we actually would perform division by $$0$$ if $$a=0$$.)

A remark on printing: If you try to print the current page directly from your browser, you will probably end up with an empty grey rectangle instead of the picture. Use the PDF-file or take a screen shot.

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