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 \(a=0\).)
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