 Solution

The quadratic form has symmetric matrixNormalized eigenvectors for $3$ and $-2$ are $(2,1)/\sqrt{5}$ and $(-1,2)/\sqrt{5}$. They form the columns of the rotation matrixThus the new $x^{\prime }$ axis is rotated counterclockwise by $\theta =\arctan (1/2)$, withThe opposite signs of the eigenvalues classify each positive level as a hyperbola. For the original level six its standard equation is $x^{\prime 2}/2-y^{\prime 2}/3=1$; for the final level one it is $3x^{\prime 2}-2y^{\prime 2}=1$. The printed change of level therefore changes the scale, but not the principal axes or asymptotes.

Setting the homogeneous quadratic part to zero gives $y^{\prime }=\pm \sqrt{3/2}{x}^{\prime }$. In the original coordinates put $y=mx$ in $2x^2+4xy-y^2=0$, obtainingThese are the slopes in the original $(x,y)$ axes; $\pm \sqrt{3/2}$ are the slopes in the rotated axes, so they must not be interchanged.