= Solution
The <quadratic form> has <symmetric matrix>
$$
M=\begin{pmatrix}2&2\\2&-1\end{pmatrix},\qquad
\det(M-\lambda I)=\lambda^2-\lambda-6=(\lambda-3)(\lambda+2).
$$
Normalized <eigenvectors> for $3$ and $-2$ are $(2,1)/\sqrt5$ and $(-1,2)/\sqrt5$. They form the columns of the <rotation matrix>
$$
Q=\frac1{\sqrt5}\begin{pmatrix}2&-1\\1&2\end{pmatrix},\qquad
\binom xy=Q\binom{x'}{y'},\qquad
Q^TMQ=D=\boxed{\begin{pmatrix}3&0\\0&-2\end{pmatrix}}.
$$
Thus the new $x'$ axis is rotated counterclockwise by $\theta=\arctan(1/2)$, with
$$
\boxed{\sin\theta=\frac1{\sqrt5},\qquad f=3x'^2-2y'^2.}
$$
The opposite signs of the <eigenvalues> classify each positive level as a <hyperbola>. For the original level six its standard equation is $x'^2/2-y'^2/3=1$; for the final level one it is $3x'^2-2y'^2=1$. The printed change of level therefore changes the scale, but not the principal axes or <asymptotes>.
Setting the homogeneous quadratic part to zero gives $y'=\pm\sqrt{3/2}\,x'$. In the original coordinates put $y=mx$ in $2x^2+4xy-y^2=0$, obtaining
$$
\boxed{m=2\pm\sqrt6.}
$$
These 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.
Back to article page
