Transform 2x^2+4xy-y^2=12 to canonical form by rotating the…

Transform $2x^2+4xy-y^2=12$ to canonical form by rotating the axes, and sketch the curve.

Step 1: Eigenvalues

\[ M=\begin{pmatrix}2&2\\2&-1\end{pmatrix},\qquad \lambda^2-\lambda-6=0\ \Longrightarrow\ \lambda_1=3,\ \lambda_2=-2. \]

The eigenvectors are $(2,1)$ for $\lambda_1$ and $(-1,2)$ for $\lambda_2$.

Step 2: Rotation

We have $\cos\varphi=\tfrac2{\sqrt5}$ and $\sin\varphi=\tfrac1{\sqrt5}$, so $\varphi=\arctan\tfrac12\approx26.57^\circ$: $x=\frac{2x'-y'}{\sqrt5},\qquad y=\frac{x'+2y'}{\sqrt5}$.

Then $2x^2+4xy-y^2=3x'^2-2y'^2$ (the $x'y'$ term cancels), so

\[3x'^2-2y'^2=12\quad\Longleftrightarrow\quad\frac{x'^2}{4}-\frac{y'^2}{6}=1\]

Step 3: Elements

A hyperbola with the following elements:

• $a=2$, $b=\sqrt6$, $c=\sqrt{10}$ and $e=\tfrac{\sqrt{10}}2$;
• vertices $\pm\left(\tfrac4{\sqrt5},\tfrac2{\sqrt5}\right)$;
• foci $\pm\sqrt{10}\,\tfrac{(2,1)}{\sqrt5}=\pm(2\sqrt2,\sqrt2)$.

The asymptotes come directly from the quadratic part: \[2x^2+4xy-y^2=0\ \Longrightarrow\ y^2-4xy-2x^2=0\ \Longrightarrow\ y=(2\pm\sqrt6)\,x .\]

This agrees with the source's $\tfrac{2(1-\sqrt6)}{4+\sqrt6}=2-\sqrt6$ and $\tfrac{2(1+\sqrt6)}{4-\sqrt6}=2+\sqrt6$, simplified.