Reduce the equation 3x^2-2xy+3y^2+2x+2y-1=0 to canonical…
Reduce the equation $3x^2-2xy+3y^2+2x+2y-1=0$ to canonical form, and find the centre and eccentricity.
Explanation of Method
A translation and a rotation reduce the given second-degree equation to an ellipse.
Step 1: Centre (translation)
\[ \frac{\partial}{\partial x}:\ 6x-2y+2=0,\qquad \frac{\partial}{\partial y}:\ -2x+6y+2=0\ \Longrightarrow\ x_0=y_0=-\tfrac12. \]
The new free term is $F'=\tfrac12(Dx_0+Ey_0)+F=\tfrac12(-1-1)-1=-2$. With $x=u-\tfrac12$, $y=v-\tfrac12$: $3u^2-2uv+3v^2=2$.
Step 2: Rotation
$A=C$, so $\varphi=45^\circ$. The matrix $\begin{pmatrix}3&-1\\-1&3\end{pmatrix}$ has eigenvalues $2$ (direction $(1,1)$) and $4$ (direction $(1,-1)$): \[u=\frac{u'-v'}{\sqrt2},\quad v=\frac{u'+v'}{\sqrt2}\ \Longrightarrow\ 2u'^2+4v'^2=2.\]
\[u'^2+\frac{v'^2}{1/2}=1,\qquad O'\left(-\tfrac12,-\tfrac12\right),\qquad e=\frac{\sqrt2}{2}\]
Here $a=1$, $b=\tfrac{1}{\sqrt2}$, $c=\sqrt{a^2-b^2}=\tfrac1{\sqrt2}$, $e=\tfrac ca=\tfrac{\sqrt2}2$. The major axis lies on $y=x$, and the foci are $O'\pm c\cdot\tfrac{(1,1)}{\sqrt2}$, i.e. $(0,0)$ and $(-1,-1)$.
Result
\[\boxed{\,u'^2+\frac{v'^2}{1/2}=1,\qquad O'\left(-\tfrac12,-\tfrac12\right),\qquad e=\frac{\sqrt2}{2}\,}\]
Remark
Note: the matrix $P=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ used in the source has $\det P=-1$. It is a reflection (in the line at $22.5^\circ$), not a rotation. The resulting equation is still correct, because a reflection is also an isometry. For a genuine rotation, use $P=\tfrac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ as above.