How to turn the ellipse $x^2 - xy + y^2 - 3y - 1 = 0$ to the canonical form using an isometric transformation? There is an exam problem I'm having trouble with, it is as follows:

Turn the equation $x^2 - xy + y^2 - 3y -1 = 0$ into the canonical form using an isometric transformation and write down the transformation.

Comparing to the general equation for conic sections I see that this is a rotated ellipse. I assume that the solution would be to multiply a rotation matrix (to align the ellipse's axes to the $x$ and $y$ axes) and a translation matrix (to bring the ellipse's center to the origin), and then somehow apply the result to the ellipse.
How would I go about doing this? How do I get the coordinates of the ellipse's center and the angle of its rotation to construct the matrices, and then how would I apply the transformation to the ellipse itself?
 A: First center the equation by definting the function $f(x,y) = x^2 - xy + y^2 - 3y - 1 = 0$ and solving the following system of equations for $(x,y)$
$$\left. \begin{aligned}
  \frac{\partial f(x,y)}{\partial x} & = 2x-y = 0 \\
  \frac{\partial f(x,y)}{\partial y} & = -x+2y-3 = 0 \\
\end{aligned} \right\} \pmatrix{x & y} = \pmatrix{1 & 2} $$
So now we reset the coordinates to $\pmatrix{x&y} \rightarrow \pmatrix{x+1&y+2}$
The new equation is $g(x,y) = x^2-x y + y^2 -4 = 0$
Now to re-orient the conic. Use $\pmatrix{x&y} \rightarrow \pmatrix{x \cos \theta - y \sin \theta & x \sin\theta + y \cos \theta}$ and set the coefficient of $x y$ equal to zero. 
$$ g = \left( 1- \frac{\sin 2 \theta}{2} \right) x^2 + \left( 1 + \frac{\sin 2 \theta}{2} \right) y^2 +2 \left( -\frac{\cos 2\theta}{2} \right) x y -4 =0 $$
$$ \left. -\frac{1}{2}\cos(2 \theta)=0 \right\} \theta =\pm \frac{\pi}{4} $$
$$\boxed{ d(x,y) = \frac{1}{2} x^2 + \frac{3}{2} y^2  -4 =0 }$$
A: To keep it understandable albeit inelegant I'll do the passage in two steps
First the translation
$x^2 - xy + y^2 - 3y -1 = 0$
we look for a new centre $(h;\;k)$ so we substitute $x=x'+h;\;y=y'+k$
$-(h+x) (k+y)+(h+x)^2+(k+y)^2-3 (k+y)-1=0$
$x'^2-x' y'+y^2+x' (2 h-k)+y' (-h+2 k-3) +h^2-h k+k^2-3 k-1=0$ 
To have no first degree terms we put $2h-k=0;\;-h+2 k-3=0$ which gives
$h=1;\;k=2$ and we plug these values in the previous equation
$x'^2 - x' y' + y'^2=4$
Now we want to get rid of the $x'y'$ term. To do so we have to rotate the axis using these equations
$\begin{aligned}x'&=X\cos \theta -Y\sin \theta \\y'&=X\sin \theta +Y\cos \theta \end{aligned}$
$(X \sin \theta+Y \cos \theta)^2-(X \cos \theta-Y \sin \theta) (X \sin \theta+Y \cos \theta)+(X \cos \theta-Y \sin \theta)^2=4$
collecting terms
$X^2 \left(\sin ^2\theta+\cos ^2\theta-\sin \theta \cos \theta\right)+X Y \left(\sin ^2\theta-\cos ^2\theta\right)+Y^2 \left(\sin ^2\theta+\cos ^2\theta+\sin \theta \cos \theta\right)=4$
as we want the term $XY$ off we set $\sin ^2\theta-\cos ^2\theta=0$
$\tan^2\theta=1\to \theta=\pm \dfrac{\pi}{4}$
if we want the major axis of the ellipse to be horizontal we choose $\theta=\dfrac{\pi}{4}$ and substitute this value in the last equation
$\dfrac{X^2}{8}+\dfrac{3 Y^2}{8}=1$
the equations of the roto-translation altogether are 
$\begin{aligned}x&=X\cos \frac{\pi}{4}-Y\sin \frac{\pi}{4} +1\\y&=X\sin \frac{\pi}{4}+Y\cos \frac{\pi}{4} +2\end{aligned}$
or
$\begin{aligned}x&=\frac{\sqrt 2}{2}(X-Y) +1\\y&=\frac{\sqrt 2}{2}(X+Y) +2\end{aligned}$
hope this helps
A: You have to find an orthonormal basis of eigenvectors associated to the matrix of the quadratic part of the equation:
$$Q=\begin{bmatrix}
1& -\dfrac12\\ -\dfrac12&1
\end{bmatrix}$$
The characteristic polynomial is $\;\chi_Q(\lambda)=(\lambda-1)^2-\dfrac14$, whence the eigenvalues $\lambda_1=\dfrac12$, $\;\lambda_2=\dfrac32$, and the corresponding eigenvectors $(1,1)$ and $(-1,1)$.  
Normalise these vectors to obtain an orthonormal basis, and the change of basis matrix
$$u=\biggl(\dfrac1{\sqrt 2},\dfrac1{\sqrt 2}\biggr),\qquad v=\biggl(-\dfrac1{\sqrt 2},\dfrac1{\sqrt 2}\biggr),\qquad P=\begin{bmatrix}
\dfrac1{\sqrt2}& -\dfrac1{\sqrt2}\\ \dfrac1{\sqrt2}&\dfrac1{\sqrt2}
\end{bmatrix}.$$
Usibg this matrix, we obtain the equation of the conic in the new coordinate system $(x',y')$:
$$\frac12x'^2+\dfrac32y'^2-\dfrac3{\sqrt2}(x'+y')=1$$
Let's centre this equation rewriting it as
\begin{align}
&\phantom{\iff}\;\frac12\bigl(x'^2-3\sqrt 2x'\bigr)+\frac32\bigl(y'^2-\sqrt2 y')=1\\
&\iff\;\frac12\biggl(x'-\frac{3\sqrt 2}2\biggr)^2+\frac32\biggl(y'-\frac{\sqrt 2}2\biggr)^2=1+\frac94+\frac34=4
\end{align}
So, setting 
\begin{cases}
X=x'-\dfrac{3\sqrt2}2=\dfrac{\sqrt2}2(x+y-3),\\
Y=y'-\dfrac{\sqrt 2}2=\dfrac{\sqrt2}2(-x+y-1),
\end{cases}
we obtain the equation in canonical form:
$$\frac{X^2}8+\frac{3Y^2}8=1.$$
A: $$\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2-3y-1=0$$ or
$$\left(x-\frac{y}{2}\right)^2+3\left(\frac{y}{2}-1\right)^2=4$$ or
$$\frac{\left(x-\frac{y}{2}\right)^2}{4}+\frac{\left(\frac{y}{2}-1\right)^2}{\frac{4}{3}}=1.$$
Now, let $x'=x-\frac{y}{2}$ and $y'=\frac{y}{2}-1$...
