Conic sections in standard form I'm trying to convert the equation
$$x^2 +2y^2 +4x-4y+4=0$$
into its standard form by choosing a new set of axes.
Yet, when I go down the conventional route, there is no xy term so $$cot2{\theta}={(a-c)/{b}}$$ doesn't work. 
I've simplified it but it turns out as 
$$(x+2)^2 +2(y-1)^2 =1$$ 
So, it is a circle in standard form already?

Note that 
$$(x+2)^2+2(y-1)^2=1$$
is not correct and that it is not a circle.
We have
$$x^2 +2y^2 +4x-4y+4=0$$
$$\iff (x^2+4x+4)+2(y^2-2y+1)-2=0$$
$$\iff (x+2)^2+2(y-1)^2=2$$
$$\iff \frac{(x+2)^2}{\left(\sqrt 2\right)^2}+\frac{(y-1)^2}{1^2}=1$$
which is an ellipse.
