Reduce to its canonical form the conic $\Gamma:=x^2+4xy+4y^2+2x-4y$ I have to reduce to its canonical form the conic $\Gamma:=x^2+4xy+4y^2+2x-4y$.
Studying the orthogonal Invariants of the matrix $A$=$\begin{pmatrix}
1 & 2 & 1\\
2 & 4 & -2\\
1 & -2 & 0
\end{pmatrix}$ I found that $\Gamma$ is a parabola. The problem is that I'm not able to find a translation $t_{\underline{v}}$ : $x=x'+h$ , $y=y'+k $ in order to get rid of the linear terms in $x$ and $y$ (I get a system with no solutions).
Also I don't understand why, differently than other exercises I did before this one, (I had to find the canonical form of an ellipse and a hyperbole) when I try to use the method of eigenvalues, studying  the submatrix $Q$, I find $0$ and $5$ as eigenvalues, so the determinant of the diagonal matrix is zero but the determinant of my initial matrix $A$ is $-16$, so I can't write $\Gamma$ in the form $\alpha x^2+\beta y^2 +\gamma=0.$
Am I missing something, or doing any stupid mistake?
Thank you in advance.
 A: The equation can be written $(x+2y)^2+2x-4y=0,$ but this only gives you the direction of the axis: $x+2y=0$ is parallel to the axis. 
To get the axis-vertex tangent form that e.g. $y=x^2$ is in (axis $x=0$, vertex tangent $y=0$), you can write the equation as $(x+2y+c_1)^2=c_2(2x-y+c_3),$ where we choose $2x-y+c_3$ since it's perpendicular to $x+2y+c_1.$ Now comparing coefficients with the original equation we get $(x+2y-\frac35)^2+\frac85(2x-y-\frac9{40})=0.$
Now to get a coordinate change that conserves distances ($\sqrt{5}=\sqrt{1^2+2^2}$), we need $x'=\frac{x+2y-\frac35}{\sqrt{5}}$ and $y'=\frac{2x-y-\frac9{40}}{\sqrt{5}},$ and this gets the final $y'=-\frac{5\sqrt{5}}{8}x'^2.$
A: To answer your last question first, a parabola is a different sort of beast from an ellipse or hyperbola. Its canonical equation is $Y=X^2$: you can’t eliminate the linear terms as you can for those other types of conics, and there’s only one second-degree term. So, trying to eliminate those linear terms was doomed to failure, as you’ve found. Also, $\det Q=0$ and $\det A\ne0$ identifies this conic as a parabola.  
Not all is lost, though. I would proceed as Jan-Magnus Økland did, by writing the quadratic part as the square of a linear term—indeed, $\det Q=0$ tells you that it is a perfect square—and then computing the three constants to put the equation in the form $(x-2y-c_1)^2=c_2(2x-y+c_3)$. The isometry that you’re looking for can be extracted from that—the parenthesized terms are equations of the parabola’s axis and tangent at the vertex, respectively.  
However, since you’ve already computed eigenvectors and eigenvalues of the submatrix $Q$, you can work out the rotation and translation directly. The eigenvectors still give you the directions of the principal axes, and the one with an eigenvalue of zero corresponds to the parabola’s axis direction. No doubt you’ve found that $(-2,1)$ is in the null space of $Q$, so the rotation $$R=\frac1{\sqrt5}\begin{bmatrix}1&2\\-2&1\end{bmatrix}$$ will make the parabola’s axis vertical. Applying this rotation and rearranging things a bit gives us the equation $$y' = \frac58\sqrt5x'^2-\frac34x'-1.$$ I trust that you know how to find the vertex of this parabola for the remaining translation.
You could instead find the vertex first: From the already-computed axis direction, the equation of the tangent at the vertex has the form $2x-y+c=0$. Find a $c$ such that this line intersects the untransformed parabola in exactly one point—that point is its vertex.
