<s>[INST] Canonical form of a quadratic form 
Find the canonical form of $2x^2+y^2+3y=0$

What are the way to approach this? 
I know that the eigenvalues are positive and that
$$2x^2+y^2+3y=0\iff2x^2+(y+\frac{3}{2})^2-\frac{9}{4}=0$$
For: $x'=2x$, $y'=y+\frac{3}{2}$
We get: $$x'^2+y'^2-\frac{9}{4}=0$$
How do I categorize this quadratic form?
 [/INST] As you noticed the equation 
$$2x^2+y^2+3y=0\tag{1}$$
is equivalent to
\begin{equation*}
2x^{2}+\left[ y-\left( -\frac{3}{2}\right) \right] ^{2}-\left( \frac{3}{2}
\right) ^{2}=0.\tag{2}
\end{equation*}
Dividing by $(3/2)^2 $ and writting $2$ as $1/(\sqrt{2}/2)^2$, we obtain
\begin{equation*}
\frac{x^{2}}{\left( \frac{3}{4}\sqrt{2}\right) ^{2}}+\frac{\left[ y-\left( -\frac{3}{2}\right) \right] ^{2}}{\left( \frac{3}{2}\right) ^{2}}=1,\tag{3}
\end{equation*}
which is the equation of the shifted ellipse centered at $\left( h,k\right)
=\left( 0,-\frac{3}{2}\right) $ and semiaxes $a=\frac{3}{4}\sqrt{2}$ and $ b=\frac{3}{2}>a$:
\begin{equation*}
\frac{\left( x-h\right) ^{2}}{a^{2}}+\frac{\left( y-k\right) ^{2}}{b^{2}}=1.\tag{4}
\end{equation*}
$\qquad\quad\qquad\quad\qquad\quad$ 
Comments:


*

*The general equation of a conic is
\begin{equation*}
Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0
\end{equation*}
If $B^{2}-4AC<0$, the conic is an ellipse. Equation $(1)$ is the particular case
\begin{equation*}
A=2,\quad C=1,\quad E=3,\quad B=D=F=0;
\end{equation*}
with $B^{2}-4AC=-4\times 2\times 1=-8<0$.

*You only get 
$$x'^2+y'^2-\frac{9}{4}=0$$
for $x'=\sqrt{2}x$, $y'=y+\frac{3}{2}$.

</s>