Canonical form of a quadratic form

Question

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?

Answer 1

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:

1. 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 $B^{2}-4AC=-4\times 2\times 1=-8<0$.
2. You only get $$x'^2+y'^2-\frac{9}{4}=0$$ for $x'=\sqrt{2}x$, $y'=y+\frac{3}{2}$.