Reduction of a $2$ dimensional quadratic form I'm given a matrix $$A = \begin{pmatrix}-2&2\\2&1\end{pmatrix}$$ and we're asked to sketch the curve $\underline{x}^T A \underline{x} = 2$ where I assume $x = \begin{pmatrix}x\\y \end{pmatrix}$. Multiplying this out gives $-2 x^2+4 x y+y^2 = 2$. 
Also, I diagonalised this matrix by creating a matrix, $P$, of normalised eigenvectors and computing $P^T\! AP = B$. This gives $$B = \begin{pmatrix}-3&0\\0&2\end{pmatrix}$$ and so now multiplying out  $\underline{x}^T B \underline{x} = 2$ gives $-3x^2 + 2y^2 = 2$.
Plugging these equations into Wolfram Alpha gives different graphs, can someone please explain what I'm doing wrong? Thanks! 

A quadratic form is "equivalent" to it's diagonalized form only in the sense that these quadratic forms share an equivalence class, but the resulting polynomials are of course different (as you showed), otherwise we wouldn't need the equivalence classes!
