# 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!

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Care to explain why you think they should be the same? – nbubis Apr 13 '12 at 15:57
Oh I thought the reduction of a quadratic form to one which had a diagonal matrix in an orthonomal basis gave essentially the same quadratic form just in a 'nicer' way? – user26069 Apr 13 '12 at 16:11

Suppose $x$ is a point on $x^T A x = 2$. Then, $x^T P P^T A P P^T x = 2$ $\Rightarrow x^T P B P^T x=2$. Thus the curve $x^T B x =2$ is just the curve $x^T A x = 2$ rotated by $P^T$.