Find the isometry that takes the conic section
$\qquad 3x^2-4xy+3y^2+\sqrt2 x+\sqrt2 y = 0 \quad (*)$
to standard form.
Solution. First, $3x^2-4xy+3y^2=\pmatrix{x & y}A \pmatrix{x \\ y}$, where $A =\pmatrix{3 & -2 \\ -2 & 2}$.
$A$ has eigenvalues $1,5$ and corresponding eigenvectors $\pmatrix{1 \\ 1},\pmatrix{1 \\ -1}$.
Let $\pmatrix{x \\ y}=Q\pmatrix{X \\ Y}$, where $Q =\frac{1}{\sqrt{2}}\pmatrix{1 & 1 \\ 1 & -1}$. Then, the LHS of $(*)$ becomes
$3x^2-4xy+3y^2+\sqrt2 x+\sqrt2 y$
$=\pmatrix{X & Y}Q^TAQ \pmatrix{X \\ Y} + 2X$
$=X^2+5Y^2+2X$
...
I don't quite understand the particular choice of $Q =\frac{1}{\sqrt{2}}\pmatrix{1 & 1 \\ 1 & -1}$.
From the statement $\pmatrix{x \\ y}=Q\pmatrix{X \\ Y}$, $Q$ could be any $2\times 2$ matrix whose image is a superset of the conic section (?).
It seems that we specifically choose $Q$ to be orthogonal in order for it to be an isometry, and its columns are the eigenvectors of $A$, which makes matrix multiplications easier when $A$ is diagonalised.
Also, the conic's equation is an ellipse slanted at a $45$-degree angle to the $x$-axis, and I noticed that the columns of $Q$ are the standard basis vectors of $\mathbb R^2$ rotated by exactly that angle. But I'm not sure if it's some sort of rotation or just a coincedence.
What exactly is this method doing to the conic section?
Would some other orthogonal (or even non-orthogonal, if we didn't require an isometry) choice of $Q$ also work here?