Use an orthogonal transformation and a translation to express in standard form the curve given by the equation
You will want to rotate the axes here.
$\displaystyle x = x'~cos(\theta) - y'~sin(\theta)$
$\displaystyle y = x'~sin(\theta) + y'~cos(\theta)$
Then you want to pick a value for $\displaystyle \theta$ such that the x'y' term has a coefficient of 0. Then its just a matter of completing the square on x' and y'.
-Dan