Question 1066181
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The leading (highest degree terms) are


8x^2 - 16xy + 8y^2 + . . . = 8(x-y)^2 + . . . 


After orthogonal transformation u = {{{(x+y)/sqrt(2)}}}, v = {{{(x-y)/sqrt(2)}}} you will get this curve in the form 


v^2 + {lover degree terms} = 0,


which is a parabola.