Categorizing the conic section given by $x^2-4xy+y^2+8x+2y-5=0$ 
Find the conic section of $x^2-4xy+y^2+8x+2y-5=0$

I came to the following using diagonalization and bi-linear form
$$-\left(x-\frac{5}{\sqrt{2}}\right)^2+\left(\sqrt{3}y-\frac{\sqrt{3}}{\sqrt{2}}\right)^2-19=0$$
So I can conclude (using eigenvalues)that $\lambda_{1}$ have the same sign as $k=-19$ so it is an empty set?
 A: Kindly check the following website about the classification of conic sections.
https://www.ck12.org/analysis/Classifying-Conic-Sections/lesson/Classifying-Conic-Sections-ALG-II/
Using the determinant, this section turns out to be a hyperbola.
You can also classify the conic section by noting that a hyperbola can be represented in either of the following forms.
$$\frac{(x-x_0)^2}{a^2} - \frac{(y-y_0)^2}{b^2} = 1,$$
or
$$\frac{(y-y_0)^2}{b^2} - \frac{(x-x_0)^2}{a^2} = 1.$$
A: First thing is to remove the offset and center the curve on the origin. The easiest way I do this is to take $f(x,y) = x^2-4xy+y^2+8x+2y-5$ and solve the system $$\left. \begin{align} \frac{\partial f}{\partial x} & = 0 \\ \frac{\partial f}{\partial y} & = 0 \end{align} \right\} \begin{aligned} x &= 2 \\ 
y &= 3 \end{aligned}$$
So the center is $(2,3)$ and the centered curve is found with the substitution
$$ \pmatrix{x \\y} \rightarrow \pmatrix{x+2 \\y+3} $$
$$f(x+2,y+3) = x^2 - 4 x y + y^2 -6 = 0$$
Now we need to rotate the coordinates such as to make the coefficient of $x y$ zero.
$$ \pmatrix{x \\y} \rightarrow \pmatrix{x \cos \theta - y \sin\theta \\ x \sin \theta + y \cos \theta} $$ which yields the curve
$$ x^2 +y^2 -2 (x^2-y^2) (\sin 2\theta) -4 x y (\cos 2 \theta) + 6 =0 $$
and the corresponding condition $\cos 2\theta =0 \Rightarrow \theta = \pm \frac{\pi}{4} $. Using the angle $\theta=\frac{\pi}{4}$ the above is
$$ -x^2 +3 y^2 + 6 = 0 $$ or in $a x^2+b y^2 =1$ form:
$$ \boxed{ \frac{1}{6}x^2 - \frac{1}{2}y^2 = 1 }$$
Immediately I see this as a hyperbola laying along the x-axis since $a>0$ and $b<0$. 
