Classification Of Conic Section 
Classify $$x^2-4xy+y^2+8x+2y-5=0$$

So the eigenvalues are $\lambda_{1}=3,\lambda_{2}=-1$ so the eigenvectors are $v_{1}=\begin{pmatrix} \frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix}$ and $v_{2}=\begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix}$
So We have $$\begin{pmatrix} x &y \end{pmatrix}\begin{pmatrix} 3 &0\\ 0&-1\end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}+\begin{pmatrix} 8 &2 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}-5=0$$
$$3x^2-y^2+\frac{6}{\sqrt{2}}x+\frac{10}{\sqrt{2}}y-5=0$$
$$3(x+\frac{1}{\sqrt{2}})^2-\frac{3}{2}-(y-\frac{5}{\sqrt{2}})+\frac{25}{2}-5=0$$
$$3x''^2-y''^2=-6$$
$$-\frac{1}{2}x''^2+\frac{1}{6}y''^2=1$$ Which is?
But in the answer they got 
$$\frac{1}{6}x''^2-\frac{1}{2}y''^2=1$$ which is hyperbola as it is in the form of $\frac{x''^2}{a^2}-\frac{y''^2}{b^2}=1$

You don't have to find the reduced equations to classify the conic:


*

*$\;\begin{vmatrix}1&-2&4\\-2&1&1\\4&1&-5\end{vmatrix}\ne 0$, so the conic is non-degenerate.

*The quadratic form $x^2-4xy+y^2$ has signature $(1,1)$:
$$x^2-4xy+y^2=(x-2y)^2-3y^2.$$


As a conclusion, the conic is a hyperbola. 
For an ellipse, the signature would be $(2,0)$.
