Q:

Find the eccentricity of a conic

Find the eccentricity $e$ of the conic $$S \equiv 39x^2+11y^2-96xy+14x+2y-34=0.$$
My try:
Comparing with general second degree conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ we have
$a=39$, $b=11$, $2h=-96$, $2g=14$, $2f=2$ and $c=-34$
The determinant $$\Delta=\begin{vmatrix}
39 &  -48&7 \\ 
 -48& 11 & 1\\ 
 7&1  & -34
\end{vmatrix} \ne 0$$ and
$h^2-ab \gt 0$, and hence the conic is a Hyperbola. 
Now since there is $xy$ term one way is to rotate the axes to make $xy$ term zero by angle $$\tan(2\theta)=\frac{-96}{28}.$$ But the algebra is very tedious.  Is there any other approach? 

A:

The formula derived in this answer is
$$
e^2=\frac{2\sqrt{(A{-}C)^2+B^2}}{\sigma(A{+}C)+\sqrt{(A{-}C)^2+B^2}}
$$
Setting $A=39,B=-96,C=11,D=14,E=2,F=-34$ gives
$$
\begin{align}
e^2&=\frac{2\cdot100}{-50+100}=4\\
e&=2
\end{align}
$$
since $\sigma=\operatorname{sign}\left(F\!\left(B^2-4AC\right)+\left(AE^2{-}BDE{+}CD^2\right)\right)=\operatorname{sign}\left(-250000\right)=-1$

