Finding Eccentricity from the rotating ellipse formula I see that from a normal ellipse formula, we can acquire the eccentricity via this formula here.
However, for this formula (1):

$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$

When parameter $B = 0$, we would have normal ellipse, and the formula from the link above can be used. 
But when $B ≠ 0$, we will have a tilting ellipse, and its eccentricity will change as well. In fact, if we are to find the current eccentricity of the given formula (1), what would be the formula for the eccentricity in this case?

If $S(p,q)$ be a focus, $L: ax+by+c=0$ be the directrix and $e$ be the eccentricity of a conic with  $P(x,y)$ is any point on the conic,
$$e=\frac{\sqrt{(x-p)^2+(y-q)^2}}{\frac{|ax+by+c|}{\sqrt{a^2+b^2}}}$$
On simplification, $$\{a^2(1-e^2)+b^2\}x^2-2abe^2xy+\{a^2+(1-e^2)b^2\}y^2+()x+()y+()=0$$
Comparing with the given equation, $a^2(1-e^2)+b^2=A, a^2+(1-e^2)b^2=C$ and $-2abe^2=B$
So, we have three unknowns $a,b,e$ with three equations which can be solved easily.
On elimination of $a,b,$ we get $$(B^2-4AC)e^4-4e^2\{B^2+(A-C)^2\}+4\{B^2+(A-C)^2\}=0$$
The value of discriminant is $4^2\{B^2+(A-C)^2\}^2-4(B^2-4AC)4\{B^2+(A-C)^2\}$
$=16\{B^2+(A-C)^2\}\{B^2+(A-C)^2-(B^2-4AC)\}=16(A+C)^2\{B^2+(A-C)^2\}$
So, $e^2=\frac{4\{B^2+(A-C)^2\}\pm4(A+C)\sqrt{B^2+(A-C)^2}}{2(B^2-4AC)}=\frac{2\sqrt{B^2+(A-C)^2}\{\sqrt{B^2+(A-C)^2}\pm(A+C)\}}{B^2-4AC}$ 
For the '-' sign, this is the same value we reached at in the 1st answer as $B^2-4AC=B^2+(A-C)^2-(A+C)^2=\{\sqrt{B^2+(A-C)^2}+A+C\}\{\sqrt{B^2+(A-C)^2}-A-C\}$
