<s>[INST] 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?
 [/INST] As the translation does not change the eccentricity of a conic, let's us define $x=X+h,y=Y+k$ to get $AX^2+BXY+CY^2=1$ 
Using Rotation of axes as eccentricity is one of the invariants in rotation,
$$x'^2(A\cos^2t+B\sin t\cos t+C\sin^2t)+x'y'(B\cos2t-(A-C)\sin2t)+y'^2(A\sin^2t-B\sin t\cos t+C\cos^2t)=1$$
To remove the $x'y'$ term, $\tan 2t=\frac{B}{A-C}\implies \frac{\sin2t}{B}=\frac{\cos2t}{A-C}=\pm\frac1{\sqrt{B^2+(A-C)^2}}$
If we take the $'+'$ sign, $\sin2t=\frac B{\sqrt{B^2+(A-C)^2}},\cos2t=\frac {A-C}{\sqrt{B^2+(A-C)^2}}$
Comparing with the standard form we get 
$a^2=\frac1{A\cos^2t+B\sin t\cos t+C\sin^2t}$
$=\frac2{A(1+\cos2t)+B\sin2t+C(1-\cos2t)}$
$=\frac{2\sqrt{B^2+(A-C)^2}}{(A+C)\sqrt{B^2+(A-C)^2}+(A-C)^2+B^2}$
$=\frac2{A+C+\sqrt{B^2+(A-C)^2}}$
Similarly, $b^2=\frac2{A+C-\sqrt{B^2+(A-C)^2}}>a^2$
So, $b$ is the semi-major axis, $a$  is the semi-minor axis.
Consequently, the equation of the conic becomes $$\frac{x'^2}{\frac2{A+C+\sqrt{B^2+(A-C)^2}}}+\frac{y'^2}{\frac2{A+C-\sqrt{B^2+(A-C)^2}}}=1$$ 
$e^2=1-\frac{a^2}{b^2}=1-\frac{A+C-\sqrt{B^2+(A-C)^2}}{A+C+\sqrt{B^2+(A-C)^2}}=\frac{2\sqrt{B^2+(A-C)^2}}{A+C+\sqrt{B^2+(A-C)^2}}$
If $A+C-\sqrt{B^2+(A-C)^2}=0$ i.e., $4AC=B^2,e^2=1,$ the conic becomes a Parabola.
If $A+C-\sqrt{B^2+(A-C)^2}>0$ i.e., $4AC>B^2,e^2<1,$ the conic becomes an ellipse.
If $A+C-\sqrt{B^2+(A-C)^2}<0$ i.e., $4AC<B^2,e^2>1,$ the conic becomes a hyperbola.
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