<s>[INST] rotation of conic sections In the discriminant test of conic sections(rotations), why we're checking with $B^2-4AC$.
How $B^2-4AC=B'^2-4A'C'$, where $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is changed to $A'x^2+C'y^2+D'x+E'y+F'=0$ using rotations by angle alpha.
 [/INST] We can find from here, the condition for the general 2nd degree equation to represent a pair of straight lines.
If that condition is not satisfied we can apply the following logic.
A conic is the locus of a point $(x,y)$ which maintains constant ratio(called eccentricity$(e)$) of the distances from a fixed point(called focus $(h,k)$ (say),) and a fixed line(called directrix) $lx+my+n=0$(say).
So, $$e=\frac{(x-h)^2+(y-k)^2}{\frac{lh+mk+n}{\sqrt{l^2+m^2}}}$$
On squaring and rearrangement we get,
$$\{m^2+(1-e^2)l^2\}x^2-2lme^2xy+\{l^2+(1-e)^2m^2\}y^2+(..)x+(..)y+(..)=0$$
Comparing with the original equation, 
   $A=m^2+(1-e^2)l^2,B=-2lme^2, C=l^2+(1-e)^2m^2$
So, $B^2-4AC=4(e^2-1)(l^2+m^2)^2$
Now, 
for ellipse $0\le e<1, B^2-4AC=4(e^2-1)(l^2+m^2)^2<0$
for parabola $e=1, B^2-4AC=4(e^2-1)(l^2+m^2)^2=0$
for hyperbola $e>1, B^2-4AC=4(e^2-1)(l^2+m^2)^2>0$
for circle, $A=C\implies m^2+(1-e^2)l^2=l^2+(1-e)^2m^2\implies e=0$,the $xy$ must be absent in the general equation and the focus co0incides with the centre.

Using Rotation of axes,
$A'=A\cos^2\alpha+B\sin\alpha \cos\alpha +C\sin^2\alpha$
$\implies 2A'=(A+C)+(A-C)\cos2\alpha+B\sin2\alpha$
$B'=B(\cos^2\alpha-\sin^2\alpha)-2(A-C)\sin\alpha \cos\alpha=B\cos2\alpha-(A-C)\sin2\alpha$
$C'=A\sin^2\alpha-B\sin\alpha \cos\alpha +C\cos^2\alpha$
$\implies 2C'=A+C-\{B\sin2\alpha+(A-C)cos2\alpha\}$
$2A'\cdot 2C'-B'^2$
$=(A+C)^2-\{B\sin2\alpha+(A-C)cos2\alpha\}^2-\{B\cos2\alpha-(A-C)\sin2\alpha\}^2$
$=(A+C)^2-B^2\{\sin^22\alpha+\cos^22\alpha\}-2B(A-C)\{2\sin2\alpha\cos2\alpha-2\sin2\alpha\cos2\alpha\}-(A-C)^2\{\sin^22\alpha+\cos^22\alpha\}$
$=(A+C)^2-(A-C)^2-B^2$
$$\implies 4A'C'-B'^2=4AC-B^2$$ (This can be established by eliminating $\sin2\alpha, \cos2\alpha$ from the three equations.)
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