Any second-degree curve equation can be written as
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}$$ or $$ax^2+2hxy+by^2+2gx+2fy+c=0\tag{2}$$ where $$A,B,C,D,E,E,a,b,c,f,g,h\in\mathbb R$$
To find type of conic and nature of conic we use $\Delta$ and is given by $$\Delta=\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}$$ $$=abc+2fgh-af^2-bg^2-ch^2\tag{3}$$
If $\Delta$ is $0$, it represents a degenerate conic section. Otherwise, it represents a non-degenerate conic section.
Also, the type of conic section that the above equation represents can be found using the discriminant of the equation, which is given by $B^2-4AC$.
Conditions regarding the quadratic discriminant are as follows:
If $\Delta=0$:
$\bullet$ If $h^2-ab\gt0$, the equation represents two distinct real lines.
$\bullet$ If $h^2-ab=0$, the equation represents parallel lines.
$\bullet$ If $h^2-ab\lt0$, the equation represents non-real lines.
If $\Delta\neq0$:
$\bullet$ If $B^2-4AC\gt0$, it represents a hyperbola and a rectangular hyperbola $(A+C=0)$.
$\bullet$ If $B^2-4AC=0$, the equation represents a parabola.
$\bullet$ If $B^2-4AC\lt0$, the equation represents a circle $(A=C,B=0$) or an ellipse $(A\neq C)$. For a real ellipse, $\Big(\frac\Delta{a+b}\lt0\Big)$.
So for the given case, the equation of conic is (after putting all the points in the general equation of conic and finding all the coefficient) $$ax^2-axy+4ay^2-4a=0$$
Comparing above equation with equation (1) and (2) we get the values of required coefficient as $A=a$, $B=-a$ and $C=4a$ so we get the value of $B^2-4AC=-3a^2$ and also from equation (3) we get $\Delta=-15a^2$. Since $\Delta\neq0$ and $B^2-4AC\lt0$ and also $A\neq C$, so as we know, this is the condition of an ellipse.
$\therefore$ Conic $ax^2-axy+4ay^2-4a=0$ is a real ellipse.
In fact, a conic has 4 foci. We can see this if we look at a canonical ellipse,which is wide and short, and start making it smaller in the direction of the x-axis. The two foci get closer, until we reach a circle when they collapse to one point. Then, if we continue they start to have a different trajectory - up and down. This cannot be, and what really happens is that two foci escape to the complex part of the plane, while the other two arrive from it.
A purely geometro algebraic definition of foci (well not exactly, because it depends on the projective coordinates you chose!) is the followin:
take the points I=[1:i:0] and J=[1:-i:0], and a conic C. By bezout's theorem there are two lines through I that are tangent to C and likewise with J. The two pair of lines intersect at 4 points, which are precisely the foci.
Remark: The fact that there are two tangents from a point to C does not follow immediately from Bezout's, "tangency" is not a linear condition. However, it does follow if first we use projective duality with respect to C.
Best Answer
$2Ax+By+D=0,By+2Cy+E=0$ has solution $(x,y)=\left(\frac{2CD-BE}{B^2-4AC},-\frac{BD-2AE}{B^2-4AC}\right)$,
provided $B^2-4AC\not=0$, thus the conic should not be a parabola. Geometrically, the formula breaks down for a parabola because it intends to identify the center of the conic but the center of a parabola does not exist.
Then substituting $(x,y)\to \left(x+\frac{2CD-BE}{B^2-4AC},y-\frac{BD-2AE}{B^2-4AC}\right),$ we get $$A\left(x+\frac{2CD-BE}{B^2-4AC}\right)^2+B\left(x+\frac{2CD-BE}{B^2-4AC}\right)\left(y-\frac{BD-2AE}{B^2-4AC}\right)+C\left(y-\frac{BD-2AE}{B^2-4AC}\right)^2+D\left(x+\frac{2CD-BE}{B^2-4AC}\right)+E\left(y-\frac{BD-2AE}{B^2-4AC}\right)+F=0$$ which expands to $$Ax^2+Bxy+Cy^2+F+\frac{-AE^2+BDE-CD^2}{4AC-B^2}=0$$