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.
There are really two cases for the non-zero determinant.
Consider $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 + 1 = 0.$
In one case we have
$$\begin{bmatrix}2A& B &D\\B&2C&E\\D&E&2F\end{bmatrix}
= \begin{bmatrix}2 & 0 &0\\0&2&0\\0&0&-2\end{bmatrix}
$$
and in the other we have
$$\begin{bmatrix}2A& B &D\\B&2C&E\\D&E&2F\end{bmatrix}
= \begin{bmatrix}2 & 0 &0\\0&2&0\\0&0&2\end{bmatrix}.
$$
Non-zero determinants both times, but one case is an ellipse (in fact, a circle) and the other is the empty set.
Moreover, $x^2 - 1 = 0$ leads to a zero determinant (the only non-zero entries are $2A$ and $2F$) and it is a pair of parallel lines, but if we just change $F$ from $-1$ to $1$ we get the empty set you exhibited.
I tried to tease apart the cases in a slightly different way in https://math.stackexchange.com/a/2096865
-- in summary, for non-zero discriminant one can transform the equation into one that is symmetric around the origin (and then the sign of the constant term tells whether the solution set is empty),
and for the zero discriminant there is a case that reduces to a quadratic equation in some combination of $x$ and $y$, and the discriminant of that equation tells you whether the solution is two parallel lines, one line (the two parallel lines merged into one), or empty.
Best Answer
There are two ways to prove this:
Formal: You can show, through a bunch of ugly computation, that the expression $B^2-4AC$ is invariant under rotation. So, consider when $B=0$ (in other words, when the conic section's directrix is parallel to one of the axes). It is easy to see that for a hyperbola $-4AC$ is positive, for an ellipse $-4AC$ is negative, and for a parabola $-4AC$ is $0$. For a better worded explanation, go to this link.
Very Informal But Intuitive: Take the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. Imagine if $x$ and $y$ were very large numbers. We can forget about $Dx+Ey+F$ because it becomes insignificant when compared to $Ax^2+Bxy+Cy^2=0$. Now, divide by $x^2$:
$$A+B\left(\frac{y}{x}\right)+C\left(\frac{y}{x}\right)^2=0$$
We notice that we now have a quadratic in $\frac{y}{x}$.
This next part is a little hard to explain in words (and my English is sort of bad) but I will try my best.
The number of solutions to this equation represents the number of ways in which the graph of the equation "zooms off towards infinity." Imagine zooming out really far from a graph of a hyperbola. You would only see an "X" formed by two lines (these lines are the asymptotes of the hyperbola). If you solve for $\frac{y}{x}$ in the above equation, you would be solving for the slopes of those lines. Imagine zooming out really far from a graph of a parabola. You would only see one line (the axis of symmetry for the parabola). If you solve for $\frac{y}{x}$ in the above equation, you would be solving for the slope of that line. If you zoomed out really far from a graph of an ellipse, you would see a point.
So, if $A+B\left(\frac{y}{x}\right)+C\left(\frac{y}{x}\right)^2=0$ has two solutions for $\frac{y}{x}$, the equation is a hyperbola. One solution means parabola. Zero solutions means ellipse or circle. The number of solutions corresponds to the sign of $B^2-4AC$.
I sort of like this informal proof because it explains why the discriminant of a conic looks like that of a quadratic.