Demonstration of an imaginary ellipse

Question

I'm studying the classfication of non-degenerate conics based on the determinant of the matrix associated to it $A_Q$. I didn't understand, in the case of a non-degenerate ellipse (det $A_{33}>0$, minor of $A_Q$, and det $A_Q \neq 0$) why we have an imaginary ellipse if $(A+C)detA_Q > 0$. I tried looking to some examples, also from Wikipedia, $x^2+y^2+10=0$ has no real solutions. How can I demonstrate that $\{(x,y)^T \in \mathbb{R}^2 : ax^2 + bxy + cy^2 + dx + ey + f = 0\}$ is empty?

Answer 1

When $D=b^2-4ac\neq 0$, you can translate the center $(\frac{2cd-be}{b^2-4ac},-\frac{bd-2ae}{b^2-4ac})$ to the origin without changing the form of the conic:

$$ax^2+bxy+cy^2+f+\frac{-ae^2+bde-cd^2}{4ac-b^2}=0\tag1$$ or $$ax^2+bxy+cy^2+\frac{\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{vmatrix}}{\begin{vmatrix}a&b/2\\b/2&c\end{vmatrix}}=0\tag2$$

(For $D>0,$ equation $(1)$, hence the original equation, represents a hyperbola (or a pair of lines when $f+\frac{-ae^2+bde-cd^2}{4ac-b^2}=0$) since then $ax^2+bxy+cy^2=a(x+y(b+\sqrt{D})/2a)(x+y(b-\sqrt{D})/2a);$ compare with $xy=k.$)

When $D<0,$ equation$(1)$ can represent an ellipse and $ax^2+bxy+cy^2=a(x+y(b+i\sqrt{-D})/2a)(x+y(b-i\sqrt{-D})/2a)$ which are complex lines that meet in the real point $(0,0).$ or in other words $ax^2+bxy+cy^2=0$ is the origin (or in the original conic just the center). So if in addition to $D<0$ $(a+c)>0$ then $f+\frac{-ae^2+bde-cd^2}{4ac-b^2}>0$ gives $ax^2+bxy+cy^2$ positive and there are no real points. And if $(a+c)<0$ then $f+\frac{-ae^2+bde-cd^2}{4ac-b^2}<0$ gives $ax^2+bxy+cy^2$ negative and there are no real points.

So the condition $\operatorname{trace}(\begin{pmatrix}a&b/2\\b/2&c\end{pmatrix})\cdot\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{vmatrix}>0$ is the correct one.