How to determine if arbitrary point lies inside or outside a conic Given the general equation of a conic 
$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $, is there a way to determine if an arbitrary point $(x_1,y_1)$ lies inside or outside of the conic (ex. parabola or ellipse)?
 A: Let $g(x,y) = Ax^2+Bxy+Cy^2+Dx+Ey+F$.  I think the condition you want is that $(x_1,y_1)$ is "inside" the conic if and only if
$$ g(x_1,y_1) \quad\text{has the same sign as}\quad
\left|\begin{matrix}
2A & B & D \\
B & 2C & E \\
D & E & 2F
\end{matrix}\right| $$
(If that determinant is zero, the conic is degenerate — see this question — so the idea of "inside" doesn't make sense.)  This works for ellipses and hyperbolas at least, and I think it works for parabolas too, though I don't see how to prove it now.
The idea is that the conic $g(x,y)=0$ divides the plane into two regions $g(x,y)<0$ and $g(x,y)>0$, one of them being "inside" and the other "outside"; to find out which is which, test the centre $(x_c,y_c)$ of the conic (which is inside if it's an ellipse and outside if it's a hyperbola).  It turns out that
$$ \left|\begin{matrix}
2A & B & D \\
B & 2C & E \\
D & E & 2F
\end{matrix}\right|
= 2g(x_c,y_c)
\left|\begin{matrix}
2A & B \\
B & 2C
\end{matrix}\right|
\tag{$\ast$} $$
The determinant on the RHS is the (negative of the) discriminant of the conic; if the conic is an ellipse, that determinant is positive, so $g(x_c,y_c)$ and the determinant on the LHS have the same sign (which is what we want, because the centre is inside), whereas if the conic is a hyperbola then the determinant on the RHS is negative, so $g(x_c,y_c)$ and the determinant on the LHS have opposite signs (which is what we want, because the centre is outside).  
To show ($\ast$), use the fact that $\nabla g(x_c,y_c)=0$; since
$$ \frac{\partial g}{\partial x}(x,y) = 2Ax+By+D
\qquad\text{and}\qquad
\frac{\partial g}{\partial y}(x,y) = Bx+2Cy+E
\text{ ,} $$
we get that
$$ \left[\begin{matrix} 2A & B & D \\ B & 2C & E \end{matrix}\right]
\left[\begin{matrix} x_c \\ y_c \\ 1 \end{matrix}\right]
= \left[\begin{matrix} 0 \\ 0 \end{matrix}\right] $$
and so
$$ g(x_c,y_c) = \frac12
\left[\begin{matrix} x_c & y_c & 1 \end{matrix}\right]
\left[\begin{matrix}
2A & B & D \\
B & 2C & E \\
D & E & 2F
\end{matrix}\right]
\left[\begin{matrix} x_c \\ y_c \\ 1 \end{matrix}\right]
= \frac12 (Dx_c+Ey_c+2F) $$
whence
$$ \left|\begin{matrix}
2A & B & D \\
B & 2C & E \\
D & E & 2F
\end{matrix}\right|
= \det\left(
\left[\begin{matrix}
2A & B & D \\
B & 2C & E \\
D & E & 2F
\end{matrix}\right]
\left[\begin{matrix}
1 & 0 & x_c \\
0 & 1 & y_c \\
0 & 0 & 1
\end{matrix}\right]\right)
= \left|\begin{matrix}
2A & B & 0 \\
B & 2C & 0 \\
D & E & 2g(x_c,y_c)
\end{matrix}\right|
$$
which gives ($\ast$).
