An example of a degenerate conic is a pair of intersecting lines. You get this when the plane cutting the cone passes through the vertex of the cone.
e.g. x + 3y -2 = 0 and 2x - y + 1 = 0
Multiply those two expressions together:
(x + 3y -2)(2x - y + 1) = 0
2x^2 - 3y^2 + 5xy - 3x + 5y - 2 = 0
If you look at
http://mathworld.wolfram.com/ConicSectionDiscriminant.html
and replace x1 by x, x2 by y, and x3 by 1, you have the form you want, and the discriminant described there is
abc + 2fg - af^2 - bg^2 - ch^2, where for the example I've just given
a = 2, b = -3, c = -2, h = 5/2, g = -3/2, f = 5/2 and so
2*(-3)*(-2) + 2*(5/2)*(-3/2) - 2*25/4 + 3*9/4 + 2*25/4
= 11.25
I thought I remembered that for a degenerate conic the discriminant is zero, but obviously this is not so.
For a circle, a = b = 1, c = -(radius)^2, h=g=f=0
and so discriminant = -r^2. For a general circle,
b = a and h = 0 and g^2 + f^2 > (a^2)c (otherwise it's degenerate), and discriminant = (a^2)c + 2fg - af^2 - ag^2
For the central ellipse
(x^2)/(a^2) + (y^2)/(b^2) = 1,
we have to replace a by b^2, b by a^2, c by (a^2)(b^2), and
g = f = h
and discriminant
= (a^4)(b^4)
Another example of a degenerate conic occurs when the plane not only passes through the vertex of the cone, but also is tangential to the cone. The section is then not a pair of lines, but one line. You could see the form of the equation by squaring the general linear expression:
e.g.
(2x - 4y - 5)^2
= 4x^2 + 16y^2 - 16xy - 20x -40y + 25
Thus a = 4, b = 16, h = -8, g = -10, f = -20, c = 25
and the discriminant is
4*16*25 + 2*(-20)*(-10) - 4*20^2 - 16*10^2 - 25*8^2
= -2800
I don't remember a rule of thumb for determining from the equation what sort of conic section it is. There's a matrix method, but it takes a while.
re YOUR ADDITIONAL QUESTION:
Ax^2 + Cy^2 + Dx + Ey + F = 0
can certainly be a conic section: It can be an ellipse if A, C have the same sign, a circle if they are equal, or a hyperbola if they have opposite signs. Of course, it can also have no real points satisfying it. Because the xy term is missing, we can move F to the right side and complete the squares on the left:
A[x^2 + (D/A)x + (D^2)/(4A^2)]
+ C[y^2 + (E/C)y + (E^2)/(4C^2)]
......... = (D^2)/(4A^2) + (E^2)/(4C^2) - F
If A and C are both positive and the right side is negative, there are no real points on it.
If the right side is zero, then it represents a degenerate conic. The equation could be something like
5(x-2)^2 + (y+4)^2 = 0, which is degenerate because it is just one point, (2, -4)
or 5(x-2)^2 - (y+4)^2 = 0, which is a pair of straight lines intersecting at (2, -4)