# Degenerate conics

I was studying about the discriminant of a conic and got to the case where it equals 0. The book I'm referring to says that such a case means that the equation represents a parabola, a pair of parallel lines,a line or has no graph. However, I observed that there is one more possible case-all points on the plane. I'd be grateful if my findings could be verified or discarded with reason.

• Degenerate conics - I'm tempted to flag this post as offensive. ;-) – Lucian Aug 29 '14 at 14:07

It depends on the definition of conic. If you simply ask a conic to be the set of points in the affine plane satistfying a polynomial equation of degree $\leq 2$ in coordinates, say, $(x, y)$, then yes, the conic satisfying the equation $0 = 0$, i.e., the whole plane, is trivially is a conic, but this isn't very interesting. Probably it would be a good convention not to permit this case, or when one goes to write down theorems about conics, one would frequently have to make an exception for this case, which would be ugly and detract from results.

• Essentially, that's what I got. Assuming all the coefficients to be zero. So thanks. But I wonder why such a thing isn't mentioned in books or online. – Student Aug 29 '14 at 12:53
• If your book is careful, it will either discuss this case explicitly, or it will formulate its definition of conic to avoid this issue. Either way, it's probably not discussed much because it is "trivial", but that's not a reason not to discuss it at all. – Travis Aug 29 '14 at 13:04

For $Ax^2+Bxy+Cy^2+Dx+Ey+k=0$, let $p=B^2-4AC$.

If $p\lt 0$, ellipse, circle, point or no curve.

If $p=0$, parabola, 2 parallel lines, 1 line or no curve.

If $p\gt 0$, hyperbola or 2 intersecting lines.