the conic discriminant test
Faced with a tangled equation like 5x^2 - 4xy + 8y^2 + ... = 0, you want to know quickly: is this an ellipse, a parabola, or a hyperbola — without rotating axes, completing squares, or plotting anything? The conic discriminant test answers in one short calculation, using just three of the six coefficients.
For the general equation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, compute the discriminant B^2 - 4AC. The sign decides the type: if B^2 - 4AC < 0 the conic is an ellipse (and a circle when also B = 0 and A = C); if B^2 - 4AC = 0 it is a parabola; if B^2 - 4AC > 0 it is a hyperbola. The reason it works is that B^2 - 4AC is unchanged by rotating or shifting the axes — it is a rotation-invariant of the equation — so it reports the conic's true type no matter how tilted or off-center the equation is written. (Memory aid: the same sign pattern as the curves' standard signs — both squares same sign gives the closed ellipse, opposite signs the open hyperbola, one square missing the parabola.)
This test is the fast classifier that turns a fearsome second-degree equation into an instant verdict. It is the algebraic counterpart of the geometric tilting in the cone — the angle of the slice and the sign of B^2 - 4AC are two faces of the same fact — and it is exactly the eccentricity classification (e < 1, e = 1, e > 1) read off algebraically.
For 5x^2 - 4xy + 8y^2 - 36 = 0: A = 5, B = -4, C = 8, so B^2 - 4AC = 16 - 160 = -144 < 0 — an ellipse (here a tilted one). For xy = 1, write x^2(0) + xy(1) + y^2(0): A = 0, B = 1, C = 0, so B^2 - 4AC = 1 > 0 — a hyperbola. For x^2 - 4xy + 4y^2 + x = 0: B^2 - 4AC = 16 - 16 = 0 — a parabola.
Negative means ellipse, zero means parabola, positive means hyperbola.
The discriminant names the type but cannot detect a degenerate case: B^2 - 4AC > 0 might be a real hyperbola or just a pair of crossing lines. After classifying, still check whether the conic has degenerated to a point, a line, or no real points.