How this instrument works
Any conic — ellipse, parabola, or hyperbola — can be written as the general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0. The discriminant Δ = B² − 4AC ignores the linear and constant terms entirely and looks only at the three coefficients attached to the squared and cross terms, because those three alone decide the curve's basic shape: whether it closes into a loop, opens once, or opens twice.
The reason this one sign survives no matter how the conic is tilted on the page is that Δ is a rotational invariant. Writing the quadratic part as a 2×2 symmetric matrix M = [[A, B/2], [B/2, C]], one finds Δ = −4·det(M), and rotating the axes to remove the xy term conjugates M by an orthogonal matrix, which leaves its determinant — and so Δ — completely unchanged. Whatever angle the curve sits at, B² − 4AC comes out the same.
The test says nothing about degeneracy. A negative, zero, or positive Δ still classifies the quadratic part correctly even when the full equation collapses to a single point, a pair of crossing lines, or no real locus at all — that outcome is decided by D, E, and F, not by A, B, and C. A circle is simply the tidy special case of the ellipse branch, where Δ is negative and additionally A equals C with B equal to zero.
- Write your equation in general form, Ax² + Bxy + Cy² + Dx + Ey + F = 0, and read off only the three quadratic coefficients.
- Enter the x² coefficient into A (coefficient of x²).
- Enter the cross-term coefficient into B (coefficient of xy) — use 0 if there is no xy term.
- Enter the y² coefficient into C (coefficient of y²).
- Read Discriminant, B² − 4AC: negative places the curve with the ellipses, zero with the parabolas, positive with the hyperbolas.
Worked example — the equation of a circle
Start from the equation of a circle centred at the origin, x² + y² = r². Rearranged as x² + 0·xy + y² − r² = 0, the quadratic coefficients are A (coefficient of x²) = 1, B (coefficient of xy) = 0, and C (coefficient of y²) = 1, regardless of how large r is. Feed those three numbers in and Discriminant, B² − 4AC returns 0² − 4(1)(1) = −4 before the radius ever enters the calculation.
A negative discriminant of −4 sorts the circle among the ellipses, which is exactly right — a circle is the special ellipse whose two axis lengths are equal, so both squared terms sit on the same side of the ledger. Change the shape without changing the sign: 4x² + 9y² − 36 = 0, an ellipse rather than a circle, has A = 4 and C = 9, giving Δ = 0² − 4(4)(9) = −144. Still negative, still an ellipse, just no longer round.
Questions
What does the discriminant B² − 4AC actually measure?
It measures the sign of the product of the eigenvalues of the conic's quadratic part. When B² − 4AC is negative those eigenvalues share a sign, producing a bounded, closed curve — an ellipse. When it is positive they have opposite signs, producing the two unbounded branches of a hyperbola. Zero means one eigenvalue vanishes, the degenerate direction that gives a parabola its single open arm.
Why is a circle treated as a kind of ellipse rather than its own case?
Because the discriminant test only sees A, B, and C, and a circle's equation, x² + y² = r², has A = C and B = 0 — the same negative-Δ signature as every other ellipse. That extra condition, A = C with B = 0, narrows the ellipse family down to circles specifically; drop it and you get ordinary ellipses, ovals stretched more along one axis than the other.
How does this discriminant differ from the b² − 4ac of the quadratic formula?
They share a formula but not a subject. The quadratic-formula discriminant b² − 4ac tests one equation in one variable, ax² + bx + c = 0, and its sign says whether the two roots are real or complex. This discriminant tests one equation in two variables, Ax² + Bxy + Cy² + ... = 0, and its sign says whether the resulting curve is an ellipse, parabola, or hyperbola — a different question that happens to reuse the same arithmetic shape.
Does a zero or positive discriminant guarantee an actual parabola or hyperbola?
Only for the quadratic part. Δ classifies the shape of Ax² + Bxy + Cy², but the full curve can still degenerate depending on D, E, and F — Δ = 0 can describe a genuine parabola or a pair of parallel lines, and Δ > 0 can describe a genuine hyperbola or a pair of crossing lines. Ruling out degeneracy takes a further check on the full 3×3 matrix of coefficients, not just A, B, and C.
What's the most common mistake when applying this test by hand?
Forgetting the cross term. An equation with no xy in it, like 2x² + 5y² − 10 = 0, still has B = 0 — some students skip straight to −4AC and get the sign wrong when A and C carry opposite signs, since 0² − 4AC then reduces to just −4AC. Write B in explicitly, even as a zero, before multiplying anything out.