Conic Sections
Identify any conic from its general equation — including rotated ones — with centre, foci and axes.
4x² + 9y² − 36 = 0
B² − 4AC = -144
between 0 and 1
axes are already aligned
| Reduced form | 4x′² + 9y′² − 36 = 0 | after rotating, and centring where there is a centre |
| Centre | (0, 0) | |
| Semi-major axis a | 3 | |
| Semi-minor axis b | 2 | |
| Centre to focus c | 2.23607 | c² = a² − b² |
| Vertices | (3, 0) (-3, 0) (0, 2) (0, -2) | |
| Foci | (2.2361, 0) (-2.2361, 0) | |
| Asymptotes | — | only a hyperbola has them |
| Determinant | -1296 | zero means the conic is degenerate |
The curve
x from -3.30equal scale on both axesto 3.30
Every second-degree equation in two variables draws a conic, and one number tells you which. The discriminant B² − 4AC is negative for an ellipse, zero for a parabola and positive for a hyperbola — whatever rotation or shift the other terms apply.
One equation, four curves
Slice a double cone with a plane and what you get depends only on the angle of the slice: a circle, an ellipse, a parabola or a hyperbola. Algebraically that family is exactly the set of curves satisfying Ax² + Bxy + Cy² + Dx + Ey + F = 0, and the six coefficients between them decide the shape, where it sits and which way it points.
The work splits cleanly. The three quadratic coefficients A, B and C fix the shape and the orientation; D and E only move the curve around; F only scales it. That is why the discriminant uses A, B and C alone — shifting a hyperbola cannot turn it into an ellipse.
What the cross term does
A non-zero B rotates the curve off the axes, and it is the case most calculators refuse. The standard treatment is to rotate the coordinate system until the term vanishes, which happens at the angle θ satisfying cot 2θ = (A − C) ÷ B. In the rotated frame the equation has no xy term and completing the square works as usual; the results are then rotated back. This page does exactly that, so xy = 1 comes out correctly as a hyperbola at 45° with the axes as its asymptotes.
< 0 ellipse · = 0 parabola · > 0 hyperbola
- 1 Read off A, B and C. The coefficients of x², xy and y². Everything about the shape lives in these three.
- 2 Compute B² − 4AC. For 4x² + 9y² − 36 = 0 that is 0 − 4(4)(9) = −144, so it is an ellipse.
- 3 Check for degeneracy before going further. If the 3×3 determinant of the equation is zero, the “conic” is really a point, one or two lines, or nothing. x² − y² = 0 is two crossing lines, not a hyperbola.
- 4 Rotate the cross term away if B is non-zero. Turn the axes by θ with cot 2θ = (A − C) ÷ B, which leaves an equation in x′ and y′ alone.
- 5 Complete the square to find the centre and axes. 4x² + 9y² = 36 becomes x²/9 + y²/4 = 1: a = 3, b = 2, c = √5, so e = √5/3 ≈ 0.745.
Telling them apart
Eccentricity measures how far the conic departs from a circle.
| Conic | B² − 4AC | Eccentricity | Foci | Standard form |
|---|---|---|---|---|
| Circle | Negative, with A = C and B = 0 | 0 | Both at the centre | (x − h)² + (y − k)² = r² |
| Ellipse | Negative | Between 0 and 1 | Two, inside | (x−h)²/a² + (y−k)²/b² = 1 |
| Parabola | Zero | Exactly 1 | One | (x − h)² = 4p(y − k) |
| Hyperbola | Positive | Greater than 1 | Two, outside | (x−h)²/a² − (y−k)²/b² = 1 |
The focus is what actually defines them
The classification above is convenient but not fundamental. What unites all four curves is the focus–directrix property: a conic is the set of points whose distance to a fixed point, divided by the distance to a fixed line, is a constant. That constant is the eccentricity, and it alone determines the shape — 0 gives a circle, values below 1 an ellipse, exactly 1 a parabola, above 1 a hyperbola.
The other defining properties follow from it and are worth knowing because they are what the tool checks itself against. On an ellipse the distances to the two foci always add to 2a; on a hyperbola their difference is always 2a; on a parabola the distance to the focus equals the distance to the directrix. These are not approximations — they hold at every point, which makes them a proper test of whether computed foci are right.
One boundary to note. When the 3×3 determinant vanishes the equation still has solutions, but they no longer form a conic: x² + y² = 0 is a single point and x² − y² = 0 is a pair of crossing lines. These are the degenerate cases, and this page names them rather than reporting an eccentricity for a curve that is not there.