General second-degree equation Ax²+Bxy+Cy²+Dx+Ey+F=0 — discriminant classification: Visual walkthrough

We use only these ideas from elsewhere: Rotation of Axes, Quadratic Forms and Eigenvalues, and the shapes themselves — ellipse, parabola, hyperbola, and their collapsed forms. Everything else is built here.

Step 1 — What does the equation even draw?

WHAT: We plot every point $(x, y)$ that satisfies the equation. The set of all such points is the conic.

WHY: Before we ask "what shape?", we must remember the curve is nothing but a collection of points that balance the equation to zero. No magic — just a balance condition.

PICTURE: Below, the red curve is every $(x, y)$ making $3 x^{2} + 4 x y + 3 y^{2} - 12 = 0$ true. Notice it is a closed loop — an ellipse — but tilted: its long axis does not line up with the $x$ or $y$ axis. That tilt is caused entirely by the $B x y = 4 x y$ term.

Figure — General second-degree equation Ax²+Bxy+Cy²+Dx+Ey+F=0 — discriminant classification

Step 2 — Split the equation into three jobs

WHAT: We mentally cut the six-term equation into these three pieces.

WHY: If shifting ( $D, E$ ) and the constant ( $F$ ) never change the family of shape, then to classify we can throw them away and stare only at the shape group $A x^{2} + B x y + C y^{2}$ . That is a huge simplification — from six numbers down to three.

PICTURE: The same red ellipse, then the identical shape after we set $D = E = 0$ (removing the shift). It moved to the origin but is the same ellipse, same tilt, same size-of-loop feeling. Shifting slides; it does not reshape.

Step 3 — The shape group is a "quadratic form": its level curves ARE the conics

WHAT: We look at the contour lines of the surface $z = A x^{2} + B x y + C y^{2}$ . Each contour is one conic.

WHY: Whether those contour loops close up or fly off to infinity is the whole question. So we need to know: does $Q$ ever return to the same value in a closed loop, or does it keep growing off in some direction? This is exactly what quadratic forms answer.

PICTURE: A bowl-shaped surface (for an ellipse-type $Q$ ) sliced by horizontal planes; the slices, seen from above, are nested closed loops — ellipses. Next to it, a saddle surface whose slices are open, two-piece curves — hyperbolas.

Recall Why "level curve"?

A level curve of $Q$ is where the height $z = Q (x, y)$ stays fixed ::: it's a contour line; the shape of that contour (loop vs open branches) is the conic family.

Step 4 — Rotate the axes to hunt for what stays fixed

WHAT: We spin our coordinate grid by an angle $θ$ (see Rotation of Axes) and re-measure every point.

WHY: The tilt of the ellipse in Step 1 came from the $B x y$ term. If we can turn our grid to line up with the ellipse's own axes, the annoying cross term will vanish, and the equation becomes a shape we recognise instantly. We rotate not to change the curve — the curve is fixed — but to change our description of it.

PICTURE: The red ellipse sits still. A black dashed grid $(x^{'}, y^{'})$ turns until its axes point along the ellipse's long and short directions. In that grid, the ellipse looks perfectly upright.

Step 5 — After rotating, the three coefficients transform

WHAT: We substitute the Step-4 formulas into $A x^{2} + B x y + C y^{2}$ and collect the $x^{'2}$ , $x^{'} y^{'}$ , $y^{'2}$ terms. Out pop $A^{'}, B^{'}, C^{'}$ .

WHY: We want to see how the three numbers change when we rotate — because only by seeing the change can we spot a combination of them that refuses to change. The term $B^{'}$ is our target: if we can drive it to $0$ , the cross term dies.

PICTURE: A before/after strip: the coefficient trio $(A, B, C)$ feeds through the rotation "machine" and comes out as $(A^{'}, B^{'}, C^{'})$ . The $B^{'}$ dial is highlighted in red — it's the one we're about to zero out.

Step 6 — Choose the exact angle that kills the cross term

WHAT: We solve $B^{'} = 0$ for $θ$ . There is always such an angle.

WHY tan and not something else? We have an equation of the form $B cos 2 θ = (A - C) sin 2 θ$ . Dividing both sides by $cos 2 θ$ turns it into $\frac{sin 2 θ}{cos 2 θ} = \frac{B}{A - C}$ , and that left side is by definition $tan 2 θ$ . So $tan$ is the tool that answers "for which angle is this ratio met?" We then read $θ$ off via $arctan$ — the inverse question "which angle has this tan?"

PICTURE: A right triangle built from $sin 2 θ$ (opposite) and $cos 2 θ$ (adjacent). The red hypotenuse's slope is $\frac{B}{A - C}$ ; the angle at the base is $2 θ$ . Below it, the special case $A = C$ where the triangle stands vertical and $θ = 4 5^{\circ}$ .

Step 7 — The invariant: $B^{'2} - 4 A^{'} C^{'} = B^{2} - 4 A C$

WHAT: Plug the Step-5 formulas for $A^{'}, B^{'}, C^{'}$ into $B^{'2} - 4 A^{'} C^{'}$ and simplify. Everything with $θ$ cancels.

WHY: This is the punchline. A quantity that doesn't care about tilt must be describing the shape itself, not our viewing angle. So $B^{2} - 4 A C$ is a genuine fingerprint of the conic family — a discriminant that survives rotation. (It is checked term-by-term in the VERIFY block.)

PICTURE: A dial labelled $θ$ that the reader can imagine spinning: $A^{'}, B^{'}, C^{'}$ all wobble as $θ$ turns, but the readout $B^{'2} - 4 A^{'} C^{'}$ (in red) stays frozen on one value — a flat horizontal line while the individual coefficients oscillate.

Step 8 — In the straightened frame, read the shape off directly

WHAT: With no cross term, only $A^{'} x^{'2} + C^{'} y^{'2}$ remain in the shape group. We test the signs of $A^{'}$ and $C^{'}$ .

WHY: Now the shape is transparent:

$A^{'}, C^{'}$ same sign ( $A^{'} C^{'} > 0$ ): both squares push the same way → a closed loop = ellipse. Then $- 4 A^{'} C^{'} < 0$ , so $B^{2} - 4 A C < 0$ . ✓
$A^{'}, C^{'}$ opposite signs ( $A^{'} C^{'} < 0$ ): squares fight → open two-branch = hyperbola. Then $- 4 A^{'} C^{'} > 0$ , so $B^{2} - 4 A C > 0$ . ✓
One of them is $0$ ( $A^{'} C^{'} = 0$ ): only a single square survives → parabola. Then $- 4 A^{'} C^{'} = 0$ , so $B^{2} - 4 A C = 0$ . ✓

PICTURE: Three upright curves side by side in the straightened $(x^{'}, y^{'})$ frame — a red ellipse ( $A^{'}, C^{'}$ both $+$ ), a red hyperbola ( $A^{'}$ + and $C^{'}$ $-$ ), and a red parabola (only $A^{'}$ survives, $C^{'} = 0$ ) — each captioned with the sign of $A^{'} C^{'}$ and the matching sign of $B^{2} - 4 A C$ .

Step 9 — The degenerate escape hatch

WHAT: We add one final check — the $3 \times 3$ determinant $Δ$ .

WHY: An ellipse can shrink to a point, a hyperbola to two crossing lines, a parabola to parallel lines. These limiting cases still satisfy the equation but aren't proper curves. $Δ \neq = 0$ certifies you have a genuine, non-collapsed conic.

PICTURE: A red ellipse shrinking through smaller and smaller loops until it becomes a single red dot ( $Δ \to 0$ ), shown as a fading sequence — the "loop collapses to a point" limit made visible.

The one-picture summary

This single diagram compresses the whole journey: the tilted red conic → rotate by the angle $tan 2 θ = \frac{B}{A - C}$ → the cross term dies → read the sign of $A^{'} C^{'}$ → which is fixed forever by the rotation-proof number $B^{2} - 4 A C$ , with $Δ$ standing guard against collapse.

Recall Feynman retelling — explain the whole walkthrough in plain words

Picture an oval drawn on paper, but slanted. Its slant is the fault of the $x y$ term. I don't like slants, so I spin my graph paper until the oval sits upright — there's always exactly one spin angle that does it, and $tan 2 θ = \frac{B}{A - C}$ tells me it (if the two stretches $A$ and $C$ are equal, the spin is just $4 5^{\circ}$ ). Once upright, the equation only has $x^{'2}$ and $y^{'2}$ , no mixing. Now the shape is obvious: if both squared terms pull the same way, I get a closed loop (ellipse); if they fight, I get two flying-apart branches (hyperbola); if one term vanishes, I get the in-between parabola. Here's the beautiful part: I never actually have to spin the paper. The combination $B^{2} - 4 A C$ comes out identical no matter how I spin — it's frozen. So I just compute it once from the original numbers: negative means loop, positive means split, zero means the in-between. Last, I glance at the big determinant $Δ$ : if it's zero, my loop secretly shrank to a dot or my branches secretly straightened into plain lines — a collapsed, degenerate case, not a real conic.