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

WHAT is this equation?

WHY does $B^2 - 4AC$ decide the shape?

The magic is that $B^2-4AC$ is a rotation invariant: rotate the axes to kill the $xy$ term, and this number doesn't change. So it captures something intrinsic to the shape, not to the tilt.

Derivation from first principles — rotation invariance

Step 1 — Rotate the axes by angle $\theta$. Replace $x = x'\cos\theta - y'\sin\theta,\qquad y = x'\sin\theta + y'\cos\theta.$ Why this step? A tilted conic becomes axis-aligned in some rotated frame; we want to find that frame.

Step 2 — Substitute into $Ax^2+Bxy+Cy^2$. Collecting terms gives new coefficients $A', B', C'$: $A' = A\cos^2\theta + B\cos\theta\sin\theta + C\sin^2\theta,$ $C' = A\sin^2\theta - B\cos\theta\sin\theta + C\cos^2\theta,$ $B' = B\cos 2\theta - (A-C)\sin 2\theta.$ Why this step? We want to see how $A,B,C$ transform, so we can hunt for a quantity that stays fixed.

Step 3 — Choose $\theta$ to eliminate the cross term. Set $B'=0$: $B\cos 2\theta = (A-C)\sin 2\theta \;\Rightarrow\; \tan 2\theta = \frac{B}{A-C}.$ Why this step? Once $B'=0$, the equation has no $xy$: it's a standard, un-tilted conic we recognize instantly.

Step 4 — Compute $B'^2 - 4A'C'$. After algebra (verified below): $B'^2 - 4A'C' = B^2 - 4AC.$ Why this step? This proves $B^2-4AC$ is invariant under rotation — it's a property of the shape itself, independent of tilt.

Step 5 — In the rotated frame ($B'=0$) read off the shape. Equation looks like $A'x'^2 + C'y'^2 + \dots = 0.$

• $A'C' > 0$ (same sign): $\;A'x'^2+C'y'^2=\text{const}$ → ellipse. Then $B'^2-4A'C' = -4A'C' < 0$. ✓
• $A'C' < 0$ (opposite signs): → hyperbola. Then $-4A'C' > 0$. ✓
• $A'C' = 0$ (one is zero): only one squared term → parabola. Then $-4A'C' = 0$. ✓

Since $B^2-4AC = B'^2-4A'C' = -4A'C'$, the sign of $B^2-4AC$ tells us the sign of $A'C'$ — hence the shape. QED.

HOW to fully classify (including degenerate cases)

A conic can collapse: an ellipse into a point, a hyperbola into two crossing lines, a parabola into parallel lines. To catch these, use the full $3\times3$ determinant: $\Delta = \begin{vmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{vmatrix}.$

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Flashcards

Connections

• Rotation of Axes — the transformation that removes the $xy$ term.
• Quadratic Forms and Eigenvalues — $B^2-4AC$ relates to the sign of eigenvalues of $\left(\begin{smallmatrix}A&B/2\\B/2&C\end{smallmatrix}\right)$.
• Ellipse Standard Equation, Parabola Standard Equation, Hyperbola Standard Equation — the axis-aligned targets.
• Circle as Special Ellipse — $A=C,\ B=0$.
• Discriminant of a Quadratic — the 1-variable cousin $b^2-4ac$ (don't confuse!).
• Degenerate Conics — point, line, line-pairs when $\Delta=0$.

Concept Map

Hinglish (regional understanding)