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

Before we start, a quick reminder of the two tools, stated in plain words:

Prerequisite tours if any of these feel shaky: Discriminant of a Quadratic, Degenerate Conics, Rotation of Axes, Quadratic Forms and Eigenvalues.

The scenario matrix

Every classification problem falls into exactly one of these cells. The final column names the example that hits it.

#	Cell (the scenario)	What decides it	Example
1	$Δ_{2} < 0$ , $A \neq = C$ or $B \neq = 0$ → tilted ellipse	$B^{2} - 4 A C < 0$ , not a circle	Ex 1
2	$Δ_{2} < 0$ , $A = C, B = 0$ → circle (limiting ellipse)	equal squares, no cross term	Ex 2
3	$Δ_{2} = 0$ → parabola (perfect square quadratic part)	$B^{2} - 4 A C = 0$	Ex 3
4	$Δ_{2} > 0$ → hyperbola (indefinite, factors into 2 lines)	$B^{2} - 4 A C > 0$ , $Δ \neq = 0$	Ex 4
5	Degenerate, $Δ_{2} > 0$ → two intersecting lines	$Δ_{2} > 0$ but $Δ = 0$	Ex 5
6a	Degenerate, $Δ_{2} = 0$ → parallel distinct lines	$Δ_{2} = 0$ , $Δ = 0$ , distinct roots	Ex 6
6b	Degenerate, $Δ_{2} = 0$ → coincident line	equal roots	Ex 6b
6c	Degenerate, $Δ_{2} = 0$ → empty (no real points)	negative constant	Ex 6c
7	Degenerate, $Δ_{2} < 0$ → single point (collapsed ellipse)	$Δ_{2} < 0$ and $Δ = 0$	Ex 7
7b	Imaginary ellipse, $Δ_{2} < 0$ → empty set	$Δ \neq = 0$ but no real points	Ex 7b
8	Zero-input edge: $A = C = 0$ (only $x y$ term) → rectangular hyperbola	$B \neq = 0, A = C = 0$	Ex 8
9	Word problem — searchlight cone → identify shape	build the equation first	Ex 9
10	Exam twist — a parameter $k$ ; find $k$ for a parabola	solve $B^{2} - 4 A C = 0$	Ex 10
11	Not-a-conic edge — $A = B = C = 0$ → single line or no locus	linear / constant only	Ex 11

The same logic reads more naturally as a decision tree — follow the arrows from the top. The two diamonds are the two numbers ( $Δ_{2}$ then $Δ$ ); each leaf is one row of the table above.

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

Ex 1 — Cell 1: tilted ellipse

Worked example Classify

$5 x^{2} - 4 x y + 5 y^{2} - 12 = 0$ . Forecast: Same coefficient on $x^{2}$ and $y^{2}$ but there IS an $x y$ term. Guess: ellipse or circle? Commit before reading.

Step 1. Read $A = 5, B = - 4, C = 5, D = 0, E = 0, F = - 12$ . Why this step? Classification always starts by matching the equation to the template letter-for-letter.

Step 2. Compute the discriminant: $B^{2} - 4 A C = (- 4)^{2} - 4 (5) (5) = 16 - 100 = - 84 < 0.$ Why this step? Negative discriminant ⇒ ellipse family.

Step 3. Circle or ellipse? A circle needs $A = C$ and $B = 0$ . Here $A = C = 5$ but $B = - 4 \neq = 0$ . So it is a tilted (rotated) ellipse, not a circle. Why this step? The parent note warns: negative alone ≠ circle. The cross term is a tilt, so the two axes of the oval have different lengths in the $x, y$ frame.

Step 4. Confirm it's a genuine conic: $Δ = 5 - 2 0 - 2 50 00 - 12 = - 12 (5 \cdot 5 - (- 2) (- 2)) = - 12 (25 - 4) = - 252 \neq = 0.$ Why this step? $Δ \neq = 0$ guarantees the ellipse hasn't collapsed to a point.

Verify (deriving the standard form). Since $A = C$ , the tilt formula $tan 2 θ = B / (A - C)$ has a zero denominator, so $2 θ = 9 0^{\circ}$ and $θ = 4 5^{\circ}$ . At $4 5^{\circ}$ , $cos θ = sin θ = \frac{1}{2}$ , so from the Rotation of Axes rule the new coefficients are $A^{'} = A cos^{2} θ + B cos θ sin θ + C sin^{2} θ = 5 \cdot \frac{1}{2} + (- 4) \cdot \frac{1}{2} + 5 \cdot \frac{1}{2} = \frac{5 - 4 + 5}{2} = 3,$ $C^{'} = A sin^{2} θ - B cos θ sin θ + C cos^{2} θ = 5 \cdot \frac{1}{2} - (- 4) \cdot \frac{1}{2} + 5 \cdot \frac{1}{2} = \frac{5 + 4 + 5}{2} = 7.$ With $B^{'} = 0$ the equation becomes $3 x^{'2} + 7 y^{'2} - 12 = 0$ , i.e. $3 x^{'2} + 7 y^{'2} = 12$ — a genuine ellipse with semi-axes $12/3 = 2$ (along $x^{'}$ ) and $12/7$ (along $y^{'}$ ). Notice $A^{'} C^{'} = 21 > 0$ (same sign, closed loop) and $B^{'2} - 4 A^{'} C^{'} = - 84 = B^{2} - 4 A C$ , matching the invariance. Ellipse. ✓

Related: Ellipse Standard Equation, Circle as Special Ellipse.

Ex 2 — Cell 2: the circle (limiting ellipse)

Worked example Classify

$2 x^{2} + 2 y^{2} - 8 x + 4 y - 1 = 0$ . Forecast: No $x y$ term, equal squares. What special ellipse is this?

Step 1. $A = 2, B = 0, C = 2, D = - 8, E = 4, F = - 1$ . Why this step? Read every coefficient off the equation first — the whole classification rests on getting $A, B, C$ right.

Step 2. $B^{2} - 4 A C = 0 - 4 (2) (2) = - 16 < 0$ ⇒ ellipse family. Why this step? Establish the family before checking the special case.

Step 3. Special-case test: $A = C = 2$ and $B = 0$ . Both hold ⇒ circle. Why this step? Equal squared coefficients with no cross term means the level curve is equally "stretched" in every direction — the definition of a round circle.

Step 4. Sanity: divide by $2$ and complete the square: $x^{2} + y^{2} - 4 x + 2 y - \frac{1}{2} = 0 \Rightarrow (x - 2)^{2} + (y + 1)^{2} = \frac{1}{2} + 4 + 1 = \frac{11}{2} .$ Why this step? A positive right-hand side ( $11/2 > 0$ ) confirms a real circle of radius $11/2$ .

Verify: Radius $^{2} = 11/2 = 5.5 > 0$ , centre $(2, - 1)$ . Circle. ✓

Ex 3 — Cell 3: the parabola

Worked example Classify

$9 x^{2} - 12 x y + 4 y^{2} + x - 2 = 0$ . Forecast: Is the degree-2 part a perfect square? If yes — which family?

Step 1. $A = 9, B = - 12, C = 4$ . Why this step? Pull the three degree-2 coefficients out; $B$ is the coefficient of $x y$ , so $B = - 12$ , not the coefficient of $x$ .

Step 2. $B^{2} - 4 A C = 144 - 4 (9) (4) = 144 - 144 = 0$ ⇒ parabola. Why this step? A zero discriminant is the borderline between ellipse and hyperbola — exactly the parabola case.

Step 3. The tell-tale sign: $9 x^{2} - 12 x y + 4 y^{2} = (3 x - 2 y)^{2}$ , a perfect square. Why this step? When the quadratic form squashes to a single squared expression, one direction has zero curvature — that's the borderline shape between loop and split, i.e. a parabola. See Parabola Standard Equation.

Step 4. Confirm non-degenerate. With $D = 1, E = 0, F = - 2$ : $Δ = 9 - 6 1/2 - 6 40 1/2 0 - 2 .$ Expand along the last row: $\frac{1}{2} ((- 6) (0) - 4 (1/2)) + (- 2) (9 \cdot 4 - (- 6) (- 6)) = \frac{1}{2} (- 2) - 2 (36 - 36) = - 1 \neq = 0.$ Why this step? $Δ \neq = 0$ ⇒ genuine parabola, not a pair of parallel lines.

Verify: $Δ = - 1 \neq = 0$ and $Δ_{2} = 0$ together ⇒ parabola. ✓

Ex 4 — Cell 4: the hyperbola

Ex 5 — Cell 5: degenerate, two intersecting lines

The figure below draws exactly this collapse and is worth reading line by line. The blue line is $y = x - 1$ ; feed it a point like $(2, 1)$ and the original equation gives $0$ . The yellow line is the mirror $y = - (x - 1)$ ; the point $(0, 1)$ sits on it and also satisfies the equation. The two lines are what the hyperbola of Cell 4 becomes when it degenerates — its two branches have flattened onto their own asymptotes. The red dot marks $(1, 0)$ , the single crossing point: it is the "waist" the shrinking hyperbola pinched down to, and it lies on both lines simultaneously (set $x = 1$ in either equation to get $y = 0$ ).

Ex 6 — Cell 6a: degenerate, parallel distinct lines

Ex 6b — Cell 6b: degenerate, coincident line

Ex 6c — Cell 6c: degenerate, empty set

Ex 7 — Cell 7: degenerate, single point

Ex 7b — Cell 7b: imaginary (empty) ellipse

Worked example Classify

$4 x^{2} + 9 y^{2} - 8 x + 8 = 0$ . Forecast: Same ellipse family as Ex 7, but nudge the constant up. Point, real oval, or nothing?

Step 1. $A = 4, B = 0, C = 9, D = - 8, E = 0, F = 8$ .

Step 2. $B^{2} - 4 A C = - 144 < 0$ ⇒ ellipse family.

Step 3. Complete the square: $4 (x - 1)^{2} - 4 + 9 y^{2} + 8 = 4 (x - 1)^{2} + 9 y^{2} + 4 = 0 \Rightarrow 4 (x - 1)^{2} + 9 y^{2} = - 4.$ Why this step? The left side is a sum of squares, so it is $\geq 0$ ; it can never equal $- 4$ . There are no real points — an imaginary (empty) ellipse.

Step 4. Note $Δ = 40 - 4 090 - 4 08 = 9 (32 - 16) = 144 \neq = 0$ . So $Δ_{2} < 0$ with $Δ \neq = 0$ passes the "genuine conic" test, yet the real graph is empty. Why this step? Lesson: $Δ \neq = 0$ guarantees non-degenerate, but for the ellipse family a further sign check decides real-vs-imaginary. The rule: an ellipse-family curve is real only when $A$ and $Δ$ have opposite signs. Here $A = 4 > 0$ and $Δ = 144 > 0$ share the same sign, so the oval lives only in complex coordinates — hence the empty real graph. (In Ex 1 the real ellipse had $A = 5 > 0$ and $Δ = - 252 < 0$ : opposite signs, so it was genuinely visible.)

Verify: $4 (x - 1)^{2} + 9 y^{2} \geq 0 > - 4$ for all real $x, y$ ⇒ no solutions. Empty (imaginary ellipse). ✓

Ex 8 — Cell 8: zero-input edge, pure $x y$ term

Worked example Classify

$x y = 6$ . Forecast: Just a product equals a constant. This is the classic "reciprocal" curve — which family?

Step 1. Rewrite as $x y - 6 = 0$ . Then $A = 0, B = 1, C = 0, D = 0, E = 0, F = - 6$ . Why this step? The $x y$ term has coefficient $1$ , so $B = 1$ ; there are NO $x^{2}$ or $y^{2}$ terms, so $A = C = 0$ .

Step 2. $B^{2} - 4 A C = 1 - 4 (0) (0) = 1 > 0$ ⇒ hyperbola. Why this step? Even with $A = C = 0$ the discriminant formula still applies verbatim; a positive value forces the hyperbola family.

Step 3. Because $A = C$ ( $= 0$ here) the axes lie at $4 5^{\circ}$ : rotating gives $\frac{x ^{'2}}{12} - \frac{y ^{'2}}{12} = 1$ , a rectangular hyperbola (equal semi-axes, perpendicular asymptotes = the $x, y$ axes themselves). Why this step? This is the limiting "pure cross-term" input the discriminant handles just as smoothly as any other.

Step 4. $Δ = 0 1/2 0 1/2 00 00 - 6 = - 6 (0 - \frac{1}{4}) = \frac{3}{2} \neq = 0$ ⇒ genuine (non-degenerate) hyperbola.

Verify: $Δ_{2} = 1 > 0$ , $Δ = 1.5 \neq = 0$ . Curve passes through $(2, 3), (3, 2), (- 2, - 3)$ — two branches. Rectangular hyperbola. ✓

The figure shows the two branches of $x y = 6$ . Notice the green dots at $(2, 3), (3, 2), (- 2, - 3)$ that we checked — one branch sits in the top-right, its mirror in the bottom-left. The dashed red lines are the $x$ - and $y$ -axes: they are the asymptotes the branches hug but never touch, which is exactly why $A = C = 0$ gives a rectangular hyperbola.

Ex 9 — Cell 9: word problem (searchlight cone)

Worked example A searchlight sits on the ground and points upward at a slant. The rim of its light-cone traces the equation

$16 x^{2} - 24 x y + 9 y^{2} - 40 x - 30 y = 0$ on a wall. What conic is the bright boundary? Forecast: A slanted cone cut by a plane — the parent's flashlight picture. Loop, park, or split?

Step 1. $A = 16, B = - 24, C = 9, D = - 40, E = - 30, F = 0$ . Why this step? Translate the physical curve into coefficients before touching geometry.

Step 2. $B^{2} - 4 A C = 576 - 4 (16) (9) = 576 - 576 = 0$ ⇒ parabola family.

Step 3. Check $Δ$ to be sure it's a real parabola and not a line-pair: $Δ = 16 - 12 - 20 - 12 9 - 15 - 20 - 15 0 .$ Expand along the last row: $- 20 ((- 12) (- 15) - 9 (- 20)) - (- 15) (16 (- 15) - (- 12) (- 20)) + 0$ $= - 20 (180 + 180) + 15 (- 240 - 240) = - 20 (360) + 15 (- 480) = - 7200 - 7200 = - 14400 \neq = 0.$ Why this step? $Δ \neq = 0$ confirms a genuine parabola, matching the physical intuition of a beam edge that skims the wall.

Step 4. Quadratic part $= (4 x - 3 y)^{2}$ , the perfect-square hallmark. Why this step? Factoring the degree-2 group into a single square is the algebraic fingerprint of a parabola — it double-confirms the $Δ_{2} = 0$ verdict and shows the beam-edge direction $4 x = 3 y$ .

Verify: $Δ_{2} = 0$ , $Δ = - 14400 \neq = 0$ ⇒ the boundary is a parabola. ✓

Ex 10 — Cell 10: exam twist (find the parameter)

Ex 11 — Cell 11: not-a-conic edge ( $A = B = C = 0$ )

Worked example Classify (a)

$2 x + 3 y - 6 = 0$ and (b) $0 \cdot x + 0 \cdot y + 5 = 0$ . Forecast: No degree-2 terms at all. Is the discriminant even meaningful here?

Step 1. For both, $A = B = C = 0$ . The general-equation requirement $(A, B, C) \neq = (0, 0, 0)$ is violated, so these are not second-degree equations — the conic classification simply does not apply. Why this step? The discriminant test is defined only when a quadratic part exists; recognising when you have left that regime is itself a skill.

Step 2. (a) $2 x + 3 y - 6 = 0$ is linear ⇒ a single straight line. (b) $5 = 0$ is false for every $(x, y)$ ⇒ the empty set (no locus). A constant like $0 = 0$ would instead be the whole plane. Why this step? These are the true edge cases below "conic": one non-zero linear equation gives a line; a purely non-zero constant gives nothing.

Verify: (a) $(3, 0)$ and $(0, 2)$ both satisfy $2 x + 3 y - 6 = 0$ — two points determine the line. (b) $5 \neq = 0$ always ⇒ no points. Line / empty. ✓

Wrap-up

Recall The whole decision, in order

Compute $B^{2} - 4 A C$ for the family, then $Δ$ for reality. ::: $B^{2} - 4 A C < 0$ ellipse (circle if $A = C, B = 0$ ), $= 0$ parabola, $> 0$ hyperbola — but if $Δ = 0$ it has degenerated to a point / line(s), and even with $Δ \neq = 0$ an ellipse-family curve can be empty (imaginary) when $A$ and $Δ$ share a sign.