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

Intuition Why the sign of

$A$ vs. $Δ$ tells real from empty (read once, use in L3–L5) Suppose we have rotated and shifted so the equation is $A^{'} x^{'2} + C^{'} y^{'2} = G$ with no cross or linear terms. For an ellipse $A^{'}$ and $C^{'}$ share a sign — say both $> 0$ . Then the left side $A^{'} x^{'2} + C^{'} y^{'2}$ can only be $\geq 0$ . So real points exist only if $G \geq 0$ ; if $G < 0$ the equation demands "positive $=$ negative," which is impossible — the locus is empty (imaginary ellipse). The $3 \times 3$ determinant $Δ$ packages exactly this "leftover constant" $G$ together with the shape. It turns out that for the ellipse case, $A$ and $Δ$ having the same sign forces $G < 0$ (empty), while opposite signs force $G > 0$ (real). That is the rule we will apply: a real ellipse needs $sign (A) \neq = sign (Δ)$ . You do not have to reprove this each time — just check the two signs.

Level 1 — Recognition

(Read off $A, B, C$ , compute one number, name the family.)

Recall Solution 1.1

What we do: pick the degree-2 coefficients. $A = 5, B = 2, C = 5$ . Compute the shape number: $Δ_{2} = B^{2} - 4 A C = 2^{2} - 4 (5) (5) = 4 - 100 = - 96 < 0.$ Negative ⇒ ellipse family. Is it a circle? A circle needs $A = C$ and $B = 0$ . Here $A = C = 5$ but $B = 2 \neq = 0$ , so it is a tilted ellipse, not a circle. Answer: ellipse.

Recall Solution 1.2

$A = 1, B = - 3, C = 1$ . $Δ_{2} = B^{2} - 4 A C = (- 3)^{2} - 4 (1) (1) = 9 - 4 = 5 > 0.$ Positive ⇒ hyperbola.

Recall Solution 1.3

$A = 9, B = - 12, C = 4$ . $Δ_{2} = B^{2} - 4 A C = (- 12)^{2} - 4 (9) (4) = 144 - 144 = 0.$ Zero ⇒ parabola. (Sanity check: $9 x^{2} - 12 x y + 4 y^{2} = (3 x - 2 y)^{2}$ , a perfect square — the fingerprint of a parabola.)

Level 2 — Application

(One extra factor to handle: unlisted terms, or a missing squared term.)

Recall Solution 2.1

Fit it to the template: $A = 0, B = 0, C = 1$ (the $- 8 x$ is a $D$ -term, irrelevant to shape). $Δ_{2} = B^{2} - 4 A C = 0 - 4 (0) (1) = 0.$ Zero ⇒ parabola — exactly right, $y^{2} = 8 x$ is the sideways parabola. Compare with the standard form $y^{2} = 4 p x$ , where $p$ is the focal length (the distance from vertex to focus). Matching $4 p = 8$ gives $p = 2$ . See Parabola Standard Equation.

Recall Solution 2.2

Rewrite as $x y - 1 = 0$ . The only degree-2 term is $x y$ , so $A = 0, B = 1, C = 0$ . $Δ_{2} = B^{2} - 4 A C = 1^{2} - 4 (0) (0) = 1 > 0.$ Positive ⇒ hyperbola. (Indeed $x y = 1$ is the rectangular hyperbola you already know.)

Recall Solution 2.3

$A = 4, B = 0, C = 1$ . $Δ_{2} = B^{2} - 4 A C = 0 - 4 (4) (1) = - 16 < 0.$ Negative ⇒ ellipse. Because $B = 0$ but $A = 4 \neq = C = 1$ , it's an axis-aligned (untilted) ellipse, not a circle. See Ellipse Standard Equation.

Level 3 — Analysis

(Now you must also check $Δ$ — the discriminant $Δ_{2}$ alone can lie. It can lie in two ways: collapse to lines/points, OR leave no real points at all.)

Recall Solution 3.1

$A = 1, B = 2, C = 1$ . $Δ_{2} = 4 - 4 = 0$ ⇒ parabola family. Now the degenerate test. $D = 0, E = 0, F = - 4$ : $Δ = 110110 00 - 4 .$ Expand along the last column: only the $- 4$ entry survives, times the top-left $2 \times 2$ minor: $Δ = - 4 (1 \cdot 1 - 1 \cdot 1) = - 4 (0) = 0.$ $Δ = 0$ ⇒ degenerate. Look directly: $x^{2} + 2 x y + y^{2} = (x + y)^{2} = 4 \Rightarrow x + y = \pm 2$ : two parallel lines, not a parabola. See Degenerate Conics.

Recall Solution 3.2

$A = 1, B = 0, C = - 1$ . $Δ_{2} = 0 - 4 (1) (- 1) = 4 > 0$ ⇒ hyperbola family. $D = 2, E = 0, F = 1$ : $Δ = 101 0 - 1 0 101 .$ Expand along the middle row (mostly zeros): the only entry is $- 1$ in position $(2, 2)$ , with sign $+$ , giving $- 1$ times the minor $1111 = 0$ . So $Δ = - 1 (0) = 0$ . $Δ = 0$ ⇒ degenerate. Directly: $x^{2} + 2 x + 1 - y^{2} = (x + 1)^{2} - y^{2} = 0 \Rightarrow y = \pm (x + 1)$ : two intersecting lines.

Recall Solution 3.3

$A = 2, B = 5, C = - 3$ . $Δ_{2} = 25 - 4 (2) (- 3) = 25 + 24 = 49 > 0$ ⇒ hyperbola family. $D = 1, E = 1, F = 1$ . The symmetric matrix (halve the off-diagonals): $Δ = 2 2.5 0.5 2.5 - 3 0.5 0.5 0.5 1 .$ Cofactor expansion along the top row: $Δ = 2 [(- 3) (1) - (0.5) (0.5)] - 2.5 [(2.5) (1) - (0.5) (0.5)] + 0.5 [(2.5) (0.5) - (- 3) (0.5)]$ $= 2 (- 3 - 0.25) - 2.5 (2.5 - 0.25) + 0.5 (1.25 + 1.5) = 2 (- 3.25) - 2.5 (2.25) + 0.5 (2.75)$ $= - 6.5 - 5.625 + 1.375 = - 10.75 \neq = 0.$ $Δ \neq = 0$ ⇒ genuine, non-degenerate hyperbola.

Recall Solution 3.4

$A = 1, B = 0, C = 1$ . $Δ_{2} = 0 - 4 = - 4 < 0$ ⇒ ellipse family. Degenerate test with $D = 0, E = 0, F = 4$ : $Δ = 100010004 = 4 \neq = 0.$ So the discriminant $Δ_{2}$ says "ellipse" and $Δ \neq = 0$ says "non-degenerate." Yet $x^{2} + y^{2} = - 4$ has no real solutions — a sum of squares can never be negative. The real locus is empty. Applying the real-vs-empty rule (from the intuition box at the top): here $A = 1 > 0$ and $Δ = 4 > 0$ have the same sign, so the ellipse is imaginary (empty) — exactly matching the direct observation $x^{2} + y^{2} = - 4$ . A real ellipse would need $A$ and $Δ$ to have opposite signs. Compare with Exercise 2.3, where the ellipse is genuinely real.

Level 4 — Synthesis

(Find the parameter value that switches the shape, or use rotation.)

Recall Solution 4.1

$A = 1, B = k, C = 1$ , so $Δ_{2} = B^{2} - 4 A C = k^{2} - 4.$ This flips sign as $k$ crosses $\pm 2$ :

Ellipse when $k^{2} - 4 < 0 \Rightarrow - 2 < k < 2$ .
Parabola when $k^{2} - 4 = 0 \Rightarrow k = \pm 2$ .
Hyperbola when $k^{2} - 4 > 0 \Rightarrow k < - 2$ or $k > 2$ .

The boundary $k = \pm 2$ is exactly where the quadratic part becomes a perfect square $(x \pm y)^{2}$ — the shape "parks" at a parabola while transitioning between loop and split. The figure below plots the three representative curves $x^{2} + k x y + y^{2} = 1$ for $k = 0, 2, 3$ so you can see the loop open into a parabola-edge and then split into two branches.

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

Intuition How to read the figure above

Each curve is the level set $x^{2} + k x y + y^{2} = 1$ for one value of $k$ — think "which points make the quadratic part equal $1$ ?"

Teal, $k = 0$ ( $Δ_{2} = - 4 < 0$ ): a closed loop (here a circle). Follow the red arrow to it — you can walk all the way around and return. That "you never escape to infinity" is what ellipse means.
Plum, $k = 2$ ( $Δ_{2} = 0$ ): the loop has broken open. The quadratic collapses to $(x + y)^{2} = 1$ , i.e. the two parallel lines $x + y = \pm 1$ — this is the exact borderline (parabola family) between closed and split. Notice how the plum curve no longer wraps around; it runs off to infinity in two directions.
Orange, $k = 3$ ( $Δ_{2} = 5 > 0$ ): the curve has split into two separate branches heading to opposite infinities — the signature of a hyperbola.

So reading left-to-right by $k$ , you watch one number $Δ_{2}$ go negative → zero → positive and the geometry answer back: loop → open → split. That is the entire classification theorem, drawn.

Recall Solution 4.2

Why $tan 2 θ = \frac{B}{A - C}$ ? When we rotate the axes by $θ$ the new cross-coefficient turns out to be $B^{'} = B cos 2 θ - (A - C) sin 2 θ$ (derived in the parent note). We want $B^{'} = 0$ so the tilted conic becomes axis-aligned and readable. Setting $B^{'} = 0$ gives $B cos 2 θ = (A - C) sin 2 θ$ , and dividing by $cos 2 θ$ yields $tan 2 θ = \frac{B}{A - C}$ . That is the whole origin of the "magic" formula — it is just the condition "kill the cross term." See Rotation of Axes.

Step 1 — angle. Here $A = C = 3$ , so $A - C = 0$ and $tan 2 θ$ is infinite ⇒ $2 θ = 9 0^{\circ} \Rightarrow θ = 4 5^{\circ}$ . (When $A = C$ , the tilt is always $4 5^{\circ}$ .) Step 2 — new quadratic coefficients with $cos 4 5^{\circ} = sin 4 5^{\circ} = \frac{1}{2}$ : $A^{'} = A cos^{2} θ + B cos θ sin θ + C sin^{2} θ = 3 \cdot \frac{1}{2} + 4 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} = \frac{3}{2} + 2 + \frac{3}{2} = 5,$ $C^{'} = A sin^{2} θ - B cos θ sin θ + C cos^{2} θ = \frac{3}{2} - 2 + \frac{3}{2} = 1, B^{'} = 0.$ Step 3 — what happens to $D, E, F$ ? Rotation only mixes the degree-2 and degree-1 terms among themselves; it never touches the constant. Concretely, here $D = E = 0$ (no linear terms), and a rotation of purely degree-2 $+$ constant terms sends $D x + E y$ to $D^{'} x^{'} + E^{'} y^{'}$ with $D^{'} = D cos θ + E sin θ = 0$ and $E^{'} = - D sin θ + E cos θ = 0$ — still no linear part. The constant $F = - 20$ is a plain number with no $x$ or $y$ in it, so substituting the rotation formulas cannot change it: $F^{'} = F = - 20$ . (Rotation moves the axes' directions, not the origin, so the constant term is untouched.) Step 4 — equation: $5 x^{'2} + y^{'2} - 20 = 0 \Rightarrow \frac{x ^{'2}}{4} + \frac{y ^{'2}}{20} = 1$ : an ellipse with semi-axes $2$ and $20$ . Invariance check: original $Δ_{2} = 16 - 36 = - 20$ ; rotated $Δ_{2}^{'} = 0 - 4 (5) (1) = - 20$ . ✓ Identical, as promised.

Level 5 — Mastery

(Build the equation, or combine every tool at once.)

Recall Solution 5.1

Take a positive-definite quadratic part equal to zero: $x^{2} + y^{2} = 0$ . The only real solution is $x = y = 0$ — a single point. $A = 1, B = 0, C = 1$ : $Δ_{2} = 0 - 4 (1) (1) = - 4 < 0$ ⇒ ellipse family. Degenerate test with $D = E = F = 0$ : $Δ = 100010000 = 0.$ $Δ = 0$ ⇒ degenerate ellipse = a point. This is the ellipse "shrunk to its centre." See Degenerate Conics and Circle as Special Ellipse.

Recall Solution 5.2

Family: $A = 4, B = - 4, C = 1$ . $Δ_{2} = 16 - 16 = 0$ ⇒ parabola family. Note $4 x^{2} - 4 x y + y^{2} = (2 x - y)^{2}$ . Degenerate test: $D = 8, E = - 4, F = 3$ , matrix uses $B /2 = - 2, D /2 = 4, E /2 = - 2$ : $Δ = 4 - 2 4 - 2 1 - 2 4 - 2 3 .$ Expand along the top row: $Δ = 4 [(1) (3) - (- 2) (- 2)] - (- 2) [(- 2) (3) - (- 2) (4)] + 4 [(- 2) (- 2) - (1) (4)]$ $= 4 (3 - 4) + 2 (- 6 + 8) + 4 (4 - 4) = 4 (- 1) + 2 (2) + 0 = - 4 + 4 = 0.$ $Δ = 0$ ⇒ degenerate parabola. Which kind? Substitute $u = 2 x - y$ : the equation is $u^{2} + 8 x - 4 y + 3 = 0$ . Since $8 x - 4 y = 4 (2 x - y) = 4 u$ , it becomes $u^{2} + 4 u + 3 = 0 \Rightarrow (u + 1) (u + 3) = 0 \Rightarrow 2 x - y = - 1$ or $2 x - y = - 3$ : two parallel lines.

Recall Solution 5.3

Both share $A = 1, B = - 6, C = 9$ : $Δ_{2} = 36 - 36 = 0$ ⇒ parabola family, and $x^{2} - 6 x y + 9 y^{2} = (x - 3 y)^{2}$ . Case (i): $x^{2} - 6 x y + 9 y^{2} - 4 = 0$ . Then $D = E = 0, F = - 4$ : $Δ = 1 - 3 0 - 3 90 00 - 4 = - 4 (1 \cdot 9 - (- 3) (- 3)) = - 4 (9 - 9) = 0.$ $Δ = 0$ , and $(x - 3 y)^{2} = 4 \Rightarrow x - 3 y = \pm 2$ : two parallel lines. Case (ii): $x^{2} - 6 x y + 9 y^{2} = 0$ . Now $F = 0$ too: $Δ = 1 - 3 0 - 3 90 000 = 0.$ Again $Δ = 0$ , but $(x - 3 y)^{2} = 0 \Rightarrow x - 3 y = 0$ : a single line, counted twice (a "double line"). The two parallel lines of case (i) have merged into one as the constant $F$ shrank to $0$ . Lesson: in the parabola family ( $Δ_{2} = 0, Δ = 0$ ) a degenerate curve is two parallel lines, which coincide into one double line exactly when the perfect square equals $0$ (no separating constant).

Recall Solution 5.4

False. $Δ \neq = 0$ is exactly the condition for a genuine, non-degenerate conic. Two parallel lines (and a single double line) are degenerate configurations and always give $Δ = 0$ (as Exercises 3.1, 5.2 and 5.3 all showed). So $Δ_{2} = 0$ with $Δ \neq = 0$ forces a genuine parabola — no line-pair possible. Rule summary:

$Δ \neq = 0, Δ_{2} < 0$ → ellipse type (real if $A, Δ$ opposite signs; imaginary/empty if same signs — Exercise 3.4); $Δ_{2} = 0$ → real parabola; $Δ_{2} > 0$ → real hyperbola.
$Δ = 0$ → degenerate: point, single (double) line, two parallel lines, or two intersecting lines.

Active Recall

Recall Quick self-quiz

Discriminant sign of $x^{2} - 3 x y + y^{2} = 1$ ? ::: $Δ_{2} = 9 - 4 = 5 > 0$ → hyperbola $k$ -range making $x^{2} + k x y + y^{2} + \dots$ an ellipse type? ::: $- 2 < k < 2$ $A = C$ , $B \neq = 0$ : rotation angle to kill $x y$ ? ::: $4 5^{\circ}$ $Δ_{2} < 0$ , $Δ \neq = 0$ , but no real points — what is it? ::: an imaginary (empty) ellipse; here $sign (A) = sign (Δ)$ $Δ_{2} = 0$ and $Δ = 0$ — what shapes are possible? ::: a single double line or two parallel lines