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

Before we start, one shared vocabulary reminder so no symbol is unearned.

Definition Two more symbols we will reuse below

Primed coefficients $A^{'}, B^{'}, C^{'}$ : when you spin the axes by an angle $θ$ (Rotation of Axes), the same curve is described by new coefficients. Writing $x = x^{'} cos θ - y^{'} sin θ$ and $y = x^{'} sin θ + y^{'} cos θ$ and collecting terms, the quadratic part becomes $A^{'} x^{'2} + B^{'} x^{'} y^{'} + C^{'} y^{'2}$ where $A^{'} = A cos^{2} θ + B cos θ sin θ + C sin^{2} θ,$ $C^{'} = A sin^{2} θ - B cos θ sin θ + C cos^{2} θ,$ $B^{'} = B cos 2 θ - (A - C) sin 2 θ .$ So $A^{'}, B^{'}, C^{'}$ are just " $A, B, C$ seen from a tilted pair of axes." Look at the top of the figure below: same curve, two grids.
The coordinate vector $v$ : shorthand for the point's coordinates stacked in a column, $v = [x y]$ . With the symmetric matrix $M = [A B /2 B /2 C]$ , the product $v^{⊤} M v$ multiplies out to exactly $A x^{2} + B x y + C y^{2}$ — the quadratic part in one compact package.

Below is a single reference picture for the geometry these questions keep pointing at — the three shapes, an eigenvector pair, and what "rotating the axes" does.

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

Spot the error

"For

2 x^{2} + 3 x - 5 x y + y^{2} = 0

A = 2

and

B = 3

because

B

multiplies

x

Error —

B

is the coefficient of the cross term $x y$ , so

B = - 5

; the

3 x

is part of the linear term

D = 3

. Confusing this with

b

a x^{2} + b x + c

is the classic slip.

x^{2} - y^{2} = 4

has

B^{2} - 4 A C = 4 > 0

, so it's a hyperbola — no need to check

Δ

The verdict is right here, but the reasoning is incomplete: you must check

Δ

, because the same discriminant sign with

F = 0

gives

x^{2} - y^{2} = 0

, which degenerates to two crossing lines.

"To build the matrix I plug in

A, B, C, D, E, F

straight down the diagonals."

Error — off-diagonal entries are halved:

B /2, D /2, E /2

, because

B x y = \frac{B}{2} x y + \frac{B}{2} y x

splits symmetrically to keep the matrix symmetric.

B^{2} - 4 A C = 0

so it's a parabola — a nice open curve, guaranteed."

Not guaranteed — zero discriminant is the parabola family, but if

Δ = 0

it can degenerate into two parallel lines or a single line (e.g.

x^{2} = 0

"The angle that removes the

x y

term is

tan θ = \frac{B}{A - C}

Error — it is

tan 2 θ = \frac{B}{A - C}

(double angle), because

B^{'}

came out proportional to

B cos 2 θ - (A - C) sin 2 θ

"Since

9 x^{2} - 6 x y + y^{2} = (3 x - y)^{2}

is a square,

B^{2} - 4 A C < 0

and it's an ellipse."

Error — a perfect square gives

B^{2} - 4 A C = 0

(here

36 - 36

), so it's the parabola/degenerate case, not an ellipse.

A = 0

and

C = 0

can't happen in a real conic since there's no

x^{2}

y^{2}

Error —

A = C = 0

with

B \neq = 0

leaves

B x y + \dots = 0

, still a valid second-degree equation. Then

B^{2} - 4 A C = B^{2} > 0

: a hyperbola whose axes lie along the lines

y = \pm x

Why questions

Why does the sign of

B^{2} - 4 A C

, not its magnitude, decide the shape?

Because

B^{2} - 4 A C = - 4 A^{'} C^{'}

in the rotated frame (with

A^{'}, C^{'}

the rotated coefficients above), and it's whether

A^{'}, C^{'}

have the same sign, opposite sign, or one zero that makes a closed loop, two branches, or a single-axis curve — a purely sign-driven distinction.

Why do the linear terms

D x + E y

never change the conic's type?

They only complete-the-square into shifts of

x

and

y

, moving the centre; the degree-2 core

A, B, C

that governs curvature and openness is untouched.

Why must we halve

B

D

E

in the

3 \times 3

matrix?

So the matrix is symmetric and the quadratic form

v^{⊤} M v

(with

v = [x y]

) reproduces the original equation; a cross term

B x y

must be shared equally between the

(1, 2)

and

(2, 1)

entries.

Why is a circle just a special ellipse in this classification?

A circle has equal stretching in all directions, i.e.

A = C

and

B = 0

, giving

B^{2} - 4 A C = - 4 A^{2} < 0

— the same ellipse condition, just with no tilt and equal semi-axes. The circle is the equal-axes case of the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

with

a = b

Why does eliminating the

x y

term make the shape "obvious"?

With

B^{'} = 0

the equation reduces to

A^{'} x^{'2} + C^{'} y^{'2} + \dots = 0

— a standard un-tilted conic such as $\tfrac{x'^2}{a^2}+\tfrac{y'^2}{b^2}=1$, $y'^2=4px'$, or $\tfrac{x'^2}{a^2}-\tfrac{y'^2}{b^2}=1$ — which you read off directly. The angle that achieves

B^{'} = 0

tan 2 θ = \frac{B}{A - C}

(see Rotation of Axes).

Why can we phrase the whole test in terms of eigenvalues?

The quadratic form's matrix

M = [A B /2 B /2 C]

has eigenvalues whose product equals

det M = A C - \frac{B ^{2}}{4} = - \frac{1}{4} (B^{2} - 4 A C)

; same-sign eigenvalues → ellipse, opposite → hyperbola, one zero → parabola (details in Quadratic Forms and Eigenvalues).

Why does a positive discriminant correspond to the quadratic form being "indefinite"?

Positive

B^{2} - 4 A C

means the eigenvalues have opposite signs, so

A x^{2} + B x y + C y^{2}

takes both positive and negative values — its level curves open into two branches, a hyperbola.

Edge cases

What conic is

x^{2} + y^{2} + 1 = 0

B^{2} - 4 A C = - 4 < 0

(ellipse family) but no real point satisfies it — an imaginary ellipse with no real graph. Sign alone would mislead; there are simply no real solutions.

What does

B^{2} - 4 A C > 0

with

Δ = 0

describe?

A degenerate hyperbola: two intersecting straight lines, like

x^{2} - y^{2} = 0

giving

y = \pm x

What does

B^{2} - 4 A C < 0

with

Δ = 0

describe (real case)?

A degenerate ellipse shrunk to a single point, e.g.

x^{2} + y^{2} = 0

satisfied only at the origin.

What does

B^{2} - 4 A C = 0

with

Δ = 0

describe?

A degenerate parabola: two parallel lines (e.g.

x^{2} - 1 = 0 \Rightarrow x = \pm 1

), a single line, or no real locus.

What if

A = B = C = 0

Then it isn't a second-degree equation at all — only

D x + E y + F = 0

remains, a straight line. The definition requires

(A, B, C) \neq = (0, 0, 0)

What shape is

x y = 1

Here

A = C = 0, B = 1

, so

B^{2} - 4 A C = 1 > 0

: a hyperbola, with its branches along the diagonals

y = \pm x

— even with no

x^{2}

y^{2}

terms.

What happens to

B^{2} - 4 A C

if you scale the whole equation by a nonzero constant

k

Every degree-2 coefficient scales by

k

, so

B^{2} - 4 A C

scales by

k^{2} > 0

— its sign is unchanged, so the classification is unaffected.

Recall One-line summary of the whole trap-set

Sign of $B^{2} - 4 A C$ ::: names the family (negative → loop, zero → parabola, positive → split), and $Δ \neq = 0$ ::: confirms the conic hasn't collapsed. Miss the second check and every degenerate case ambushes you.

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

True or false — justify

Spot the error

Why questions

Edge cases