The General Quadratic Equation

Try this

Press the seven examples in turn. Which readings decide the “type”? Which readings, once they hit $0$ , make the curve “break down” into lines, a point or the empty set?
Choose “ellipse” and drag only $F$ . Does $B^{2} - 4 A C$ change? And $Δ$ ? What happens in the figure as $Δ$ passes through $0$ ?
Choose “hyperbola” and drag $F$ until $Δ = 0$ . What do the two branches turn into?
The dashed lines are the curve's axes of symmetry; their angle with the $x$ -axis is exactly the rotation angle. Drag $B$ : how do the axes of symmetry turn?

Rotate and then translate the axes for the general quadratic equation $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ , and only three kinds of result are possible. Since $B^{2} - 4 A C$ is invariant under rotation and does not move at all under translation, it decides the “type” directly:

$B^{2} - 4 A C$	Type	General case ( $Δ \neq = 0$ )	Special cases ( $Δ = 0$ etc.)
$< 0$	elliptic	ellipse (or no points)	a point
$> 0$	hyperbolic	hyperbola	two crossing lines
$= 0$	parabolic	parabola	two parallel lines, one double line, or no points

Here $Δ = A \frac{B}{2} \frac{D}{2} \frac{B}{2} C \frac{E}{2} \frac{D}{2} \frac{E}{2} F$ , which is also invariant under rotation and translation of the axes.

Predict first

Which type is $4 x^{2} + 4 x y + y^{2} - 3 x + 5 y - 2 = 0$ ?

elliptic typehyperbolic typeparabolic type

Choose an answer first; then you can see the result.

Why: simplifying step by step

Discuss and simplify 3x² + 2xy + 3y² − 8x − 8y = 0

Step 1 / 6

Find the type
$B^{2} - 4 A C = 4 - 36 = - 32 < 0$ : elliptic type, so the result can only be an ellipse, a point, or no points.

$B^{2} - 4 A C = - 32 < 0$
Check for degeneracy
Rotate the axes
Translate the axes
Change one constant
A bit bigger

The general procedure: if $B = 0$ , translating the axes is enough; if $B \neq = 0$ , first rotate to remove the $x y$ -term, then translate. For the elliptic and hyperbolic types you can also translate to the center first and then rotate. The center comes from solving $2 A h + B k + D = 0$ , $B h + 2 C k + E = 0$ , and this is often less work.

If Δ = 0

Break it

Counterexample lab: remove one condition

Conclusion: The sign of $B^{2} - 4 A C$ decides whether the curve is an ellipse, a hyperbola or a parabola

$Δ \neq = 0$ : $B^{2} - 4 A C > 0$ , a genuine hyperbola.

Practice

Exercise 1Basic

Find $B^{2} - 4 A C$ for $3 x^{2} - 5 x y + 2 y^{2} + x - 7 = 0$ and decide its type. (Enter the number.)

Exercise 2Basic

$k > 0$ , and the equation $x^{2} + k x y + 4 y^{2} - 1 = 0$ is of parabolic type. Find $k$ and describe its graph.

Exercise 3Advanced

After simplification, $x^{2} - 2 x y + y^{2} - 2 x - 6 y + 1 = 0$ becomes $Y^{2} = 2 pX$ . Find $p^{2}$ .

Exercise 4Challenge

For which $k$ does $x^{2} - x y - 2 y^{2} + x + 4 y + k = 0$ represent two lines?

Try this

Why: simplifying step by step

Discuss and simplify 3x² + 2xy + 3y² − 8x − 8y = 0

If Δ = 0

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Taking half the $xy$-coefficient as $B$