Discriminant (Quadratic Curve)

Given a general Quadratic Curve

$$Ax^2+Bxy+Cy^2+Dx+Ey+F = 0,$$ (1)

the quantity $X$ is known as the discriminant, where

$$X \equiv B^2-4AC,$$ (2)

and is invariant under Rotation. Using the Coefficients from Quadratic Equations for a rotation by an angle $\theta$,

$$A' = {\textstyle{1\over 2}}A[1+\cos(2\theta)]+ {\textstyle{1\over 2}}B\sin(2\theta)+ {\textstyle{1\over 2}}C[1-\cos(2\theta)]$$

$$= {A+C\over 2} + {B\over 2} \sin(2\theta )+ {A-C\over 2}\cos(2\theta )$$ (3)

$$B' = G\cos\left({2\theta +\delta - {\pi\over 2}}\right)= G\sin(2\theta +\delta)$$ (4)

$$C' = {\textstyle{1\over 2}}A[1-\cos(2\theta)]- {\textstyle{1\over 2}}B\sin(2\theta)+{\textstyle{1\over 2}})C[1+\cos(2\theta )]$$

$$= {A+C\over 2} - {B\over 2} \sin(2\theta) + {C-A\over 2}\cos(2\theta).$$ (5)

Now let

$$G \equiv \sqrt{B^2+(A-C)^2}$$ (6)

$$\delta \equiv \tan^{-1}\left({B\over C-A}\right)$$ (7)

$$\delta_2 \equiv \tan^{-1}\left({A-C\over B}\right)= -\cot^{-1}\left({B\over C-A}\right),$$ (8)

and use

$$\cot^{-1}(x) = {\textstyle{1\over 2}}\pi-\tan^{-1}(x)$$ (9)

$$\delta_2 = \delta-{\textstyle{1\over 2}}\pi$$ (10)

to rewrite the primed variables

$$A' = {A+C\over 2}+{\textstyle{1\over 2}}G\cos(2\theta+\delta)$$ (11)

$$B' = B\cos(2\theta)+(C-A)\sin(2\theta) = G\cos(2\theta+\delta_2)$$ (12)

$$C' = {A+C\over 2}-{\textstyle{1\over 2}}G\cos(2\theta+\delta).$$ (13)

From (11) and (13), it follows that

$$4A'C' = (A+C)^2-G^2\cos(2\theta+\delta).$$ (14)

Combining with (12) yields, for an arbitrary $\theta$

$$X \equiv B'^2-4A'C'$$

$$= G^2\sin^2(2\theta +\delta )+G^2\cos^2(2\theta +\delta ) -(A+C)^2$$

$$= G^2-(A+C)^2 = B^2+(A-C)^2-(A+C)^2$$

$$= B^2-4AC,$$ (15)

which is therefore invariant under rotation. This invariant therefore provides a useful shortcut to determining the shape represented by a Quadratic Curve. Choosing $\theta$ to make $B' = 0$ (see Quadratic Equation), the curve takes on the form

$$A'x^2+C'y^2+D'x+E'y+F = 0.$$ (16)

Completing the Square and defining new variables gives

$$A'x'^2+C'y'^2=H.$$ (17)

Without loss of generality, take the sign of $H$ to be positive. The discriminant is

$$X = B'^2-4A'C' = -4A'C'.$$ (18)

Now, if $-4A'C' < 0$, then $A'$ and $C'$ both have the same sign, and the equation has the general form of an Ellipse (if $A'$ and $B'$ are positive). If $-4A'C' > 0$, then $A'$ and $C'$ have opposite signs, and the equation has the general form of a Hyperbola. If $-4A'C' = 0$, then either $A'$ or $C'$ is zero, and the equation has the general form of a Parabola (if the Nonzero $A'$ or $C'$ is positive). Since the discriminant is invariant, these conclusions will also hold for an arbitrary choice of $\theta$, so they also hold when $-4A'C'$ is replaced by the original $B^2-4AC$. The general result is

1. If $B^2-4AC < 0$, the equation represents an Ellipse, a Circle (degenerate Ellipse), a Point (degenerate Circle), or has no graph.

2. If $B^2-4AC > 0$, the equation represents a Hyperbola or pair of intersecting lines (degenerate Hyperbola).

3. If $B^2-4AC = 0$, the equation represents a Parabola, a Line (degenerate Parabola), a pair of Parallel lines (degenerate Parabola), or has no graph.