How is it that the Conic Discriminant tells us the Type of Conic?

This section is optional in the Precalculus course.
There are no exercises in this section.

This section addresses a question about conics which may have arisen as you studied Identifying Conics by the Discriminant:

The purpose of this section is not to give a complete answer to this question.
Instead, the purpose is to give enough information—at a level within the grasp of an ambitious Precalculus student—to make the answer plausible.

Here's the situation (summarized from Identifying Conics by the Discriminant):

Why is this true?
That is, why does $\,B^2 - 4AC\,$ give this information?
Keep reading!

Here's the overview of what we'll do in this optional section:

At this point, you'll be able to say:

Okay $\,\ldots\,$ so $\,B^2 - 4AC\,$ really does identify the conic.
But—where did it come from? How did someone think to look at this particular expression?
We'll give a motivation for this, too!

The details follow:

Equivalent Definitions of Conics

There are equivalent definitions of conics:

Summary: Standard Forms of Conics

The table below summarizes the standard forms for the conics—they'll all be derived in subsequent sections.
These standard forms position the conics in an $\,xy\,$-coordinate system so that the simplest possible equations result.

CONIC STANDARD FORM(S) COMPUTATION OF DISCRIMINANT, $\,B^2 - 4AC\,$
With all three terms on the same side of the equation:
$\,A\,$ is the coefficient of the $\,x^2\,$ term
$\,B\,$ is the coefficient of the $\,xy\,$ term
$\,C\,$ is the coefficient of the $\,y^2\,$ term
ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ $\displaystyle A = \frac{1}{a^2}\,$,   $B = 0\,$,   $\displaystyle C = \frac{1}{b^2}\,$ $$ \begin{align} B^2 - 4AC \ \ &=\ \ 0^2 - 4\cdot\frac{1}{a^2}\cdot\frac{1}{b^2}\cr &= \frac{-4}{a^2\,b^2}\cr\cr &< 0 \end{align} $$
parabola $$x^2 = 4py$$ or $$y^2 = 4px$$ $A = 1\,$,   $B = 0\,$,   $C = 0\,$ $$ B^2 - 4AC \ \ =\ \ 0^2 - 4(1)(0)\ \ =\ \ 0 $$ or

$A = 0\,$,   $B = 0\,$,   $C = 1\,$ $$ B^2 - 4AC \ \ =\ \ 0^2 - 4(0)(1)\ \ =\ \ 0 $$
hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ or $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ $\displaystyle A = \frac{1}{a^2}\,$,   $B = 0\,$,   $\displaystyle C = -\frac{1}{b^2}\,$ $$ B^2 - 4AC \ \ =\ \ 0^2 - 4\cdot\frac{1}{a^2}\cdot\frac{-1}{b^2}\ \ = \frac{4}{a^2\,b^2}\ \ >\ \ 0 $$ or

$\displaystyle A = -\frac{1}{b^2}\,$,   $B = 0\,$,   $\displaystyle C = \frac{1}{a^2}\,$ $$ B^2 - 4AC \ \ =\ \ 0^2 - 4\cdot\frac{-1}{b^2}\cdot\frac{1}{a^2}\ \ = \frac{4}{a^2\,b^2}\ \ >\ \ 0 $$

The Standard Forms have the Expected Discriminants

Notice, from the table above, that the standard forms have the expected discriminants:

ellipses:$\,B^2 - 4AC < 0$
parabolas:$\,B^2 - 4AC = 0$
hyperbolas:$\,B^2 - 4AC > 0$

Every general conic equation can be transformed to a standard form

Every general conic equation can be transformed to a standard form, using only (as needed) a rotation and/or horizontal/vertical translations.

None of the standard forms have an $\,xy\,$ term.
Therefore, we need to get rid of the $\,xy\,$ term from the general conic equation.
An appropriate rotation of the graph will do the job, as shown below.

After the $\,xy\,$ term is gone, horizontal/vertical translation(s) will give the desired standard form.

Rotating a Graph

Given a graph, we seek a new equation which describes the same graph—except that the graph is rotated counterclockwise by an angle $\,\theta\,$.

The necessary ‘rotation equations’ follow easily from two things:
  • the addition formulas for sine and cosine: $$ \begin{gather} \cos(a+b) = \cos a\,\cos b - \sin a\,\sin b\cr \sin(a+b) = \sin a\,\cos b + \cos a\,\sin b \end{gather} $$
  • the sketch at right, which shows:
    • an initial point $\,(x,y)\,$
    • the new point $\,(\hat x,\hat y)\,$ after a counterclockwise rotation by angle $\,\theta\,$
From the sketch: $$\begin{align} x &= h\cos\phi\cr y &= h\sin\phi\cr \cr \hat x &= h\cos(\phi + \theta)\cr &= h(\cos\phi\,\cos\theta - \sin\phi\,\sin\theta)\cr &= (h\cos\phi)\cos\theta - (h\sin\phi)\sin\theta\cr &= x\cos\theta - y\sin\theta\cr \cr \hat y &= h\sin(\phi + \theta)\cr &= h(\sin\phi\,\cos\theta + \cos\phi\,\sin\theta)\cr &= (h\sin\phi)\cos\theta + (h\cos\phi)\sin\theta\cr &= y\cos\theta + x\sin\theta \end{align} $$

Next, the equations $$\begin{gather} \hat x = x\cos\theta - y\sin\theta\qquad (1)\cr \hat y = y\cos\theta + x\sin\theta\qquad (2) \end{gather} $$ are solved for $\,x\,$ and $\,y\,$ (in terms of $\,\hat x\,$ and $\,\hat y\,$), as follows.

To eliminate $\,y\,$:

$$\begin{gather} \end{gather} $$

Thus, we have:

To rotate a graph counterclockwise by an angle $\,\theta\,$

the Rotation Equations:
$x = \hat x\cos\theta + \hat y\sin\theta$
$y = \hat y\cos\theta - \hat x\sin\theta$

Suppose you have a graph, which is the solution set of a known equation in $\,x\,$ and $\,y\,$.

A Simple Rotation Example

Here's a simple example to illustrate the idea.
If the line $\,y = x\,$ is graphed in a coordinate system where the scales on $\,x\,$ and $\,y\,$ are identical,
then it graphs as a line that makes a $\,45^\circ\,$ angle with the positive $\,x\,$-axis.

If this line is rotated by $\,45^\circ\,$ counterclockwise, then it ‘turns into’ the $y$-axis: i.e., the line $\,x = 0\,$.

Here are the steps in transforming the equation:

Rotate the General Conic Equation to Eliminate the $\,xy\,$ Term

Next, we apply the rotation equations to the general conic equation, $\,Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\,$.
The goal is to determine an appropriate angle of rotation, $\,\theta\,$, that will eliminate the $\,xy\,$ term.

With No $\,xy\,$ Term in the General Conic Equation,
Appropriate Translation(s) Will Always Produce the Standard Forms

After an appropriate rotation, we'll have a new conic equation in $\,\hat x\,$ and $\,\hat y\,$ that has no $\,\hat x\hat y\,$ term: $$\tilde A \hat x^2 + \tilde C \hat y^2 + \tilde D \hat x + \tilde E\hat y + \tilde F = 0 $$ The other coefficients likely all have changed in the process.
This equation looks unnecessarily complicated, with all the ‘hats’ ($\,\hat{}\,$) and ‘tildes’ ($\,\tilde{}\,$).
For simplicity, let's drop them all and instead investigate the equation: $$Ax^2 + Cy^2 + Dx + Ey + F = 0 \qquad (\ddagger)$$

Use the completing the square technique to get perfect squares involving $\,x\,$ and $\,y\,$:

The Conic Discriminant is Invariant Under Translations

When the general conic equation is translated (horizontally and/or vertically), a new equation results.
However, the original equation and the new (translated) equation will have exactly the same discriminant!
To see this, replace every $\,x\,$ by $\,x - c\,$ and every $\,y\,$ by $\,y - d\,$ in the general conic equation, and then focus attention on the degree-two terms: $$ \begin{gather} A(x-c)^2 + B(x-c)(y-d) + C(y-d)^2 + D(x-c) + E(y - d) + F = 0\cr\cr Ax^2 + Bxy + Cy^2 + \hat D x + \hat Ey + \hat F = 0 \end{gather} $$ Note that the coefficients of the degree-two terms haven't changed!
Thus, the discriminant is invariant under translations.

The Conic Discriminant is Invariant Under Rotation

When the general conic equation is rotated, a new equation results.
However, the original equation and the new (rotated) equation will have exactly the same discriminant!
To see this, use the rotation equations in the general conic equation, and focus attention on the degree-two terms.
For simplicity, let $\,c := \cos\theta\,$ and $\,s := \sin\theta\,$. $$ \begin{gather} A(\hat xc + \hat ys)^2 + B(\hat xc + \hat ys)(\hat yc - \hat xs) + C(\hat yc - \hat xs)^2 + \text{(non-degree-two stuff)} = 0\cr\cr \text{(drop the hats on the variables, for simplicity)}\cr\cr A(c^2x^2 + 2csxy + s^2y^2) + B(c^2xy - csx^2 + csy^2 - s^2xy) + C(c^2y^2 - 2csxy + s^2x^2) + \text{non-degree-two terms} = 0\cr\cr x^2(Ac^2 - Bcs + Cs^2) + xy(2Acs + Bc^2 - Bs^2 - 2Ccs) + y^2(As^2 + Bcs + Cc^2) + \text{non-degree-two terms} = 0\cr\cr \end{gather} $$ For this new (rotated) equation, the degree-two coefficients are: $$ \begin{align} \hat A &= Ac^2 - Bcs + Cs^2\cr\cr \hat B &= B(c^2-s^2) + 2cs(A - C)\cr\cr \hat C &= As^2 + Bcs + Cc^2 \end{align} $$ At this point, it's an exercise in patience and being careful!
Here's the new discriminant: $$ \begin{align} \hat B^2 - 4\hat A\hat C &= \bigl(B(c^2-s^2) + 2cs(A - C)\bigr)^2 - 4(Ac^2 - Bcs + Cs^2)(As^2 + Bcs + Cc^2)\cr\cr &= B^2(c^2-s^2)^2 + 4Bcs(c^2-s^2)(A-C) + 4c^2s^2(A-C)^2 \cr &\qquad \qquad \qquad -4(A^2c^2s^2 + ABc^3s + ACc^4 - ABcs^3 - B^2c^2s^2 - BCc^3s + ACs^4 + BCcs^3 + C^2c^2s^2)\cr\cr &= B^2(c^4 - 2c^2s^2 + s^4) + 4Bcs(Ac^2 - Cc^2 - As^2 + Cs^2) + 4c^2s^2(A^2 - 2AC + C^2)\cr &\qquad \qquad \qquad -4A^2c^2s^2 - 4ABc^3s - 4ACc^4 + 4ABcs^3 + 4B^2c^2s^2 + 4BCc^3s - 4ACs^4 - 4BCcs^3 - 4C^2c^2s^2\cr\cr &= B^2c^4 \color{green}{- 2B^2c^2s^2} + B^2s^4 \color{orange}{\cancel{+ 4ABc^3s}} \color{grey}{\cancel{\bcancel{- 4BCc^3s}}} \color{purple}{\bcancel{- 4ABcs^3}} \color{red}{\bcancel{\cancel{+ 4BCcs^3}}} \color{blue}{\bcancel{+ 4A^2c^2s^2}} - 8ACc^2s^2 + \color{pink}{\cancel{4C^2c^2s^2}}\cr &\qquad \qquad \qquad \color{blue}{\bcancel{- 4A^2c^2s^2}} \color{orange}{\cancel{- 4ABc^3s}} - 4ACc^4 \color{purple}{\bcancel{+ 4ABcs^3}} \color{green}{+ 4B^2c^2s^2} \color{grey}{\cancel{\bcancel{+ 4BCc^3s}}} - 4ACs^4 \color{red}{\bcancel{\cancel{- 4BCcs^3}}} \color{pink}{\cancel{- 4C^2c^2s^2}}\cr\cr &= B^2c^4 - 4ACc^4 + 2B^2c^2s^2 - 8ACc^2s^2 + B^2s^4 - 4ACs^4 \cr\cr &= c^4(B^2 - 4AC) + 2c^2s^2(B^2-4AC) + s^4(B^2-4AC)\cr\cr &= (B^2 - 4AC)(c^4 + 2c^2s^2 + s^4)\cr\cr &= (B^2 - 4AC)(\cos^2 \theta + \sin^2\theta)^2\cr\cr &= B^2 - 4AC \end{align} $$ Whew!
So, the discriminant of the new (rotated) equation is the same as the original discriminant.
Thus, the discriminant is invariant under rotation.

How Did Someone Know to Look at This Particular Expression, $\,B^2 - 4AC\,$?

At this point, you should understand why the discriminant, $\,B^2 - 4AC\,$, can be used to identify
the type of conic you get from the equation $\,Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\,$.

However, you might still be wondering—how did someone think to look at this particular expression?
Here's one way that (say) an inquisitive student might have stumbled upon the suspected importance of $\,B^2 - 4AC\,$.

First, re-name the expression $\,Ax^2 + Bxy + Cy^2\,$ by completing the square on the first two terms.
(This re-writing requires $\,A\ne 0\,$.) $$ \begin{alignat}{2} Ax^2 + Bxy + Cy^2 &= Ax^2 + (By)x + Cy^2 &\qquad&\text{(treat $\,By\,$ as the coefficient of the $\,x\,$ term)}\cr\cr &= A(x^2 + \frac{By}{A}x \qquad) + Cy^2 &&\text{(completing the square requires the coefficient of $\,x^2\,$ to be $\,1\,$)}\cr\cr &= A\left(x^2 + \frac {By}{A}x + \left(\frac{By}{2A}\right)^2\right) + Cy^2 - A\left(\frac{By}{2A}\right)^2 &&\text{(add half the $\,x\,$ coefficient, squared; add zero in appropriate form)} \cr\cr &= A\left(x + \frac{By}{2A}\right)^2 + y^2\left(C - \frac{B^2}{4A}\right)&&\text{(write first expression as a perfect square; factor out $\,y^2\,$)}\cr\cr &= A\left(x + \frac{B}{2A}y\right)^2 + y^2\left(\frac{4AC}{4A} - \frac{B^2}{4A}\right)&&\text{(get common denominator)}\cr\cr &= A\left(x + \frac{B}{2A}y\right)^2 - \frac{y^2}{4A}(B^2 - 4AC)&&\text{(algebra)}\cr\cr \end{alignat} $$ Use this new name to explore the conic equation with only degree-two terms: $$ \begin{gather} Ax^2 + Bxy + Cy^2 = 0\cr\cr A\left(x + \frac{B}{2A}y\right)^2 - \frac{y^2}{4A}(B^2 - 4AC) = 0\cr\cr \color{green}{4A^2\left(x + \frac{B}{2A}y\right)^2} - \color{red}{y^2}(B^2 - 4AC) = 0\cr\cr \end{gather} $$ Since the green and red parts are always nonnegative, the nature of the equation
changes significantly depending upon whether $\,B^2-4AC\,$ is positive, negative, or zero!

Note: If you assume $\,C\ne 0\,$, and complete the square on $\,y\,$, you reach the same conclusion.

There are no exercises in this section.

Move on to the next lesson:
Parabolas: Definition, Reflectors/Collectors, Derivation of Equations