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):
Here's the overview of what we'll do in this optional section:
At this point, you'll be able to say:
The details follow:
There are equivalent definitions 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 $\displaystyle A = \frac{1}{b^2}\,$, $B = 0\,$, $\displaystyle C = \frac{1}{a^2}\,$ $$ B^2  4AC \ \ =\ \ 0^2  4\cdot\frac{1}{b^2}\cdot\frac{1}{a^2}\ \ = \frac{4}{a^2\,b^2}\ \ >\ \ 0 $$ 
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,
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.
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:

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 $\,y\,$:
Thus, we have:
replace every $\,x\,$ by:  $\,\hat x\cos\theta + \hat y\sin\theta\,$ 
replace every $\,y\,$ by:  $\,\hat y\cos\theta  \hat x\sin\theta$ 
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:
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.
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 $\,y\,$:
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 degreetwo terms:
$$
\begin{gather}
A(xc)^2 + B(xc)(yd) + C(yd)^2 + D(xc) + E(y  d) + F = 0\cr\cr
Ax^2 + Bxy + Cy^2 + \hat D x + \hat Ey + \hat F = 0
\end{gather}
$$
Note that the coefficients of the degreetwo terms haven't changed!
Thus, the discriminant is invariant under translations.
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 degreetwo terms.
For simplicity, let $\,c := \cos\theta\,$ and $\,s := \sin\theta\,$.
$$
\begin{gather}
A(\hat xc + \hat ys)^2 + B(\hat xc + \hat ys)(\hat yc  \hat xs) + C(\hat yc  \hat xs)^2 + \text{(nondegreetwo stuff)} = 0\cr\cr
\text{(drop the hats on the variables, for simplicity)}\cr\cr
A(c^2x^2 + 2csxy + s^2y^2)
+ B(c^2xy  csx^2 + csy^2  s^2xy)
+ C(c^2y^2  2csxy + s^2x^2) + \text{nondegreetwo terms} = 0\cr\cr
x^2(Ac^2  Bcs + Cs^2)
+ xy(2Acs + Bc^2  Bs^2  2Ccs)
+ y^2(As^2 + Bcs + Cc^2) + \text{nondegreetwo terms} = 0\cr\cr
\end{gather}
$$
For this new (rotated) equation, the degreetwo coefficients are:
$$
\begin{align}
\hat A &= Ac^2  Bcs + Cs^2\cr\cr
\hat B &= B(c^2s^2) + 2cs(A  C)\cr\cr
\hat C &= As^2 + Bcs + Cc^2
\end{align}
$$
At this point, it's an exercise in patience and being careful!
Here's the new discriminant:
$$
\begin{align}
\hat B^2  4\hat A\hat C &= \bigl(B(c^2s^2) + 2cs(A  C)\bigr)^2  4(Ac^2  Bcs + Cs^2)(As^2 + Bcs + Cc^2)\cr\cr
&= B^2(c^2s^2)^2 + 4Bcs(c^2s^2)(AC) + 4c^2s^2(AC)^2 \cr
&\qquad \qquad \qquad 4(A^2c^2s^2 + ABc^3s + ACc^4  ABcs^3  B^2c^2s^2  BCc^3s + ACs^4 + BCcs^3 + C^2c^2s^2)\cr\cr
&= B^2(c^4  2c^2s^2 + s^4) + 4Bcs(Ac^2  Cc^2  As^2 + Cs^2) + 4c^2s^2(A^2  2AC + C^2)\cr
&\qquad \qquad \qquad 4A^2c^2s^2  4ABc^3s  4ACc^4 + 4ABcs^3 + 4B^2c^2s^2 + 4BCc^3s  4ACs^4  4BCcs^3  4C^2c^2s^2\cr\cr
&= B^2c^4 \color{green}{ 2B^2c^2s^2} + B^2s^4 \color{orange}{\cancel{+ 4ABc^3s}} \color{grey}{\cancel{\bcancel{ 4BCc^3s}}} \color{purple}{\bcancel{ 4ABcs^3}} \color{red}{\bcancel{\cancel{+ 4BCcs^3}}} \color{blue}{\bcancel{+ 4A^2c^2s^2}}  8ACc^2s^2 + \color{pink}{\cancel{4C^2c^2s^2}}\cr
&\qquad \qquad \qquad \color{blue}{\bcancel{ 4A^2c^2s^2}} \color{orange}{\cancel{ 4ABc^3s}}  4ACc^4 \color{purple}{\bcancel{+ 4ABcs^3}} \color{green}{+ 4B^2c^2s^2} \color{grey}{\cancel{\bcancel{+ 4BCc^3s}}}  4ACs^4 \color{red}{\bcancel{\cancel{ 4BCcs^3}}} \color{pink}{\cancel{ 4C^2c^2s^2}}\cr\cr
&= B^2c^4  4ACc^4 + 2B^2c^2s^2  8ACc^2s^2 + B^2s^4  4ACs^4 \cr\cr
&= c^4(B^2  4AC) + 2c^2s^2(B^24AC) + s^4(B^24AC)\cr\cr
&= (B^2  4AC)(c^4 + 2c^2s^2 + s^4)\cr\cr
&= (B^2  4AC)(\cos^2 \theta + \sin^2\theta)^2\cr\cr
&= B^2  4AC
\end{align}
$$
Whew!
So, the discriminant of the new (rotated) equation is the same as the original discriminant.
Thus, the discriminant is invariant under rotation.
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, rename the expression $\,Ax^2 + Bxy + Cy^2\,$ by completing the square on the first two terms.
(This rewriting 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 degreetwo 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 $\,B^24AC\,$ is positive, negative, or zero!
Note: If you assume $\,C\ne 0\,$, and complete the square on $\,y\,$, you reach the same conclusion.