Conic sections—circles/ellipses, parabolas, hyperbolas—were introduced
in the prior section.
They arise by intersecting a plane with the surface of an infinite double cone.
In this current section,
we begin the analysis of equations representing conic sections.
There are results in this section for which you might be wondering: ‘Why is this true?’
If so, then you're encouraged to study the following optional sections:
Let $\,A\,$, $\,B\,$, $\,C\,$, $\,D\,$, $\,E\,$ and $\,F\,$ be real numbers. Consider the equation:
The $\,x^2\,$, $\,xy\,$ and $\,y^2\,$ terms are often called the degreetwo terms in this equation.
Why?
Because the exponents are $\,2\,$, or they add to two:
There is a beautiful relationship between equations of this type
and the graphs of conic sections!
Indeed, both of the following are true:
It's not particularly easy to go from an arbitrary equation of the form
($\,\dagger\,$) to its graph, using paperandpencil.
(In the next few sections, you'll learn to graph fairly simple conics.)
However,
WolframAlpha is up to the task!
If you're bored, jump up there, type in a random general conic equation,
and see the awesomeness of WolframAlpha!
Here are some WolframAlpha results
to illustrate the variety of conic sections (including some degenerate and limiting cases).
You can cutandpaste the
equations below each graphic into WolframAlpha yourself
to verify the results.
Enjoy!
5x^23xy+2y^2+3x8y7 = 0

(x1)^2 + (y+3)^2 = 16

xy = 1

y = x^2  5x + 1

7  3xy  y = 9x^2 + y^2/4

x^2 + 2xy + y^2 = 0

x^2 + y^2  2xy = 5

x^2  y^2 = 0

Or, you can play with the equation $\,Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\,$ below.
Use the sliders to set the desired values for the parameters $\,A\,$, $\,B\,$, $\,C\,$, $\,D\,$, $\,E\,$, and $\,F\,$.
The initial configuration is $\,5x^2  3xy + 2y^2 + 3x  8y  7 = 0\,$; refresh the page to reset.
You can use the navigation arrows at the bottom right to see more of the graph.
Enjoy!
Even though conics can be challenging to graph, here's some really good news.
There's a simple way to decide which type of conic is represented by a given general conic equation, as follows:
CONIC DISCRIMINANT  TYPE OF CONIC SECTION  POSSIBLE MEMORY DEVICE 
$B^2  4AC < 0$ 
ELLIPSE (possibly degenerate) If, in addition, $\,A = C\,$ then it's a circle. 
Look at the letters in the word ELLIPSE! There's an L. There's an E. There's an S. It practically spells out ‘less’! When the discriminant is less than zero, it's an ellipse. 
$$B^2  4AC = 0$  PARABOLA (possibly degenerate)  (no memory device) 
$B^2  4AC > 0$  HYPERBOLA (possibly degenerate) 
If you're HYPER, then you have a GREAT deal of energy. When the discriminant is GREATer than zero, it's a HYPERbola. 
Note the similarity to the discriminant of a quadratic equation $\,ax^2 + bx + c = 0\,$.
The quadratic discriminant is $\,b^2  4ac\,$, and helps us decide how many real solutions the quadratic equation has.
Example 1:
general conic equation in standard form
Consider the equation $\,5x^2  3xy + 2y^2 + 3x  8y  7 = 0\,$ (from the WolframAlpha example above).
It's already in the form $\,Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\,$, so just read off the required coefficients
Example 2:
general conic equation not in standard form
Consider the equation $\,y = x^2  5x + 1\,$.
It's a general conic equation, but it's not in standard form.
From prior mathematical experience, you know that
quadratic functions graph as parabolas.
However, let's check that
the ‘conic discriminant’ gives the same information.
You could put the equation in standard form: either
Example 3:
a degenerate conic—a ‘point circle’
Consider the equation $\,x^2 + y^2 = 0\,$.
The only point that makes this equation true is $\,(x,y) = (0,0)\,$.
Read off $\,A = 1\,$, $\,B = 0\,$, and $\,C = 1\,$.
Thus, $\,B^2  4AC = 0^2  4(1)(1) = 4 < 0\,$.
Since the discriminant is negative, it's an ellipse.
Since, in addition, $\,A = C\,$, it's a circle.
(A circle is a special case of an ellipse.)
This is a degenerate case: a circle of radius zero, fondly called a ‘point circle’.
Example 4:
a hyperbola
Consider the equation $\displaystyle\,y = \frac 1x\,$.
In this equation, $\,x\,$ is not allowed to equal zero.
With $\,x \ne 0\,$, multiplying both sides by $\,x\,$ gives $\,xy = 1\,$, which is now recognizable as a conic.
In the equation $\,xy = 1\,$, we have $\,A = C = 0\,$ and $\,B = 1\,$.
Thus, $\,B^2  4AC = 1 > 0\,$.
Since the discriminant is positive, it's a hyperbola.
The familiar reciprocal function is a hyperbola.
Here are additional things you should think about as you study this lesson:
The limiting case we need is an infinite cylinder. (Think of an ‘infinite cylinder’ as an infinite empty paper towel roll.) Just intersect such a cylinder (see right) with a vertical plane to get a pair of parallel lines. Here's how to generate an infinite double cone, in order to get an infinite cylinder as a limiting case:
At this point, you've got a familiar infinite double cone. Now, to get the required limiting case:
This lesson is part of a Precalculus course—it is intended to prepare you for Calculus. The central idea in Calculus is a limit. These discussions of ‘limiting cases’ is a good preview of an essential Calculus idea! 
both up and down. 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 