Each student will get individual data for the assignment.
Where to find ...
Pictures of degree two polynomials in two
variables
General information about conic sections
Pictures of degree two polynomials in three variables
Pictures of normal vectors to surfaces and how to
draw them
Pictures of normal vectors to curves and how to draw them
Nothing
This might be the most disturbing
"picture". The equation x^{2}+y^{2}+1=0 is not
satisfied by any pairs of numbers, (x,y), since the righthand side is
always positive. (Maybe this is an advertisement for complex numbers?)
One point
Look at
29x^{2}+2xy+10y^{2}+8x+12y+4=0. Then realize this is
the same as (5xy)^{2}+(2x+3y+2)^{2}=0 because
(5xy)^{2}=25x^{2}10xy+y^{2} and
(2x+3y+2)^{2}=4x^{2}+9y^{2}+4+12xy+8x+12y. I
did this in reverse, actually, and started with the squares. But the
only way that (5xy)^{2}+(2x+3y+2)^{2} can be 0 is if
(squares are never negative!) 5xy=0 and 2x+3y+2=0. The first equation
gives y=5x and when substituted in the second equation you get
2x+3(5x)+2=0 so that x=2/17 and then y=10/17. So "clearly" (not so
clearly to me!) the point (2/17,10,17) is the only solution to
29x^{2}+2xy+10y^{2}+8x+12y+4=0.
A straight line
Just x+y=0 will do.
A circle, or its squeezed relative, an ellipse
Certainly
students should recognize x^{2}+2y^{2}1=0 as an
ellipse. But you may find it difficult to believe that
11x^{2}32xy+24y^{2}+24x40y+21=0 is also an
ellipse. You could try the Maple command implicitplot(11*x^232*x*y+24*y^2+24*x40*y+21=0, x=1..6, y=2..5, scaling=constrained, thickness=2, color=black, grid=[100,100]); and see the result. Most of the options to this command should be familiar (the thickness and scaling, for example). One option which might be new is grid. The default for the implicit plot command is to check on a 25 by 25 grid (that's 625 points). Requesting more points will usually give a smoother picture. Of course, the grid=[100,100] requires 4^{2}=16 times as much computation. On my home PC, the [25,25] grid took about .064 seconds and the [100,100] grid took about .72 seconds. 
With grid=[100,100] 

No specification, so grid=[25,25]  
One simple example is x^{2}+y=0.
And, again, it may not be obvious that
25x^{2}40xy+16y^{2}+13x9y+3=0 is also a parabola, but displayed to the right is the result of this Maple command:
implicitplot(25*x^240*x*y+16*y^2+13*x9*y+3=0,
x=10..0, y=10..0, scaling=constrained, thickness=2, color=black,
grid=[100,100]); 
The model hyperbolas you have been told about include xy=1 and
x^{2}y^{2}=1.
But 7x^{2}6xy+16y^{2}+48x80y+74=0 is also the equation of a hyperbola whose graph is shown to the right. 
Step 1 Take a straight line (which will be the axis of symmetry of the cone), and take a plane perpendicular to the line. In the plane, imagine a circle whose center is the point of intersection of the line and the plane. Now fix a point on the line outside of the plane.  
Step 2 Now draw a straight line through the external point and every point on the circle. This collection of lines will form a surface. Each of the lines is called a generator of the cone. The original line is the axis of symmetry of the cone. The external point is called the vertex of the cone.  
The right circular cone looks like what is sketched to the right. Suppose this cone is sliced by another plane. A conic section is the resulting curve: the intersection of the cone and the plane. The type of the curve is determined by the angle made by the intersecting plane and the axis, and how this angle relates to the vertex angle of the cone. 

Ellipse
If the angle made by the intersecting plane and the axis is greater than the vertex angle, then the plane will only slice through one of the cone's pieces. The resulting curve is an ellipse. 

Hyperbola If the angle made by the intersecting plane and the axis is less than the vertex angle (or is parallel to the axis of symmetry), then the plane will slice through both parts of the cone's pieces. The resulting curve is a hyperbola. 

Parabola If the intersecting plane is exactly parallel to one of the generating lines (and that's darn unlikely for a random plane!) then the plane will slice through only one part of the cone. The resulting curve is a parabola. 

These geometric characterizations of the conic sections are thousands of years old. There are also definitions of the various conic sections using distance in the plane. For example, an ellipse is the collection of points the sum of whose distances to two fixed points is a constant.
Applications
If the universe were empty except for the sun (a heavy point mass) and
an asteriod (a much lighter mass) than the orbit the asteroid would
have around the sun would be a conic section. The textbook discusses
this in the section on Kepler's laws. So these curves (motion in an
inverse square law force field) will be everywhere in physics.
If B^{2}4AC<0 then the curve is an ellipse.
We looked at the ellipse
11x^{2}32xy+24y^{2}+24x40y+21=0 earlier, with A=11,
B=32, and C=24. The (32)^{2}4(11)(24) turns out to be 32.
If B^{2}4AC>0 then the curve is a hyperbola.
An example of a hyperbola earlier was
7x^{2}6xy+16y^{2}+48x80y+74=0. Here A=7, B=6,
and C=16, so B^{2}4AC=(6)^{2}4(7)(16)=484>0.
If B^{2}4AC=0 then the curve is a parabola.
I had claimed earlier that
25x^{2}40xy+16y^{2}+13x9y+3=0 was a
parabola. Here (40)^{2}4(25)(16)=0.
These algebraic statements are several hundred years also, and can be verified using high school algebra, but the "high school algebra" must be applied very diligently. Some students might see a resemblance in the discriminant analysis and consequences to the second derivative test for functions of two variables, and the resemblance is not accidental. The second derivative test actually uses these results in a somewhat disguised form, where the Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F is the second order Taylor polynomial of the function. The complexity of the three variable pictures below indicate how complicated second derivative tests in three variables must be.
What happens in three variables? Here there are more models, and things are more complicated. The general form is
Here is an elliptical cylinder: x^{2}+2y^{2}1=0.
There's no z mentioned, so we take the ellipse in the xyplane and
just move it up and down in the z direction any distance desired.
The Maple command  
Here is a cylinder over a hyperbola: x^{2}+z^{2}2y^{2}2xz12yz+9z^{2}1=0
How do I "know" this? Well, I got this by expanding the following:
The Maple command 
So far these polynomials and their graphs really come from two dimensions. What can we have that's different from cylinders and points and lines?
New, threedimensional stuff
Well, maybe this isn't too new. An ellipsoid is the threedimensional analogue of the ellipse. Here is an equation for one ellipsoid (xz)^{2}+3(xy)^{2}+2(y+z)^{2}=1 and to the right is a picture of it.  
The simplest paraboloid is z=x^{2}+y^{2}, a sort of cup. Here is a disguised elliptical paraboloid and its picture: (xz)^{2}+5z^{2}+2(y+z)=1.  
The prototype here is z=x^{2}y^{2}. The sections with z=constant are hyperbolas and the sections with x or y=constant are parabolas. This can be a bit confusing when you see one that is disguised. For example, let me just take the previous example, and change one sign: (xz)^{2}5z^{2}+2(y+z)=1 This saddle is sitting sideways.  
In three dimensions there are two different analogs of hyperbolas. The word "sheet" is used in an antique, specialized sense with surfaces: it means one connected "piece" of a surface. So a hyperboloid with one sheet is a surface with one (connected) piece, and a hyperboloid with two sheets is a surface with two (connected) pieces. The Oxford English Dictionary gives this definition with a citation from 1827: "A portion of a surface analogous to the branch of a curve." Of course this is definition #11 for "sheet" and there are many other definitions for the word.
Hyperboloid of one sheet
Hyperboloid of two sheets

The textbook has some similar pictures (see section 12.6, especially the table on page 836). The hyperboloid of two sheets always sits inside a cone, and the onesheeted version sits outside a cone. Maybe you can "see" these cones if you rotate the lines which are the asymptotes for the original hyperbola.
> F:=x^3+5*x*zy^3+y; 3 3 F := x + 5 x z  y + y  
The surface is not familiar to me. I really wrote something weird, at random. Since there are terms of degree greater than 2, certainly I don't expect anything nice or familiar.  
> subs({x=1,y=0,z=0},F); 1  
Here I'm "plugging" in values for F's variables, and verifying that F(1,0,0)=1, so that (1,0,0) is on the surface defined by F(x,y,z)=1.  
> Fx:=diff(F,x); 2 Fx := 3 x + 5 z > Fy:=diff(F,y); 2 Fy := 3 y + 1 > Fz:=diff(F,z); Fz := 5 x > subs({x=1,y=0,z=0},Fx); 3 > subs({x=1,y=0,z=0},Fy); 1 > subs({x=1,y=0,z=0},Fz); 5 > T:=spacecurve(<3*t+1,1*t,5*t>,t=0..(.3),color=blue,thickness=3, axes=normal);  
Here are computations of partial derivatives, and evaluations of the partial derivatives. The gradient of F at (1,0,0) is (3,1,5), so that the spacecurve command draws a short line segment starting at (1,0,0) in the direction of the gradient vector.  
> S:=implicitplot3d(F, x=2..2, y=2..2, z=2..2, axes=normal, labels=[x,y,z], grid=[50,50,50], color=red); > display({S,T}};  
The implicitplot3d command draws the surface. Specifying that
grid=[50,50,50] means that we should get a better picture
than the default, but Maple must check 50^{3}=125,000
points in R^{3} and this command took about 2 seconds on my
home PC. Below is the picture that resulted from displaying the normal vector and the piece of the surface together.  
What if I wanted to find the normal vector for some point that isn't as "nice" as (1,0,0)? Suppose I changed x, say, from 1 to .95, and z from 0 to .15. What y values could I get? Here's one way to have Maple approximate the answers.  
> fsolve(subs({x=.95,z=.15},F=1)); 1.289608097 > Q:={x=.95,y=1.289608097,z=.15}; Q := {x = 0.95, z = 0.15, y = 1.289608097} > subs(Q,F); 1.000000000 > m1:=subs(Q,Fx);m2:=subs(Q,Fy);m3:=subs(Q,Fz); m1 := 1.9575 m2 := 3.989267132 m3 := 4.75 > T1:=spacecurve(<m1*t+.95,m2*t1.289608097,m3*t.15>,t=0..(.3), color=blue,thickness=3, axes=normal); > display({S,T,T1}};  
The first command, fsolve, asks for numerical solutions of
F(x,y,z)=1 after substitution of the x and z values. The
fsolve command can return more than one answer. I defined Q so that I would not have to type too much later. I substituted Q in F so that I could check fsolve's answer. Generally the answer should check to within the accuracy of Digits (the default accuracy is 10 decimal places). Then I used Q and the previously computed partial derivatives to get the gradient at the new point. The spacecurve command created a new line segment in the direction <m1,m2,m3> starting at Q and finally display({S,T,T1}) showed a neat picture. Two views of the picture are below. In the lefthand view, I tried to show that the vectors were perpendicular to the surface: this view is from the side of the vectors. The other view is an attempt to show "things" in perspective.  
> F:=x^53*x*y^2+4*x+2*y; 5 2 F := x  3 x y + 4 x + 2 y > subs({x=1,y=2},F); 3 > Fx:=diff(F,x);Fy:=diff(F,y); 4 2 Fx := 5 x  3 y + 4 Fy := 6 x y + 2 > m1:=subs({x=1,y=2},Fx);m2:=subs({x=1,y=2},Fy); m1 := 3 m2 := 10  
I invented the function F(x,y)=x^{5}3xy^{2}+4x+2y. Then I verified that F(1,2)=3. I computed the partial derivatives, and now I know that the gradient of F at (1,2) is <3,10>.  
> T:=plot([3*t+1,10*t+2,t=0..(.1)],color=blue,thickness=2); > S:=implicitplot(F=3, x=5..5,y=5..5, thickness=2, scaling=constrained, color=green, grid=[50,50]); > display({S,T});  
If the option scaling=constrained is omitted from the
implicitplot instruction, the normal vector doesn't appear to
be correct. (I spent 10 minutes figuring this out, darn it! Try it
yourself. After you realize that the unconstrained view
squashes the graphs, the fact that orthogonality is not "preserved" is
not too surprising. Sigh.) Look at the picture. If x=1.72, there are two values of y which will put (1.72,y) on the curve. Maple can approximate these values, and then it can find and graph representations of the normal vectors at these points. Added 4/1/2006 I thank Ms. Caroline Pisacka for showing me some inaccuracies in past versions of the previous several commands.  
> fsolve(subs(x=1.72,F)); 2.264611822, 1.877014923 > Q1:={x=1.72,y=2.264611822};Q2:={x=1.72,y=1.877014923}; Q1 := {x = 1.72, y = 2.264611822} Q2 := {x = 1.72, y = 1.877014923} > v1:=subs(Q1,Fx);v2:=subs(Q1,Fy);w1:=subs(Q2,Fx);w2:=subs(Q2,Fy); v1 := 32.37525269 v2 := 23.37079400 w1 := 37.19109774 w2 := 19.37079401 > yn:=2.264611822; yp:=1.877014923; yn := 2.264611822 yp := 1.877014923 > Tn:=plot([v1*t1.72,v2*t+yn,t=0..(.02)],color=blue,thickness=2); > Tp:=plot([w1*t1.72,w2*t+yp,t=0..(.02)],color=blue,thickness=2); > display({S,T,Tn,Tp});  
I wrote the assignment statements (that is, using := to give values to variables) because I didn't want to keep entering the numbers or using my mouse to copy them. I admit that with two points and two normal vectors, keeping everying "straight" (correct) can be difficult. Also, I had to play with the plot statements (the range of the variable t, in particular) to get things to look good. 
Maintained by greenfie@math.rutgers.edu and last modified 10/13/2006.