First, just to clarify the nomenclature used by my text book, an equation of the form:
A(x squared) + Bxy + C(y squared) + Dx + Ey + F = 0
is a conic if A, B, and C are not simultaneously equal to 0.
The text then goes on to discuss 4 topics:
Topic 1: Identifying conics, which shows how a conic can be identified from the above equation when the Bxy term is not present, by multiplying A and C and evaluating the result.
Topic 2: Using a rotation of axes to transform equations, which relates the rotated axes with the original axes by using this theorem:
x = xprime(cos theta) - yprime(sin theta)
y = xprime(sin theta) + yprime(cos theta)
This section develops the idea of how the xy term of the general equation can be eliminated if the x and y axes are rotated through an appropriate angle. The resultant equation then takes on the form of a more recognizable conic equation, with vertices and foci and other characteristics with respect to the rotated axes.
Topic 3: Dicusses one example equation (which contains an xy term), first by a rotation to eliminate the xy term, to arrive at the recognizable equation of a hyperbola, which is then hand drawn. Then the topic gives the procedure for graphing the equation with a graphing utility, by solving the original equation using the quadratic equation.
Topic 4: Identifying Conics without a rotation of axes.
That's it. In my march through my precalculus book, I was greatly anticipating this chapter, but unless I'm missing something entirely, one important thing that I wanted to learn was not covered.
I want to be able to programmatically draw a rotated ellipse on a computer monitor. This means that I can't just hand-swag it in like the book does. Plus I want to determine what the angle of rotation is, which means I have to generate my own equations. The problem is coming up with the equation. For example, let's say I want to draw an ellipse on a set of axes that is rotated through 45 degrees. Let's say that for the transformed axis I want the origin to be in the center, and the major axis to be parallel with the xprime axis. And let's say that its vertices are (-10,0) and (10,0), and its y-intercepts are (0, 5) and (0, -5) with respect to the rotated axes.
What's the equation?
For a non-rotated ellipse, the equation would be
(x squared / 100) + (y squared / 25) = 1
For this non-rotated ellipse, I could simply create a table of x and y values, beginning with x = -10 and working in increments of 1 up to x = 10, I could calculate the y values (y = plus/minus (squareroot(25*(1 - (x squared / 100))))).
But how do I construct the table for an ellipse that has the same vertices and intercepts, but with respect to a rotated set of axes? It looks as if I need to somehow take the original non-rotated ellipse and get it into the general form, where it has an xy term due to the desired amount of rotation. Unfortunately, the text gives no guidance on how to do this.
The only thing I can think of is to use a brute force method, using the transformation theroem equations that relates x and y with xprime and yprime. By calcuating a table for a non-rotated ellipse with the vertices and y-intercepts that I want, I could then use each x and y value in the table as constants into the theorem equations. I would have two equations with two unknowns (one with a known x and unknown xprime and yprime, and one with a known y with an unknown xprime and yprime). I could calculate a table of xprimes and yprimes, with each xprime/yprime pair being calculated from each x/y pair. I'm thinking that this set of xprime/yprime pairs would give the rotated ellipse (since the theorem equations contains the rotation angle as one of the variables).
Is there another way to accomplish this? I'm thinking that there must be, because my method seems too roundabout (plus I'm not even sure if it will work). Ideally, I would like to create an equation that will give me a conic with the characteristics and rotated angle that I want. Is there a way to do this?