Originally Posted by

**spiritualfields** JakeD, using

((xprime (cos theta)) - (yprime (sin theta))) squared / a squared

+

((xprime (sin theta)) + (yprime (cos theta))) squared / b squared

= 1

seems promising. I didn't think of substituting the theorem directly into the equation. I'll start off with known vertices and intercepts, and after expanding the equation (using an angle of rotation of my choosing), I should end up with a form that I can solve with the quadratic formula, right?

My only wonder now is if I can select, as the domain for the expanded equation, the interval (-a, +a) to solve for y. Will that get me the complete rotated ellipse?

You see, when I said I wanted to programmatically solve for a rotated ellipse, I really mispoke, as "programmatically" in this forum could easily be interpreted as "how to get the TI-89 to display a rotated ellipse?" or MathCad, or Mathematica, and so on. What I really meant is that what I want to do is calculate the table of x and y values myself, that represent the rotated ellipse, and using that table of x and y values I would draw the ellipse "programattically" onto a computer screen by setting the x and y pixel points to a certain color, say red against a white background, using a looping algorithm. Figuring out the table of x and y values for a non-rotated ellipse is easy, but a little tricky when you want to calculate the x and y values for a rotated ellipse.

I'm going to try it with your suggested substitutin method first. If that doesn't work, I'll try the brute force method that I mentioned in my opening post.