Use the transformation:
(rotation around the origin and angle )
So, I have been trying to figure out how to rotate an ellipse in the form of:
My guess is that making a translate vector of T<x^2, y^2> will do the trick. My goal is simply to make the ellipse not vertical or horizontal. My reasoning is because ^2 will have moved the numbers farther away from zero more than the numbers closer to zero resulting in a rotation. Is my reasoning correct? And, how could I find the equation of an ellipse if I wanted to rotate it x degrees?
What Fernando Revilla gave is a matrix multiplication. Another way to say essentially the same thing is to use the "coordinate transformation" , where I have just written out the result of the matrix multiplication. Here, is the angle of rotation (about the origin), x and y are the original coordinates, and x' and y' are the rotated coordinates.
For example, if the original ellipse were , that transformation would give, for any angle, ,
Notice a few thing about that- all terms are still quadratic. Linear terms signal a translation of the center of the figure but rotating keeps the center at (0, 0). There is, however, an " x'y' " term that is not in the "standard" form for a conic section. That signals a rotation. If, say, the angle of rotation were , then the equation would be
However, and so that a rotation by a right angle would give math]
just switching "major axis" and "minor axis".
Typically, what one wants to do is the "other way around"- given a quadratic in two variables, say , write it in "standard form".
Writing , , .
Now, to remove the x'y' turn, which never appears in the "standard form", I have to choose to make . That's easy. Either replace with to get so that or recognize that so that we must have .
Either gives so that and the equation becomes . That shows that the graph of xy= 1 (or y= 1/x) is a hyperbola with the lines y= x and y= -x as axes, the x and y axes as asymptotes.