How do you find the coefficients for a conic in a rotated system?

1 Answer
Mar 4, 2015

This is a good question, but a little bit vague. I'll try to guess:

When you say, "a conic," do you (1) have an equation and you want to rotate the axes to get a simpler equation? Or do you (2) have a graph or some points that the curve goes through?

The general conic equation is a relation between x and y:

#A x^2 + B x y +C y^2 + Dx + Ey + F = 0#

(1) • If you want to rotate this curve to "get rid of" the xy-term, you can substitute coordinates u and v, rotated by angle #theta#, where

#u = x cos theta + y sin theta#
#v = -x sin theta + y cos theta#

or to substitute in the general conic equation,

#x = u cos theta - v sin theta#
#y = u sin theta + v cos theta#

If you use the specific angle #theta# where

#cot (2 theta) = (A-C)/B#

then you will get an equation in u and v that has no xy-term.

(2) • If you just have some points that the conic goes through, you need five points to determine a unique conic. You then have 5 equations in the 6 variables A, B, C, D, E, F.

They are scalable so this determines a unique conic. For example #x^2 + y^2 - 25# is the same circle as #3x^2 + 3y^2 - 75 = 0#.

Note: If you have six or more points, they may not all lie on the same conic.

// dansmath strikes again! //

p.s. Here's the story, and an example, from Stewart's Calculus: