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

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:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

(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

$\cot \left(2 \theta\right) = \frac{A - C}{B}$

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 $3 {x}^{2} + 3 {y}^{2} - 75 = 0$.

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:
http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS/upfiles/topics/ess_at_13_ra_stu.pdf