Rotated Conical Sections

Date: 07/15/99 at 10:30:25
From: Gregory Leigh
Subject: Rotated Conical Sections

I'd like to consider myself fully proficient in "straight" conics, 
however when it comes to that blasted XY term, I get stuck. I know and 
understand the rotation formulas used to find the angle of rotation, 
however do not know how to find the focus, directrix, latus rectum, 
etc. with such a rotated conic. I would like to know how to do this so 
as to add it to a program I am creating.

Date: 07/15/99 at 13:39:52
From: Doctor Rob
Subject: Re: Rotated Conical Sections

The equations rotating axes to make the conic axes parallel to the
new coordinate axes are:

   X = x*cos(t) + y*sin(t)
   Y = -x*sin(t) + y*cos(t)

   x = X*cos(t) - Y*sin(t)
   y = X*sin(t) + Y*cos(t)

Start with your equation in terms of x and y. Determine the angle of 
rotation t needed. Substitute for x and y using the second set of 
equations above, expand, and simplify. That should give you an 
equation without an X*Y term. Put the result into standard form. 
Compute (or read off) the parts you are interested in, using the
XY-coordinate system. If you want a dimensionless quantity, like the 
eccentricity, or a length, like the latus rectum, these are the same 
in any coordinate system, so you are done. If you want some quantity 
which depends on the coordinate system, like a vertex, take its 
XY-coordinates and use the same second set of equations above to 
compute its xy-coordinates. Do the same for any object where you have 
an equation in X and Y: substitute for X and Y their equals from the 
first set of equations above to transform it into an equation in terms 
of x and y.

Example:  Find all parts of the ellipse

   x^2 + x*y + y^2 - 7*x + 2*y + 4 = 0.

It turn out the angle here is t = -Pi/4, and the transforming
equations are

   X = x*sqrt(2)/2 - y*sqrt(2)/2
   Y = x*sqrt(2)/2 + y*sqrt(2)/2

   x = X*sqrt(2)/2 + Y*sqrt(2)/2
   y = -X*sqrt(2)/2 + Y*sqrt(2)/2

Substituting and simplifying, I got the equation

   [X-5*sqrt(2)/2)]^2/18 + [Y-3*sqrt(2)/2]^2/6 = 1

This has center (X,Y) = (5*sqrt(2)/2,3*sqrt(2)/2), semi-minor axis
b = sqrt(6), semi-major axis a = sqrt(18) = 3*sqrt(2), and 
eccentricity e = sqrt(6)/3. The vertices are at

   (11*sqrt(2)/2,3*sqrt(2)/2) and
   (-sqrt(2)/2,3*sqrt(2)/2).

The foci are at (2*sqrt(3) + 5*sqrt(2)/2,3*sqrt(2)/2) and
(-2*sqrt(3) + 5*sqrt(2)/2,3*sqrt(2)/2).

The directrices are

   X = 5*sqrt(2)/2 + 3*sqrt(3) and
   X = 5*sqrt(2)/2 - 3*sqrt(3).

The length of the latus rectum is 2*sqrt(2).  Back in the 
xy-coordinate system, you find that the center is at

   x = 5*sqrt(2)/2*sqrt(2)/2 + 3*sqrt(2)/2*sqrt(2)/2,
     = 5/2 + 3/2 = 4,
   y = -5*sqrt(2)/2*sqrt(2)/2 + 3*sqrt(2)/2*sqrt(2)/2,
     = -5/2 + 3/2 = -1,

so the center is (x,y) = (4,-1).  You can do similarly for the foci
and vertices. The directrices are

   x*sqrt(2)/2 + y*sqrt(2)/2 = 5*sqrt(2)/2 + 3*sqrt(3),
   x + y = 5 + 3*sqrt(6),

and

   x + y = 5 - 3*sqrt(6).

Get the idea?

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/