Rotated Conical SectionsDate: 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/ |
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