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August 6th, 2012, 09:57 AM  #1 
Member Joined: Jun 2012 Posts: 76 Thanks: 0  rotation of axes question
x^2 + 4xy + y^2  3 = 0 rewrite the equation in a rotated x'y' system without ans x'y' term. express the equation involving x' and y' in the standard form of a conic section. my solution is x'^2  y'^2/3 = 1 would anyone like to check this for me ? 
August 6th, 2012, 12:16 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689  Re: rotation of axes question
It would help if you would supply the equations transforming (x,y) to (x',y').

August 6th, 2012, 12:35 PM  #3 
Senior Member Joined: Jun 2011 Posts: 164 Thanks: 0  Re: rotation of axes question
first you want to find the angle you are rotating, using the formula 1/2(Tan^(1))(B/(AC)) is A does not equal to C, however, in this case, they do equal to each other, meaning the angle of the rotation is 45 degrees. let ? = the angle of rotation A'=Acos^2(?)+Bsin?cos?+Csin^2(?) B'=0 C'=Asin^2(?)Bsin?cos?+Ccos^2(?) D'=0 E'=0 F'=3 D' and E' are both = 0 because D and E are 0 so A'=1/2+4x(1/2)+1/2 C'=1/24x(1/2)+1/2 F'=3 solve each equation A'=3 C'=1 F'=3 Finally, the equation i have is: 3x'^2y'^23=0 if we divide everything by 3 (i believe that is what you did) x'^2(y'^2)/31=0, Which is exactly what you got, i believe you are correct 
August 6th, 2012, 01:22 PM  #4  
Member Joined: Jun 2012 Posts: 76 Thanks: 0  Re: rotation of axes question Quote:
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