My Math Forum Translation and Rotation of Axes Problem

 Algebra Pre-Algebra and Basic Algebra Math Forum

 March 13th, 2014, 12:59 AM #1 Newbie   Joined: Jul 2013 Posts: 21 Thanks: 0 Translation and Rotation of Axes Problem The equation of a parabola in the x''y'' plane is (y'')^2=4x''. The equation of that parabola in the x'y' plane is (y'-3)^2 = 4(x'-6). How do I get the original equation of that parabola in the xy plane before the axes rotated? (The x'y' plane was rotated to an angle of 45 degrees). Please help. This is confusing. I know that x = (x')costheta - (y')sintheta and y = (x')sintheta + (y')costheta and that's all I know The graph looks something like this: http://tinypic.com/r/332s1uv/8
 March 13th, 2014, 02:04 AM #2 Member   Joined: Mar 2013 Posts: 90 Thanks: 0 Re: Translation and rotation of axes Rewrite $x'$ and $y'$ in terms of $x$ and $y$ and substitute into the second equation. The equation in $x,\,y$ will be that of the curve in the $xy$ plane. By the way the first equation shows that the $x''y''$ plane is a translation of the $x'y'$ plane. Did you notice it?
 March 14th, 2014, 04:56 AM #3 Newbie   Joined: Jul 2013 Posts: 21 Thanks: 0 Re: Translation and Rotation of Axes Problem So I just have to use the formula x' = xcostheta + ysintheta and y' = -xsintheta + ycostheta and plug them into (y'-3)^2 = 4(x'-6), then simplify the equation and that's the equation in the xy plane? Am I right?

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