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Rotating 1/x to make a hyperbola?

  1. Nov 22, 2003 #1
    Hey everyone, I was having trouble with this question.

    The graph of 1/x is a hyperbola, but it's equation does not fit the form (x-h)/(a^2) - (y-k)/(b^2) = 1. Rotate 1/x using polar coordinates, change it back into cartesian coordinates, and write the equation in standard hyberbola notation.

    I understand how to convert to polar from cartesian and vice versa, but I don't know how to rotate graphs. Any help would be great! Thanks!
  2. jcsd
  3. Nov 22, 2003 #2
    When you have the expression in polar form, replace the angle θ with θ+φ to rotate by an angle φ (or θ-φ, depending on your sign conventions). You'll probably want to use the trig addition identities (for, say, sin(a+b) in terms of sines and cosines of a and b) to put your expression into a form that can be easily converted back to Cartesian coordinates.
  4. Nov 22, 2003 #3


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    If (x,y) are the original coordinates and (x', y') are the coordinates rotated at angle θ then

    x'= x cos(θ)+ y sin(θ);

    y'= -x sin(θ)+ y cos(θ);

    Notice that if you solve the two equations for x, y in terms of x',y', this is, inverting the change, you get

    x= x' cos(θ)- y' sin(θ)

    y= x' sin(θ)+ y' cos(θ)

    Exactly what you would get if you replace θ by -θ

    In particular, if θ= π/2, then

    x= (√(2)/2) (x'- y'); y= (√(2)/2)(x'+ y')

    xy= (1/2)(x'2- y'2)= x'2/2- y'2/2= 1.
    Last edited: Nov 22, 2003
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