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Rotating axes to elimnate xy

  1. Aug 24, 2005 #1
    I am having trouble rotating the axes to eliminate the xy-term.
    8x^2+64xy+8y^2+12x+12y+9=0

    I know Ax^2+Bxy+Cy^2+Dx+Ey+F=0
    however, the professor is looking for an equation not an answer.

    Here are my choices and I am stumped:
    40x^2 + 12 sq rt2 x + 12 sq rt 2 y + 9 = 0

    40x^2 - 24y^2 +12 sq rt2y +9 = 0

    -24y^2 + 12 sq rt2x + 12 sq rt 2y + 9 = 0

    40x^2 - 24y^ + 12 sq rt2x + 9 = 0

    I am not sure what the process is to get to this point. I can run the basic Rotation Theorem for Conics and answer the problem with an angle degree.

    PLEASE HELP!
     
  2. jcsd
  3. Aug 24, 2005 #2

    TD

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    You can also reduce these conics to their standard Euclidean form using lineair algebra, by diagonalizing the corresponding matrix.
     
  4. Aug 24, 2005 #3

    arildno

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    Introduce new variables as follows:
    [tex]u=x\cos\theta+y\sin\theta, v=-x\sin\theta+y\cos\theta[/tex]
    Here, [tex]\theta[/tex] is the angle between the positive x-axis and the positive u-axis.

    Solve for x and y in terms of u and v, and gain:
    [tex]x=u\cos\theta-v\sin\theta, y=u\sin\theta+v\cos\theta[/tex]

    Substitute these expressions for x and y in your equation, and eliminate the uv-term by making a smart choice of [tex]\theta[/tex]


    then, redefine "u" and "v" to be new "x" and "y".
    (I don't see much point in doing so, but it seems your professor wants you to do that)
     
    Last edited: Aug 24, 2005
  5. Aug 24, 2005 #4
    Can you show me how?
     
  6. Aug 24, 2005 #5

    arildno

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    Put the expressions in u and v for x into the x's place, and similar for the expression for y.

    For example, [tex]x^{2}=(u\cos\theta-v\sin\theta)^{2}=u^{2}\cos^{2}\theta-uv\sin(2\theta)+v^{2}\sin^{2}\theta[/tex]
     
  7. Aug 24, 2005 #6

    arildno

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    If you haven't done so already, we also have:
    [tex]xy=(u\cos\theta-v\sin\theta)(u\sin\theta+v\cos\theta)=\frac{u^{2}-v^{2}}{2}\sin(2\theta)+uv\cos(2\theta)[/tex]
    [tex]y^{2}=(u\sin\theta+v\cos\theta)^{2}=u^{2}\sin^{2}\theta+uv\sin(2\theta)+v^{2}\cos^{2}\theta[/tex]
    Use these expressions to your heart's content..
     
  8. Aug 24, 2005 #7

    HallsofIvy

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    You were given [tex]x=u\cos\theta-v\sin\theta, y=u\sin\theta+v\cos\theta[/tex]
    Go ahead and replace x and y in your equation by those, do the algebra and then pick θ so that the "uv" term has 0 coefficient.
     
  9. Aug 24, 2005 #8

    saltydog

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    Tony, what happen with all this? You figured the angle via:

    [tex]Cot[2\alpha]=\frac{A-C}{B}[/tex]

    so you get:

    [tex]\alpha=\frac{\pi}{4}[/tex]

    Great. Make the substitution:

    [tex]x=\overline{x}Cos[\alpha]-\overline{y}Sin[\alpha][/tex]

    [tex]y=\overline{x}Sin[\alpha]+\overline{y}Cos[\alpha][/tex]

    Substitute those expressions into the equation right. For example the [itex]8x^2[/itex] would be:

    [tex]8\left[\frac{\overline{x}}{\sqrt{2}}-\frac{\overline{y}}{\sqrt{2}}\right]^2[/tex]

    You can do the rest. Simplify to:

    [tex]40\overline{x}^2-24\overline{y}^2+12\sqrt{2}\overline{x}+9=0[/tex]

    The [itex]\overline{x}-\overline{y} [/itex] axes are just 45 degrees tilted from the x-y axes. The equation is a hyperbola.

    Right?

    Edit: Corrected angle formula: it's Cot(2a)
     
    Last edited: Aug 25, 2005
  10. Aug 25, 2005 #9
    Thank you very much. This distance learning is extremely difficult. I am a visual person and your help is truly appreciated.
     
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