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Thread: Rotated Quadratic Equations

  1. #1
    Member VitaX's Avatar
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    Rotated Quadratic Equations

    $\displaystyle x^2 + 2\sqrt{3}xy + 3y^2 + 8\sqrt{3}x - 8y + 32 = 0$

    A) Use the Discriminant test to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola.

    B) Find a suitable rotation of axis

    C) Find an equation for the graph in a $\displaystyle x'y'$ - plane

    D) Sketch the graph over top both coordinate systems with your angle of rotation displayed.

    My Work:

    A) $\displaystyle B^2 - 4AC \rightarrow (2\sqrt{3})^2 - 4(1)(3) = 0$ So, The equation is a parabola.

    B) Here's where I get somewhat confused, so bear with my work.
    $\displaystyle tan 2\alpha = \frac{B}{A-C} \rightarrow \frac{2 \sqrt{3}}{1-3} \rightarrow \frac{\sqrt{3}}{-1}$
    $\displaystyle 2 \alpha = tan^{-1} \left(\frac{-\pi}{3}\right)$
    $\displaystyle \alpha = \frac{-\pi}{6}$

    From here on I'm not really quite sure what to do. The notes I have on this don't really give me enough help to continue on. I know later on though I have to utilize these:

    $\displaystyle A'=Acos^2 \alpha + Bcos\alpha sin\alpha + Csin^2 \alpha$

    $\displaystyle B'=Bcos2\alpha + (C-A)sin2 \alpha$

    $\displaystyle C'=Asin^2 \alpha - Bcos\alpha sin\alpha + Ccos^2 \alpha$

    $\displaystyle D'=Dcos\alpha + Esin\alpha$

    $\displaystyle E'=-Dsin\alpha + Ecos\alpha$

    $\displaystyle F'=F$
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  2. #2
    MHF Contributor

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    Since you know that $\displaystyle \alpha= -\frac{\pi}{6}$, you can immediately calculate $\displaystyle cos(\alpha)=\frac{\sqrt{3}}{2}$ and $\displaystyle sin(\alpha)= -\frac{1}{2}$ and put them into those equations. What is stopping you?
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Since you know that $\displaystyle \alpha= -\frac{\pi}{6}$, you can immediately calculate $\displaystyle cos(\alpha)=\frac{\sqrt{3}}{2}$ and $\displaystyle sin(\alpha)= -\frac{1}{2}$ and put them into those equations. What is stopping you?
    So I just plug in those values into the equations I stated above and come up with the values for A',B',C',D',E' and come up with the new "shifted" equation?
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  4. #4
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    That's what the formulas say isn't it?
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  5. #5
    Member VitaX's Avatar
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    $\displaystyle A'=1\left(\frac{\sqrt{3}}{2}\right)^2 + 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\fra c{-1}{2}\right) + 3\left(\frac{-1}{2}\right)^2 \rightarrow A'=\frac{3}{4}$
    $\displaystyle B'=2 \sqrt{3}\left(\frac{1}{2}\right) + (3-1)(\frac{-\sqrt{3}}{2} \rightarrow B'=0$
    $\displaystyle C'=1\left(\frac{-1}{2}\right)^2 - 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\fra c{-1}{2}\right) + 3\left(\frac{\sqrt{3}}{2}\right)^2 \rightarrow C'=\frac{19}{4}$
    $\displaystyle D'=8\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) + (-8)\left(\frac{-1}{2}\right) \rightarrow D'=16$
    $\displaystyle E'=-8\sqrt{3}\left(\frac{-1}{2}\right) + (-8)\left(\frac{\sqrt{3}}{2}\right) \rightarrow E'=0$
    $\displaystyle F'=32$

    $\displaystyle \frac{3}{4}(x')^2 + \frac{19}{4}(y')^2 + 16x' + 32 = 0$

    Is this equation correct for the graph in the $\displaystyle x'y'$ - plane? I'm having a little trouble putting this in the correct form of a parabola (if this equation is correct that is), if someone could point me in the right direction that would be great.
    Last edited by VitaX; Apr 15th 2010 at 02:05 PM.
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  6. #6
    Member VitaX's Avatar
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    Am I supposed to utilize $\displaystyle x=x'cos\alpha - y'sin\alpha$ and $\displaystyle y=x'sin\alpha + y'cos\alpha$
    I'm so confused right now.
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