Results 1 to 6 of 6

Math Help - Rotated Quadratic Equations

  1. #1
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185

    Rotated Quadratic Equations

    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 x'y' - plane

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

    My Work:

    A) 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.
    tan 2\alpha = \frac{B}{A-C} \rightarrow \frac{2 \sqrt{3}}{1-3} \rightarrow \frac{\sqrt{3}}{-1}
    2 \alpha = tan^{-1} \left(\frac{-\pi}{3}\right)
    \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:

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

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

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

    D'=Dcos\alpha + Esin\alpha

    E'=-Dsin\alpha + Ecos\alpha

    F'=F
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,974
    Thanks
    1121
    Since you know that \alpha= -\frac{\pi}{6}, you can immediately calculate cos(\alpha)=\frac{\sqrt{3}}{2} and sin(\alpha)= -\frac{1}{2} and put them into those equations. What is stopping you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by HallsofIvy View Post
    Since you know that \alpha= -\frac{\pi}{6}, you can immediately calculate cos(\alpha)=\frac{\sqrt{3}}{2} and 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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,974
    Thanks
    1121
    That's what the formulas say isn't it?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    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}
    B'=2 \sqrt{3}\left(\frac{1}{2}\right) + (3-1)(\frac{-\sqrt{3}}{2} \rightarrow B'=0
    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}
    D'=8\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) + (-8)\left(\frac{-1}{2}\right) \rightarrow D'=16
    E'=-8\sqrt{3}\left(\frac{-1}{2}\right) + (-8)\left(\frac{\sqrt{3}}{2}\right) \rightarrow E'=0
    F'=32

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

    Is this equation correct for the graph in the 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; April 15th 2010 at 02:05 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Am I supposed to utilize x=x'cos\alpha - y'sin\alpha and y=x'sin\alpha + y'cos\alpha
    I'm so confused right now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic Equations
    Posted in the Algebra Forum
    Replies: 10
    Last Post: September 21st 2011, 05:46 PM
  2. PDF on quadratic equations
    Posted in the Peer Math Review Forum
    Replies: 7
    Last Post: July 21st 2010, 08:23 AM
  3. Replies: 1
    Last Post: June 12th 2008, 09:30 PM
  4. Quadratic equations again
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 11th 2007, 09:18 AM
  5. Quadratic equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 7th 2007, 11:10 AM

Search Tags


/mathhelpforum @mathhelpforum