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Math Help - rotation of axes

  1. #1
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    rotation of axes

    Given the equation 21x^2 + 10 root3xy + 31y^2 = 144

    a. determine the type of conic using the discriminant
    b.find the angle of rotation
    c.use rotation formulas to eliminate the xy-term and sketch

    I was assigned a bunch of problems just like this for homework. The only problem is that my book doesn't really give an example of this type of problem, nor did my teacher in class. If anyone could help my figure this out I would be sooo grateful. Thanks so much!
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  2. #2
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by kristenrae View Post
    Given the equation 21x^2 + 10\sqrt{3}xy + 31y^2 = 144

    a. determine the type of conic using the discriminant
    b.find the angle of rotation
    c.use rotation formulas to eliminate the xy-term and sketch
    a. determine the type of conic using the discriminant


    The general equation for conic sections can be written as
    Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

    Comparing the given equation with the general equation,
    A = 21, B = 10\sqrt{3},\ C = 31, D = 0, E = 0,\ \mbox{and}\ F = -144 ------- ■

    The discriminant is
    B^2 - 4AC = (100\cdot 3) - (4\cdot 21\cdot\ 31) < 0

    B^2 - 4AC < 0. Therefore, the equation represents an \boxed{\mbox{ellipse}}.

    b.find the angle of rotation


    For the general equation of a conic, the angle of rotation \theta about the origin is given by:

    \theta = \frac{\pi}{4}, if A = C
    \tan 2\theta = \frac{B}{A - C}, if A\neq C

    In this case, A\neq C
    Substituting 21 for A, 10\sqrt{3} for B, and 31 for C,

    \tan 2\theta = \frac{10\sqrt{3}}{21 - 31} = -\sqrt{3}

    Solving for \theta,

    2\theta = \arctan (-\sqrt{3}) = -60^\circ
    \implies \boxed{\theta = -30^\circ} -------●


    c.use rotation formulas to eliminate the xy-term and sketch


    If the conic section with the equation: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

    is transformed by a rotation about the origin through the angle \theta, then the equation in the new coordinates \tilde{x} and \tilde{y} has the form:

    \tilde{A}\tilde{x^2} + \tilde{C}\tilde{y^2} + \tilde{D}\tilde{x} + \tilde{E}\tilde{y} + \tilde{F} = 0

    The coordinate transformation which expresses the old coordinates in terms of the new ones is given by:

    x = \tilde{x}\cos \theta - \tilde{y}\sin \theta
    y = \tilde{x}\sin \theta + \tilde{y}\cos \theta

    Thus, putting the coordinate transformations into the original equation:

    Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

    \Rightarrow A(\tilde{x}\cos \theta - \tilde{y}\sin \theta)^2 + B(\tilde{x}\cos \theta - \tilde{y}\sin \theta)(\tilde{x}\sin \theta + \tilde{y}\cos \theta) +\ C(\tilde{x}\sin \theta + \tilde{y}\cos \theta)^2 + D(\tilde{x}\cos \theta - \tilde{y}\sin \theta) + E(\tilde{x}\sin \theta + \tilde{y}\cos \theta) + F = 0

    Opening up the brackets and expanding, we get:

    A\tilde{x}^2\cos^2 \theta + A\tilde{y}^2\sin^2 \theta - 2A\tilde{x}\tilde{y}\cos \theta\sin \theta + B\tilde{x}^2\sin \theta \cos \theta + B\tilde{x}\tilde{y}\cos^2 \theta \ - B\tilde{x}\tilde{y}\sin^2 \theta - B\tilde{y}^2\sin \theta\cos \theta + C\tilde{x}^2\sin^2 \theta + C\tilde{y}^2\cos \theta

    +\ 2C\tilde{x}\tilde{y}\sin \theta\cos \theta + D\tilde{x}\cos \theta - D\tilde{y}\sin \theta + E\tilde{x}\sin \theta + E\tilde{y}\cos \theta + F = 0

    Rearranging,
    \tilde{x}^2(A\cos^2\theta + B\sin \theta\cos \theta + C\sin^2 \theta) + \tilde{y}^2(A\sin^2\theta - B\sin \theta\cos \theta + C\cos^2 \theta) +\ \tilde{x}\tilde{y}\{\sin \theta\cos \theta(2C - 2A) + B(\cos^2 \theta - \sin^2 \theta)\} + \tilde{x}(D\cos \theta + E\sin \theta) +\ \tilde{y}(E\cos \theta - D\sin \theta) = 0

    Thus, the parameters of the new equation are:

    \tilde{A} = A\cos^2 \theta + B\cos \theta\sin \theta + C\sin^2 \theta

    \tilde{B} = 0 [\mbox{Using the identities}\ \sin 2\theta = 2\sin \theta\cos \theta\ \mbox{and}\ \cos 2\theta = \cos^2 \theta - \sin^2 \theta and using ■ and ●, we get (C - A)\sin 2\theta + B(\cos 2\theta) = -\frac{\sqrt{3}}{2}\times 10 + 10\sqrt{3}\times \frac{1}{2} = 0]

    \tilde{C} = A\sin^2 \theta - B\cos \theta\sin \theta + C\cos^2 \theta

    \tilde{D} = D\cos \theta + E\sin \theta

    \tilde{E} = E\cos \theta - D\sin \theta

    \tilde{F} = F


    In our case, applying the rotation formula,

    \boxed{x = \frac{\sqrt{3}}{2}\tilde{x} + \frac{1}{2}\tilde{y}}
    \boxed{y = -\frac{1}{2}\tilde{x} + \frac{\sqrt{3}}{2}\tilde{y}}


    \tilde{A} = \frac{63}{4} - \frac{15}{2} + \frac{31}{4} = 16

    \tilde{C} = \frac{21}{4} + \frac{93}{4} + \frac{30}{4} = 36

    \tilde{D} = 0

    \tilde{E} = 0

    \tilde{F} = -144


    and the equation gets the form:

    16\tilde{x^2} + 36\tilde{y^2} - 144 = 0

    \implies \boxed{\frac{\tilde{x}^2}{9} + \frac{\tilde{y}^2}{4} = 1},

    which is the equation of a standard ellipse.
    Attached Thumbnails Attached Thumbnails rotation of axes-rotatedellipse.jpg  
    Last edited by fardeen_gen; May 12th 2009 at 09:22 AM. Reason: Answered additional queries by OP
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  3. #3
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    Thank you so much! That was so helpful. I understand completely up until applying the rotation formula. How did you find the values of A through F? Thanks so much again!
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  4. #4
    Super Member fardeen_gen's Avatar
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    Check edited reply.
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