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Math Help - Help with Rotation of Equations

  1. #1
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    Help with Rotation of Equations

    8x^2 - 4xy + 5y^2 = 36

    @=theta

    1. Find tan 2@ and find cos 2@

    2. Find sin@ and cos@

    3. Transform the equation xy=1 by rotation, thus eliminating the xy term.
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  2. #2
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    Hello, The Real Cosby!

    You should be familiar with the "Rotation" formulas.
    Otherwise, you should not have been assigned this problem.

    The general quadratic equation is: . Ax^2 + Bxy + Cy^2 + Dx + Ey + F \;=\;0


    8x^2 - 4xy + 5y^2 \:=\: 36\;\;{\color{blue}[1]}

    1. Find \tan2\theta and \cos2\theta
    We have: . A = 8,\;B = -4,\;C = 5,\;F = -36

    Formula: . \tan2\theta \;=\;\frac{B}{A-C}

    We have: . \tan2\theta \:=\:\frac{-4}{8-5} \:=\:\boxed{-\frac{4}{3}}

    Since \theta is in Quadrant 2: . \tan2\theta \:=\:\frac{4}{-3} \:=\:\frac{opp}{adj}

    2\theta is in a right triangle with: opp = 4,\;adj = -3
    Using Pythagorus, we get: . hyp = 5

    Therefore: . \cos2\theta \:=\:\frac{adj}{hyp} \:=\:\boxed{-\frac{3}{5}}




    2. Find \sin\theta and \cos\theta
    Since 2\theta is in Quadrant 2, then \theta is in Quadrant 1.

    Identities: . \begin{Bmatrix}\sin^2\theta &=&\dfrac{1-\cos2\theta}{2} \\ \\[-4mm] \cos^2\theta &=&\dfrac{1+\cos2\theta}{2} \end{Bmatrix}


    We have: . \begin{array}{ccccccccc}\sin^2\theta &=& \dfrac{1 -(-\frac{3}{5})}{2} &=& \dfrac{4}{5}& \Rightarrow & \sin\theta &=&\boxed{\dfrac{2}{\sqrt{5}}} \\<br />
\cos^2\theta &=& \dfrac{1 + (-\frac{3}{5})}{2} &=& \dfrac{1}{5} & \Rightarrow & \cos\theta &=&\boxed{\dfrac{1}{\sqrt{5}}}\end{array}




    3. Transform the equation xy=1 by rotation, . ??
    . . thus eliminating the xy term.
    Formulas: . \begin{Bmatrix}x' &=& x\cos\theta - y\sin\theta \\ y' &=& x\sin\theta + y\cos\theta \end{Bmatrix}


    We have: . \begin{array}{ccccc}x' &=&\frac{1}{\sqrt{5}}x - \frac{2}{\sqrt{5}}y &=& \dfrac{x-2y}{\sqrt{5}} \\ \\[-4mm]<br />
y' &=&\frac{2}{\sqrt{5}}x + \frac{1}{\sqrt{5}}y &=& \dfrac{2x+y}{\sqrt{5}} \end{array}

    Substitute into [1]: . 8\left(\frac{x-2y}{\sqrt{5}}\right)^2 - 4\left(\frac{x-2y}{\sqrt{5}}\right)\left(\frac{2x+y}{\sqrt{5}}\ri  ght) + 5\left(\frac{2x+y}{\sqrt{5}}\right)^2 \;=\;36

    . . \frac{8(x^2-4xy + 4y^2)}{5} - \frac{4(2x^2-3xy-2y^2)}{5} + \frac{5(4x^2+4xy + y^2)}{5} \;=\;36

    . . \frac{8x^2 - 32xy + 32y^2 - 8x^2 + 12xy + 8y^2 + 20x^2 + 20xy + 5y^2}{5} \;=\;36

    . . \frac{20x^2 + 45y^2}{5} \;=\;36 \quad\Rightarrow\quad 4x^2 + 9y^2 \;=\;36 \quad\Rightarrow\quad\boxed{ \frac{x^2}{9} + \frac{y^2}{4} \;=\;1}\quad\text{ an ellipse!}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Drat! .I just figured out what #3 meant . . .

    It is a separate problem!


    We have: . xy \:=\:1\;\;{\color{blue}[2]}
    . . That is: . A = 0,\;B = 1,\; C= 0,\;D = 0,\;E = 0,\;F = -1

    \tan2\theta \:=\:\frac{1}{0-0} \:=\:\infty \quad\Rightarrow\quad  2\theta \:=\:\frac{\pi}{2} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{4}

    . . It is a hyperbola rotated through 45.


    \sin\theta = \frac{1}{\sqrt{2}},\;\cos\theta = \frac{1}{\sqrt{2}}

    Hence: . \begin{Bmatrix}x' &=&\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y &=& \dfrac{x-y}{\sqrt{2}} \\ \\[-4mm]<br />
y' &=&\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y &=& \dfrac{x+y}{\sqrt{2}} \end{Bmatrix}


    Substitute into [2]: . \left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\righ  t) \;=\;1 \quad\Rightarrow\quad \frac{x^2-y^2}{2} \;=\;1


    . . Therefore: . \boxed{x^2-y^2 \;=\;2}



    I need a nap . . .
    .
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  3. #3
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    Wow sir, you get a big thanks from me.

    My mind was off and I was just doing some other stuff.

    I think I need a nap too...
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