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Math Help - Rotation Formulas

  1. #1
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    Smile Rotation Formulas

    Determine the appropriate rotation formulas to use so that the new equation contains no xy-term.

    (1) x^2 - 4xy + y^2 - 3 = 0

    (2) 3x^2 - 10xy + 3y^2 - 32 = 0
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  2. #2
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    Hello, magentarita!

    I'll walk through the first one . . .


    Determine the appropriate rotation formulas to use
    so that the new equation contains no xy-term.
    Given the general quadratic: . Ax^2 + Bxy + Cy^2 + Dx + Ey + F \:=\:0

    . . the angle of rotation \theta is given by: . \tan2\theta \:=\:\frac{B}{A-C}

    . . and the new coordinates are: . \begin{Bmatrix}x &=& X\cos\theta - Y\sin\theta \\ y &=& X\sin\theta + Y\cos\theta \end{Bmatrix}



    (1)\;\;x^2 - 4xy + y^2 \:=\: 3

    We have: . A = 1,\;B = -4,\;C = 1

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

    . . Hence: . \sin\theta \:=\: -\frac{1}{\sqrt{2}},\quad \cos\theta \:=\: \frac{1}{\sqrt{2}}


    And we have: . \begin{array}{ccccc}x &=& X\left(\frac{1}{\sqrt{2}}\right) - Y\left(\text{-}\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(X + Y) \\ \\[-3mm]<br />
y&=& X\left(\text{-}\frac{1}{\sqrt{2}}\right) + Y\left(\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(\text{-}X + Y) \end{array}


    Substitute into: . x^2-4xy + y^2 \;=\;3

    . . \bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]^2 - 4\bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]\bigg[\frac{1}{\sqrt{2}}(\text{-}X+Y)\bigg] + \bigg[\frac{1}{\sqrt{2}}(\text{-}X + Y)\bigg]^2 \;=\;3

    . . \tfrac{1}{2}(X^2+2XY + Y^2) - 2(\text{-}X^2+Y^2) + \tfrac{1}{2}(X^2-2XY + Y^2) \;=\;3

    . . \tfrac{1}{2}X^2 + XY + \tfrac{1}{2}Y^2 + 2X^2 - 2Y^2 + \tfrac{1}{2}X^2 - XY + \tfrac{1}{2}Y^2 \;=\;3

    . . 3X^2 - Y^2 \;=\;3

    . . \frac{X^2}{1} - \frac{Y^2}{3} \;=\;1 \quad\hdots\quad \text{hyperbola}

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  3. #3
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    Smile Great work....

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    I'll walk through the first one . . .

    Given the general quadratic: . Ax^2 + Bxy + Cy^2 + Dx + Ey + F \:=\:0

    . . the angle of rotation \theta is given by: . \tan2\theta \:=\:\frac{B}{A-C}

    . . and the new coordinates are: . \begin{Bmatrix}x &=& X\cos\theta - Y\sin\theta \\ y &=& X\sin\theta + Y\cos\theta \end{Bmatrix}



    We have: . A = 1,\;B = -4,\;C = 1

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

    . . Hence: . \sin\theta \:=\: -\frac{1}{\sqrt{2}},\quad \cos\theta \:=\: \frac{1}{\sqrt{2}}


    And we have: . \begin{array}{ccccc}x &=& X\left(\frac{1}{\sqrt{2}}\right) - Y\left(\text{-}\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(X + Y) \\ \\[-3mm] y&=& X\left(\text{-}\frac{1}{\sqrt{2}}\right) + Y\left(\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(\text{-}X + Y) \end{array}" alt="
    y&=& X\left(\text{-}\frac{1}{\sqrt{2}}\right) + Y\left(\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(\text{-}X + Y) \end{array}" />


    Substitute into: . x^2-4xy + y^2 \;=\;3

    . . \bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]^2 - 4\bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]\bigg[\frac{1}{\sqrt{2}}(\text{-}X+Y)\bigg] + \bigg[\frac{1}{\sqrt{2}}(\text{-}X + Y)\bigg]^2 \;=\;3

    . . \tfrac{1}{2}(X^2+2XY + Y^2) - 2(\text{-}X^2+Y^2) + \tfrac{1}{2}(X^2-2XY + Y^2) \;=\;3

    . . \tfrac{1}{2}X^2 + XY + \tfrac{1}{2}Y^2 + 2X^2 - 2Y^2 + \tfrac{1}{2}X^2 - XY + \tfrac{1}{2}Y^2 \;=\;3

    . . 3X^2 - Y^2 \;=\;3

    . . \frac{X^2}{1} - \frac{Y^2}{3} \;=\;1 \quad\hdots\quad \text{hyperbola}
    Thank you for taking time to show me step by step.
    This is what I need most of the time.
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