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Thread: Rotation and Systems of Quadratic Equations

  1. #1
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    Rotation and Systems of Quadratic Equations

    x^2-10xy+y^2+1=0

    I'm supposed to rotate the axes in order to eliminate the xy-term. I'm told this problem is very simple, but I don't understand this section at all. Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by alreadyinuse View Post
    x^2-10xy+y^2+1=0

    I'm supposed to rotate the axes in order to eliminate the xy-term. I'm told this problem is very simple, but I don't understand this section at all. Any help is greatly appreciated.
    Given a general conic:
    $\displaystyle Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
    If $\displaystyle B\not = 0$ (that means it has a cross term).

    Then the required angle is:
    $\displaystyle \cot 2\theta = \frac{A - C}{B}$
    ---
    You problem is,
    $\displaystyle x^2 - 10xy+y^2 +1 =0$
    The required angle satisfies,
    $\displaystyle \cot 2\theta = \frac{ 1 -1 }{10} = 0$
    Thus,
    $\displaystyle 2\theta = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$
    ---
    When we rotate by $\displaystyle \theta$ we get the new coordinates:
    $\displaystyle x'= x\cos \theta - y\sin \theta$
    $\displaystyle y'= x\sin \theta + y\cos \theta $
    ---
    In this case $\displaystyle \theta = \frac{\pi}{4}$ thus,
    $\displaystyle x' = x\cos \frac{\pi}{4} - y\sin \frac{\pi}{4} = x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2}$
    $\displaystyle y'= x\sin \frac{\pi}{4} + y\cos \frac{\pi}{4} = x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2}$
    ---
    Substitute the new x and y coordinates:
    $\displaystyle \left( x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2} \right)^2 - 10 \left(x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2}\right)\left( x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2} \right) + \left( x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2} \right)^2+1=0$

    Expand,
    $\displaystyle \left( \frac{1}{2}x^2 - xy + \frac{1}{2}y^2 \right) - 10 \left(\frac{1}{2}x^2 - \frac{1}{2}y^2 \right) + \left( \frac{1}{2}x^2+xy+\frac{1}{2}y^2 \right)+1=0$

    Combine,
    $\displaystyle -4x^2+6y^2+1=0$
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  3. #3
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    Thanks for helping me with the last problem. I know how to solve the rest of my homework except for this one:

    xy-2y-4x=0

    Am I supposed to change the equation first?
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  4. #4
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    Quote Originally Posted by alreadyinuse View Post
    Thanks for helping me with the last problem. I know how to solve the rest of my homework except for this one:

    xy-2y-4x=0

    Am I supposed to change the equation first?
    Hint: Think of this as $\displaystyle 0x^2+1xy+0y^2 - 4x - 2y +0 =0$.
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  5. #5
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    Ah, okay. *obviously braindead*

    Thanks for all your help!
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