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Math Help - Rotation of conics (Hyperbola)

  1. #1
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    Rotation of conics (Hyperbola)

    Hey I am really stuck on this problem:

    2x^2 - 3xy - 2y^2 + 10 = 0

    I know that when AC<0 then the rotation is for a Hyperbola, 2(-2)=-4
    Now the formula for finding the degree of axis rotation is cot 2q=(A-C)/B
    so (2-(-2))/-3 = -4/3, and this is where I am stuck, I can do the rest after I have q but whatever i try does not seem to be working
    I need to solve cot q = -4/3
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  2. #2
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    Quote Originally Posted by fordyingseasons View Post
    ...
    Now the formula for finding the degree of axis rotation is cot 2q=(A-C)/B
    so (2-(-2))/-3 = -4/3, and this is where I am stuck, I can do the rest after I have q but whatever i try does not seem to be working
    I need to solve cot q = -4/3
    Is this correct?
    According to the formula you have

    \cot(2q) = -\frac43~\implies~ \tan(2q)=-\frac34~\implies~ 2q=-36.87^\circ = 143.13^\circ

    And now you can calculate q.
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  3. #3
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    Thank you, i have another problem now with that answer however, I have to plug in xcosA - ysinB and xsinB + ycosA

    In my book the give the example xcos(pie/4) - ysin(pie/4) = x(1/sqrt2)-y(1/sqrt2).
    How do i plug in 71.565?
    Last edited by fordyingseasons; June 23rd 2008 at 06:11 AM.
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