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Math Help - help with conic question

  1. #1
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    help with conic question

    Rotate the axes to eliminate the xy-term. Then write the standard form of the equation of the conic relative to x'y'-plane.

    3x^2-4xy+1=0
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    Quote Originally Posted by cityismine View Post
    Rotate the axes to eliminate the xy-term. Then write the standard form of the equation of the conic relative to x'y'-plane.

    3x^2-4xy+1=0
    Are you using eigenvalues and eigenvectors, or the the transformation formula? Either way, where are you stuck?
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  3. #3
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    This is from the conics section of a pre-calculus textbook, but the book doesn't give too many examples on solving problems like this. So I'm stuck on this one.
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    Quote Originally Posted by cityismine View Post
    This is from the conics section of a pre-calculus textbook, but the book doesn't give too many examples on solving problems like this. So I'm stuck on this one.
    No example? No formula? No theory? You've been taught nothing?

    Read this: Rotation of Axes - How Conic Sections Transform Under Rotations
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    y^2-4x^2=1

    Is that the right answer?
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  6. #6
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    Quote Originally Posted by cityismine View Post
    y^2-4x^2=1

    Is that the right answer?
    Bravo!

    And I bet it felt good doing it all yourself.
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    How do i turn this conic into rectangular form:

    r=18/(6+9cos(theta))
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    Quote Originally Posted by cityismine View Post
    How do i turn this conic into rectangular form:

    r=18/(6+9cos(theta))
    Multiply both sides of the equation by 6+9\cos(\vartheta)

    We now have 6r+9r\cos(\vartheta)=18

    Now, we convert to rectangular: 6\sqrt{x^2+y^2}+9x=18

    Thus, 6\sqrt{x^2+y^2}=18-9x

    Square both sides:

    4(x^2+y^2)=9(4-4x+x^2)

    Now we do some manipulating

    \bigg(x-\frac{36}{5}\bigg)^2-\frac{4}{5}y^2=-\frac{180}{25}+\frac{64}{25}

    \bigg(x-\frac{36}{5}\bigg)^2-\frac{4}{5}y^2=-\frac{116}{25}

    \therefore \color{red}\boxed{\frac{y^2}{\left(\frac{29}{5}\ri  ght)}-\frac{\left(x-\frac{36}{5}\right)^2}{\frac{116}{5}}=1}

    If I'm not mistaken, this is a hyperbola...

    --Chris
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    Thanks for showing me the method, I tried following along, but I got a different answer. Am I right?

    (x-3.6)^2 / 5.76 - y^2 / 7.2 = 1
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    Please post new questions on new threads.

    You're killing yourselves, here, folks.

    r = \frac{18}{6 + 9 \cos(\theta)}

    r = \frac{3}{1 + \frac{3}{2} \cos(\theta)}

    r = \frac{\frac{3}{2} (2)}{1 + \frac{3}{2} \cos(\theta)}

    Eccentricity: e = 3/2 -- This is an hyperbola
    Directrix Displacement: p = 2 -- One directrix is at x = 2

    Where is the other Directrix?

    Where are the Vertices?

    What is the length of the Minor Axis?
    Last edited by TKHunny; July 29th 2008 at 06:29 AM.
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  11. #11
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    ?? Why won't this LaTeX work?
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