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Math Help - Analalitic geometry- reducing Equation

  1. #1
    Newbie Archi.Chuckz's Avatar
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    Analalitic geometry- reducing Equation

    ^2 stands for squared
    Reduce:

    16x^2+ 24xy + 9y^2 - 10x-5y=0


    Hi there. It's going to be my finals in about 4 hours and while reviewing, I came across the equation above. I spent 2 hours trying to crack it but I just can't seem to get anywhere with the method my professor taught me. Can anyone give me a very detailed step by step on this one. Oh and if its not too much can anyone give me a similar equation example. Thanks

    By the way the final answer on the equation above is

    25x'
    ^2 - 11x' + 2y' =0.

    I need to know how its derived.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Archi.Chuckz View Post
    Reduce:
    2 2
    16x + 24xy + 9y - 10x-5y=0


    Hi there. It's going to be my finals in about 4 hours and while reviewing, I came across the equation above. I spent 2 hours trying to crack it but I just can't seem to get anywhere with the method my professor taught me. Can anyone give me a very detailed step by step on this one. Oh and if its not too much can anyone give me a similar equation example. Thanks

    By the way the final answer on the equation above is
    2
    25x' - 11x' + 2y' =0.

    I need to know how its derived.
    Is this supposed to be 16x^2 + 24xy + 9y^2 - 10x - 5y = 0?

    -Dan
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  3. #3
    Newbie Archi.Chuckz's Avatar
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    yes thats the one.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Archi.Chuckz View Post
    yes thats the one.
    And you need to transform this equation so that the x'y' term disappears?

    -Dan
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  5. #5
    Newbie Archi.Chuckz's Avatar
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    I'm not really sure on that.

    My problem thought starts in this part of the solution line.(at least in the method I was taught.)

    16(4/5x' - 3/5y')^2 + 24(4/5x' - 3/5y')(3/5x' + 4/5y') + 9(3/5x' + 4/5y')^2 - 10(4/5x' - 3/5y') - 5(3/5x' + 4/5y') = 0
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Archi.Chuckz View Post
    I'm not really sure on that.

    My problem thought starts in this part of the solution line.(at least in the method I was taught.)

    16(4/5x' - 3/5y')^2 + 24(4/5x' - 3/5y')(3/5x' + 4/5y') + 9(3/5x' + 4/5y')^2 - 10(4/5x' - 3/5y') - 5(3/5x' + 4/5y') = 0
    Okay, so expand everything:
    16 \left ( \frac{16}{25}x'^2 - \frac{24}{25}x'y' + \frac{9}{25}y'^2 \right ) + 24 \left ( \frac{12}{25}x'^2 + \frac{7}{25}x'y' - \frac{12}{25}y'^2 \right )  + 9 \left ( \frac{9}{25}x'^2 + \frac{24}{25}x'y' + \frac{16}{25}y'^2 \right )  - 10 \left ( \frac{4}{5}x' - \frac{3}{5}y' \right ) - 5 \left ( \frac{3}{5}x' + \frac{4}{5}y' \right ) = 0

    \left ( \frac{256 + 288 + 81}{25} \right ) x'^2 + \left ( \frac{- 384 + 168 + 216}{25} \right ) x'y' + \left ( \frac{144 - 288 + 144}{25} \right ) y'^2  + \left ( \frac{-40 - 15}{5} \right ) x' + \left ( \frac{30 - 20}{5} \right ) y' = 0

    \frac{625}{25}x'^2 + 0 \cdot x'y' + 0 \cdot y'^2 - \frac{55}{5}x' + \frac{10}{5}y' = 0

    25x'^2 - 11x' + 2y' = 0

    -Dan
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  7. #7
    Newbie Archi.Chuckz's Avatar
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    Wow this much simpler than the other solutions in tried. I was wondering where "288" came from in my old eqn.(stupid teacher, making simple equations complex) Thanks, now I'm off to impart this knowledge to my fellow students so I'll look smart and stuff. LOL
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