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Math Help - simplify conic equation

  1. #1
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    simplify conic equation

    Can someone show me how to simplify this equation, so it represents a conic section.

    x^2-2x+y^2-8x+1=0
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cityismine View Post
    Can someone show me how to simplify this equation, so it represents a conic section.

    x^2-2x+y^2-8x+1=0
    should the -8x be - 8y?

    do you know how to complete the square?

    write as

    x^2 - 2x + y^2 - 8y = -1

    now, complete the squares for the x's and the y's separately. then you should be able to see how to rearrange it. can you continue?
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  3. #3
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    Group and compete the square:

    I assume you have a typo and mean:

    x^{2}-2x+y^{2}-8y=-1

    (x^{2}-2x+1)+(y^{2}-8y+16)=-1+1+16

    Factor:

    (x-1)^{2}+(y-4)^{1}=16

    There....what do you have?.
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  4. #4
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    That's exactly how it is in the book, I guess it's typo in the book.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cityismine View Post
    That's exactly how it is in the book, I guess it's typo in the book.
    yes, well, it seems likely to be a typo. you can check the answer in the back of the book, if possible, to see.

    anyway, galactus gave you the answer:

    (x-1)^{2}+(y-4)^{2}=16

    what kind of conic section does this describe? maybe writing it as (x-1)^{2}+(y-4)^{2}=4^2 would help...
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  6. #6
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    The answer in the book is (1,4). So I guess that's the center of the circle.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cityismine View Post
    The answer in the book is (1,4). So I guess that's the center of the circle.
    you should not be guessing. you should be able to tell from the answer galactus gave

    the equation of a circle can be written in the form: (x - h)^2 + (y - k)^2 = r^2

    when in this form, the center is (h,k) and the radius is r
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