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Math Help - math/check+ help! conic in standard form

  1. #1
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    math/check+ help! conic in standard form

    Hi guys!

    use the graph 4x^2-y^2-8x-4y+16=0 to do the following:
    a. write the equation in standard form

    I did but could you double check my work?

    4x^2-y^2-8x-4y+16=0
    4x^2-y^2-8x-4y=-16
    4(x^2-2x) -(Y^2+4y) =-16
    4[(X^2-2x+1)-1] - [(Y^2+4y+4)-4] =-16
    4[(x-1)^2-1] - [(Y+2)^2-4] =-16
    4(x-1)^2-4 - (y+2)^2+4 =-16
    4(x-1)^2- (y+2)^2= -16

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

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

    I think I did that right! but not sure!

    there is another part to it:
    locate the center and the vertices of this curve

    The center is:
    (-2,1)

    how would I find out the vertices of the curve?

    Thanks for your help!
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  2. #2
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    Quote Originally Posted by george93 View Post
    H
    4(x-1)^2- (y+2)^2= -16
    -16

    (x-1)^2 - (y+2)^2 = 1

    -4 rtetetewt erwerwqerwerwq 16
    !
    It is correct but ugly like the Uzi.
    You have,
    4(x-1)^2-(y+2)^2=-16
    Multiply by (-1),
    (y+2)^2-4(x-1)^2=16
    Thus,
    \frac{(y+2)^2}{4^2}-\frac{(x-1)^2}{2^2}=1
    This is a hyperbola.
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