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Math Help - Parabolas-Please check

  1. #1
    Junior Member
    Joined
    Oct 2007
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    Parabolas-Please check

    Can you please check my answers.

    Find the focus and directrix of the parabola with the given equation.

    x^2=8y

    I got:
    focus: (0, 2)
    directrix: y = -2

    y^2=12x

    I got:
    focus: (3, 0)
    directrix: x = -3

    Match the equation to the graph.

    y^2=8x

    I got:



    x^2=8y

    I got:


    Find the standard form of the equation of the parabola using the information given.

    Focus: (-15, 0); Directrix: x = 15

    I got: y^2=-60x


    Vertex: (8, -9); Focus: (8, -4)
    I got: (x-8)^2=20(y+9)

    Find the vertex, focus, and directrix of the parabola with the given equation.

    (y-3)^2=4(x+2)

    I got:
    vertex: (2, -3)
    focus: (3, -3)
    directrix: x = 1


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

    I got:

    vertex: (2, -4)
    focus: (2, 0)
    directrix: y = -8
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  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by soly_sol View Post
    Can you please check my answers.

    Find the focus and directrix of the parabola with the given equation.

    x^2=8y

    I got:
    focus: (0, 2)
    directrix: y = -2

    y^2=12x

    I got:
    focus: (3, 0)
    directrix: x = -3

    Match the equation to the graph.

    y^2=8x

    I got:



    x^2=8y

    I got:


    Find the standard form of the equation of the parabola using the information given.

    Focus: (-15, 0); Directrix: x = 15

    I got: y^2=-60x


    Vertex: (8, -9); Focus: (8, -4)
    I got: (x-8)^2=20(y+9)

    Find the vertex, focus, and directrix of the parabola with the given equation.

    (y-3)^2=4(x+2)

    I got:
    vertex: (2, -3)
    focus: (3, -3)
    directrix: x = 1


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

    I got:

    vertex: (2, -4)
    focus: (2, 0)
    directrix: y = -8
    good job, everything is just fine..
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  3. #3
    Super Member
    earboth's Avatar
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    Germany
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    Quote Originally Posted by soly_sol View Post
    Can you please check my answers.

    ...

    Find the vertex, focus, and directrix of the parabola with the given equation.

    I (y-3)^2=4(x+2)

    I got:
    vertex: (2, -3)
    focus: (3, -3)
    directrix: x = 1


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

    I got:

    vertex: (2, -4)
    focus: (2, 0)
    directrix: y = -8
    Hello,

    to I:

    V(-2, 3), F(-1, 3), d: x = -3

    to II:

    V(-2, 4), F(-2, 8), d: y = 0
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  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by earboth View Post
    Hello,

    to I:

    V(-2, 3), F(-1, 3), d: x = -3

    to II:

    V(-2, 4), F(-2, 8), d: y = 0
    oh yeah, i dindn't notice all the way down. hehe
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