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Math Help - Parabolas

  1. #1
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    Parabolas

    Ok, I've been given this:

    y^2=6x

    and I have to find the vertex, focus, and directrix of this graph.

    I know how to find the vertex. In this case, it is at the origin. However, I don't have the slightest idea how to find the focus or the directrix.

    Help would be appreciated here
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  2. #2
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    Quote Originally Posted by mathgeek777 View Post
    Ok, I've been given this:

    y^2=6x

    and I have to find the vertex, focus, and directrix of this graph.

    I know how to find the vertex. In this case, it is at the origin. However, I don't have the slightest idea how to find the focus or the directrix.

    Help would be appreciated here
    The first thing to do is write the equation as x = \frac{y^2}{6}. Then you have an equation in the form x = a(y - y_0)^2 + x_0. You correctly identified that the vertex is (0, 0). Now, the focus and directrix are both a distance of \frac{1}{4a} from the vertex.

    Hence:

    Focus is (\frac{3}{2}, 0)

    Directrix is x = -\frac{3}{2}
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  3. #3
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    That's not what the back of the book says.

    The back of the book says that the focus is (-\frac{3}{2}, 0) and the directrix is x=-\frac{3}{2}
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  4. #4
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    Quote Originally Posted by mathgeek777 View Post
    That's not what the back of the book says.

    The back of the book says that the focus is (-\frac{3}{2}, 0) and the directrix is x=-\frac{3}{2}
    I fixed my formula
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    Now, the focus and directrix are both a distance of \frac{1}{4a}
    I still fail to see how that relates to the finding of the focus. I can figure out one of the terms of the focus just by looking at which direction the parabola is going, but as far as finding the other term, I am still lost.
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  6. #6
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    Quote Originally Posted by mathgeek777 View Post
    I still fail to see how that relates to the finding of the focus. I can figure out one of the terms of the focus just by looking at which direction the parabola is going, but as far as finding the other term, I am still lost.
    The focus should be "inside" the parabola. Since we have a parabola that opens to the right, the focus should be to the right of the vertex. And its distance from the vertex is \frac{1}{4a}, with a = \frac{1}{6}.

    The directrix should never touch the parabola and a perpendicular drawn to the directrix from the vertex will have length \frac{1}{4a}, which is why we must have the line x = -\frac{3}{2}.

    Edit: Your book has the wrong focus.
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