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Math Help - Locus Help

  1. #31
    Eater of Worlds
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    You can move them, but you'll get a different parabola. The parabola I posted has it's specific focus and directrix. As does any parabola.
    You can't just move the focus and directrix around and keep the same parabola.

    Let's find the equatio of a parabola with vertex at V(-4,2) and directrix at y=5.

    See, the distance from the vertex to the directrix(line y=5) is 3 units. 5-2=3.

    So, the focus is F(-4,-1). Because 2-3=-1.

    We can use (x-h)^{2}=4p(y-k)

    We know h and k, the coordinates of the vertex. So, (x+4)^{2}=4(-3)(y-2)

    (x+4)^{2}=-12(y-2)

    We want to express it as ax^{2}+bx+x

    x^{2}+8x+16=-12y+24

    12y=-x^{2}-8x+8

    y=\frac{-1}{12}x^{2}-\frac{2}{3}x+\frac{2}{3}

    There she is.

    Parabolas are used a lot in real world applications. One area is in highway construction. Vertical curves on roads(hills and valleys) are laid out using the principles of parabolas.
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  2. #32
    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    You can move them, but you'll get a different parabola. The parabola I posted has it's specific focus and directrix. As does any parabola.
    You can't just move the focus and directrix around and keep the same parabola.

    Parabolas are used a lot in real world applications. One area is in highway construction. Vertical curves on roads(hills and valleys) are laid out using the principles of parabolas.
    Ah, interesting. And how do we determine the focus of a parabola? What's the formula?
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  3. #33
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    I just showed. It's the same distance as from the vertex to the directrix.

    \frac{1}{4a}=p. |p| is the distance from the vertex to the focus.

    'a' is the coefficient of the x^2. Remember, we had -1/6 on the one I graphed. So, \frac{1}{4(-1/6)}=|\frac{3}{2}|

    It's 3/2 units from the focus to the vertex or from the vertex to the directrix.

    If p<0, then the parabola opens down. If p>0, then it opens up.

    Of course, they can be horizontal too. In that case, you have

    x=\frac{1}{4p}y^{2} instead of y=\frac{1}{4p}x^{2}

    If it's a horizontal parabola, then if p>0 it opens to the right and if p<0 it opens left.
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  4. #34
    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    I just showed. It's the same distance as from the vertex to the directrix.

    \frac{1}{4a}=p. |p| is the distance from the vertex to the focus.

    'a' is the coefficient of the x^2. Remember, we had -1/6 on the one I graphed. So, \frac{1}{4(-1/6)}=|\frac{3}{2}|

    It's 3/2 units from the focus to the vertex or from the vertex to the directrix.

    If p<0, then the parabola opens down. If p>0, then it opens up.

    Of course, they can be horizontal too. In that case, you have

    x=\frac{1}{4p}y^{2} instead of y=\frac{1}{4p}x^{2}

    If it's a horizontal parabola, then if p>0 it opens to the right and if p<0 it opens left.
    Am i correct then when i say that the x-coordinate of the focus is equal to the symmetry axis, and the y-coordinate is equal to that of the directrix?

    EDIT: Sorry, i didnt notice you edited your previous reply.
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  5. #35
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    A History Lesson

    The above curves are called conics and they were introduced by the ancient Greek Appolinius. He used an approach of cutting a cone. About 600 years later the Hellensitic Greek mathematician Pappus states them in terms of locus (like here). About 1300 years later Fermant and Descrates create analytic geometry and in which a conic is the curve:
    Ax^2+Bxy+Cy^2+De+Fy+G=0.
    About 40 years later the astronomer Johannes Kepler finallly, finally realizes an application of these conics. It was when he stated "Kepler's First Law" about the orbits of the planets following ellipses.
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