Algebra 2

The equation of a parabola is 12y=(x-1)^2-48. Identify the vertex, focus, and directrix of the parabola.

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  1. the parabola x^2 = 4py has
    vertex at (0,0)
    focus at (0,p)
    directrix at y = -p
    So. Let's rearrange things a bit.
    12y=(x-1)^2-48
    (x-1)^2 = 12(y+4)
    So, we can see that we now have
    vertex at (1,-4)
    focus at y=(-4+3) = (1,-1)
    directrix at y=(-4-3) or y = -7

    to confirm our work, see

    https://www.wolframalpha.com/input/?i=parabola+12y%3D(x-1)%5E2-48

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    oobleck
  2. 12y = (x-1)^2 - 48.
    Vertex form: Y = a(x-h)^2 + k.
    Y = (1/12)(x-1)^2 - 4.

    V(h, k) = V(1, -4).
    F(h, k+1/(4a)) = F(1, -1).
    D(h, k-1/(4a)) = D(1, -7).

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