The equation of a parabola is 12y=(x−1)^2−48
. Identify the vertex, focus, and directrix of the parabola.
To identify the vertex, focus, and directrix of the parabola, we first need to put the equation in the standard form for a parabola with a vertical axis of symmetry.
Given equation: 12y = (x-1)^2 - 48
Divide by 12 to simplify the equation: y = (1/12)(x-1)^2 - 4
This can be rewritten in the standard form: y = a(x-h)^2 + k, where (h,k) is the vertex of the parabola.
Comparing this with the given equation, we have:
a = 1/12
h = 1
k = -4
Therefore, the vertex of the parabola is (1,-4).
The focus of the parabola is given by the equation (h, k + 1/(4a)), where a = 1/12 in this case.
So, the focus is (1, -4 + 1/(4*(1/12))) = (1, -1).
The directrix of the parabola is a horizontal line at a distance of 1/(4a) units below the vertex.
So, the directrix is the line y = -4 - 1/(4*(1/12)) = -7.
Therefore, the vertex of the parabola is (1,-4), the focus is (1,-1), and the directrix is the horizontal line y = -7.