Question

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

Answer 1

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.