note: Enter your answer and show all the steps that you use to solve this problem in the space provided.The equation of a parabola is 12y=(x−1)2−48 . Identify the vertex, focus, and directrix of the parabola show work

Answer 1

To identify the vertex, focus, and directrix of the parabola, let's first write the equation in standard form by completing the square:

12y = (x-1)^2 - 48
12y = x^2 - 2x + 1 - 48
12y = x^2 - 2x - 47
y = (1/12)x^2 - (1/6)x - 47/12

Now we see that the equation is in the form y = ax^2 + bx + c, where the vertex of the parabola is given by (-b/2a, c).

So, the vertex is (-(-1/6)/(2*(1/12)), -47/12) = (1/6, -47/12)

Now, we know that the focus of the parabola is located at (h, k + 1/(4a)), and the directrix is given by y = k - 1/(4a).

Therefore, the focus of the parabola is (1/6, -47/12 + 1/(4*(1/12))) = (1/6, -47/12 + 1/3) = (1/6, -43/12)

And the directrix is y = -47/12 - 1/(4*(1/12)) = y = -47/12 - 3 = y = -59/12

Therefore, the vertex of the parabola is (1/6, -47/12), the focus is (1/6, -43/12), and the directrix is y = -59/12.