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
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.