To identify the vertex, focus, and directrix of the parabola, we need to first rewrite the equation in vertex form, which is y=a(x-h)^2+k
Given equation: 12y = (x-1)^2 - 48
Divide both sides by 12:
y = (1/12)(x-1)^2 - 4
Now we have the equation in vertex form, where h=1, k=-4, and a=1/12
Vertex(h,k): The vertex of the parabola is at the point (1,-4)
Focus: The focus of the parabola is F(h, k + 1/(4a))
In this case, a = 1/12
So, F(1, -4 + 1/(4(1/12))) = F(1, -4 + 3) = F(1, -1)
The focus of the parabola is at the point (1, -1)
Directrix: The directrix of the parabola is a horizontal line located at k - 1/(4a)
In this case, a = 1/12
So, the directrix is located at -4 - 1/(4(1/12)) = -4 - 3 = -7
The equation of the directrix is y = -7
Therefore, the vertex of the parabola is (1,-4), the focus is (1,-1), and the directrix is y = -7.
The equation of a parabola is 12y=(x−1)^2−48
Identify the vertex, focus, and directrix of the parabola. show all steps
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