To find the focus and the directrix of the given parabola [tex]\( y = -\frac{1}{12}(x-4)^2 + 2 \)[/tex], we start by identifying key information from the standard form of a parabola.
The given equation [tex]\( y = -\frac{1}{12}(x-4)^2 + 2 \)[/tex] is in the vertex form [tex]\( y = a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] represents the vertex of the parabola.
From the equation:
- [tex]\( h = 4 \)[/tex]
- [tex]\( k = 2 \)[/tex]
- [tex]\( a = -\frac{1}{12} \)[/tex]
The vertex of the parabola, therefore, is at [tex]\((4, 2)\)[/tex].
Next, we determine the coordinates of the focus and the equation of the directrix. For a parabola in the form [tex]\( y = a(x-h)^2 + k \)[/tex]:
1. The focus of the parabola is given by the point [tex]\((h, k + \frac{1}{4a})\)[/tex].
2. The directrix is given by the line [tex]\( y = k - \frac{1}{4a} \)[/tex].
We proceed by calculating each:
Focus:
To find the y-coordinate of the focus:
[tex]\[ k + \frac{1}{4a} = 2 + \frac{1}{4 \cdot -\frac{1}{12}} = 2 + \frac{1}{-\frac{1}{3}} = 2 - 3 = -1 \][/tex]
Therefore, the coordinates of the focus are:
[tex]\[ (h, k + \frac{1}{4a}) = (4, -1) \][/tex]
Directrix:
To find the y-coordinate of the directrix:
[tex]\[ y = k - \frac{1}{4a} = 2 - \frac{1}{4 \cdot -\frac{1}{12}} = 2 - \frac{1}{-\frac{1}{3}} = 2 + 3 = 5 \][/tex]
Therefore, the equation of the directrix is:
[tex]\[ y = k - \frac{1}{4a} = 5 \][/tex]
Putting it all together, we find:
- The focus of the parabola is [tex]\((4, -1)\)[/tex],
- The directrix is [tex]\( y = 5 \)[/tex].
Thus, the correct answer is:
C) Focus [tex]\((4, -1)\)[/tex], directrix is [tex]\( y = 5 \)[/tex]
