To plot the vertex, focus, and directrix for the parabola represented by the equation
[tex]\[ x = -\frac{1}{12}(y-2)^2 + 1 \][/tex]
we need to follow these steps:
### Step-by-Step Solution
1. Identify the vertex:
The equation of the parabola can be written in the form:
[tex]\[ x = a(y-k)^2 + h \][/tex]
Here, we have [tex]\( a = -\frac{1}{12} \)[/tex], [tex]\( k = 2 \)[/tex], and [tex]\( h = 1 \)[/tex]. The vertex [tex]\((h, k)\)[/tex] of the parabola is thus:
[tex]\[
\text{Vertex} = (1, 2)
\][/tex]
2. Determine the value of [tex]\(p\)[/tex]:
The value of [tex]\(p\)[/tex] for a parabola of the form [tex]\( x = a(y-k)^2 + h \)[/tex] is given by:
[tex]\[
a = \frac{1}{4p}
\][/tex]
Here, [tex]\( a = -\frac{1}{12} \)[/tex]. We can solve for [tex]\( p \)[/tex] by setting up the equation:
[tex]\[
-\frac{1}{12} = \frac{1}{4p}
\][/tex]
Solving for [tex]\( p \)[/tex]:
[tex]\[
-12 = 4p \quad \Rightarrow \quad p = -3
\][/tex]
3. Plot the focus:
The focus of the parabola is located at a distance of [tex]\( p \)[/tex] from the vertex along the axis of symmetry. Since [tex]\( p \)[/tex] is negative, the focus is to the left of the vertex. The coordinates of the focus are:
[tex]\[
\text{Focus} = \left( h + p, k \right) = (1 - 3, 2) = (-2, 2)
\][/tex]
4. Determine the directrix:
The directrix is a vertical line perpendicular to the axis of symmetry of the parabola, and it is located at a distance of [tex]\( |p| \)[/tex] from the vertex on the side opposite the focus. The equation of the directrix is:
[tex]\[
x = h - p = 1 - (-3) = 1 + 3 = 4
\][/tex]
### Summary of Key Points to Plot:
- Vertex: [tex]\((1, 2)\)[/tex]
- Focus: [tex]\((-2, 2)\)[/tex]
- Directrix: [tex]\( x = 4 \)[/tex]
Using the drawing tools, you can plot these points and the directrix on the graph. The vertex is the point [tex]\((1, 2)\)[/tex], the focus is the point [tex]\((-2, 2)\)[/tex], and the directrix is the vertical line [tex]\( x = 4 \)[/tex].
