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To graph the focus and the directrix of the parabola given by the equation [tex]\(x = -\frac{1}{8}(y-3)^2 + 1\)[/tex], you need to start by identifying the vertex form of a parabola and using the properties of the parabola. Here's a step-by-step solution:
### Step 1: Identify the vertex and orientation
The standard form of a parabola that opens horizontally is given by:
[tex]\[ x = a(y - k)^2 + h \][/tex]
Comparing this with the given equation [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex], we can see that:
- The vertex [tex]\((h, k)\)[/tex] is [tex]\( (1, 3) \)[/tex]
- Since [tex]\(a = -\frac{1}{8}\)[/tex], the parabola opens to the left.
### Step 2: Determine the value of [tex]\(4p\)[/tex]
In the standard form [tex]\(x = a(y-k)^2 + h\)[/tex], [tex]\(a\)[/tex] is related to the parameter [tex]\(p\)[/tex] by:
[tex]\[ a = \frac{1}{4p} \][/tex]
From the given parabola:
[tex]\[ -\frac{1}{8} = \frac{1}{4p} \][/tex]
Solving for [tex]\(p\)[/tex]:
[tex]\[ 4p = -8 \][/tex]
[tex]\[ p = -2 \][/tex]
Here, [tex]\( p \)[/tex] tells us the distance from the vertex to the focus and from the vertex to the directrix along the axis of symmetry.
### Step 3: Find the focus
The focus is [tex]\(p\)[/tex] units away from the vertex in the direction the parabola opens. Since the parabola opens to the left and [tex]\(p = -2\)[/tex]:
- The focus is located 2 units to the left of the vertex: [tex]\((1, 3)\)[/tex].
[tex]\[ \text{Focus} = (1 - 2, 3) = (-1, 3) \][/tex]
### Step 4: Find the directrix
The directrix is a vertical line that is [tex]\(p\)[/tex] units away from the vertex in the opposite direction from the focus. Since [tex]\(p = -2\)[/tex], the directrix is 2 units to the right of the vertex at [tex]\( x = 3 \)[/tex]:
[tex]\[ \text{Directrix:} \, x = 1 - (-2) = 3 \][/tex]
### Step 5: Plot the parabola, vertex, focus, and directrix
1. Vertex: Plot the vertex at [tex]\((1, 3)\)[/tex].
2. Focus: Plot the focus at [tex]\((-1, 3)\)[/tex].
3. Directrix: Draw the vertical line [tex]\( x = 3 \)[/tex].
4. Parabola: Sketch the parabola opening to the left with the vertex at [tex]\((1, 3)\)[/tex], passing through points that satisfy the equation [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex].
### Final Result:
- Vertex: [tex]\((1, 3)\)[/tex]
- Focus: [tex]\((-1, 3)\)[/tex]
- Directrix: [tex]\( x = 3 \)[/tex]
Make sure to label these points and line clearly on your graph to indicate the focus and the directrix of the parabola.
### Step 1: Identify the vertex and orientation
The standard form of a parabola that opens horizontally is given by:
[tex]\[ x = a(y - k)^2 + h \][/tex]
Comparing this with the given equation [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex], we can see that:
- The vertex [tex]\((h, k)\)[/tex] is [tex]\( (1, 3) \)[/tex]
- Since [tex]\(a = -\frac{1}{8}\)[/tex], the parabola opens to the left.
### Step 2: Determine the value of [tex]\(4p\)[/tex]
In the standard form [tex]\(x = a(y-k)^2 + h\)[/tex], [tex]\(a\)[/tex] is related to the parameter [tex]\(p\)[/tex] by:
[tex]\[ a = \frac{1}{4p} \][/tex]
From the given parabola:
[tex]\[ -\frac{1}{8} = \frac{1}{4p} \][/tex]
Solving for [tex]\(p\)[/tex]:
[tex]\[ 4p = -8 \][/tex]
[tex]\[ p = -2 \][/tex]
Here, [tex]\( p \)[/tex] tells us the distance from the vertex to the focus and from the vertex to the directrix along the axis of symmetry.
### Step 3: Find the focus
The focus is [tex]\(p\)[/tex] units away from the vertex in the direction the parabola opens. Since the parabola opens to the left and [tex]\(p = -2\)[/tex]:
- The focus is located 2 units to the left of the vertex: [tex]\((1, 3)\)[/tex].
[tex]\[ \text{Focus} = (1 - 2, 3) = (-1, 3) \][/tex]
### Step 4: Find the directrix
The directrix is a vertical line that is [tex]\(p\)[/tex] units away from the vertex in the opposite direction from the focus. Since [tex]\(p = -2\)[/tex], the directrix is 2 units to the right of the vertex at [tex]\( x = 3 \)[/tex]:
[tex]\[ \text{Directrix:} \, x = 1 - (-2) = 3 \][/tex]
### Step 5: Plot the parabola, vertex, focus, and directrix
1. Vertex: Plot the vertex at [tex]\((1, 3)\)[/tex].
2. Focus: Plot the focus at [tex]\((-1, 3)\)[/tex].
3. Directrix: Draw the vertical line [tex]\( x = 3 \)[/tex].
4. Parabola: Sketch the parabola opening to the left with the vertex at [tex]\((1, 3)\)[/tex], passing through points that satisfy the equation [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex].
### Final Result:
- Vertex: [tex]\((1, 3)\)[/tex]
- Focus: [tex]\((-1, 3)\)[/tex]
- Directrix: [tex]\( x = 3 \)[/tex]
Make sure to label these points and line clearly on your graph to indicate the focus and the directrix of the parabola.
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