Let's start by identifying the key components of the parabola given by the equation [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex].
### Step 1: Identify the Vertex
The vertex form of a parabola's equation [tex]\( x = a(y - k)^2 + h \)[/tex] tells us that:
- The vertex of the parabola is at [tex]\((h, k)\)[/tex].
- Here, [tex]\( h = 1 \)[/tex] and [tex]\( k = 3 \)[/tex].
So, the vertex is at the point [tex]\((1, 3)\)[/tex].
### Step 2: Determine the Focus
In the given equation [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex], the coefficient [tex]\(-\frac{1}{8}\)[/tex] can be related to the form [tex]\( \frac{1}{4a} \)[/tex] of a parabola that opens sideways.
If [tex]\( x = -\frac{1}{8}(y - 3)^2 + 1 \)[/tex] is the equation of the parabola, then comparing it to the standard form:
[tex]\[ x = a(y - k)^2 + h \][/tex]
We see that [tex]\( a = -\frac{1}{8} \)[/tex].
The distance from the vertex to the focus, [tex]\( p \)[/tex], satisfies [tex]\( \frac{1}{4a} = \frac{1}{4 \cdot (-\frac{1}{8})} = -2 \)[/tex].
Here, [tex]\( p = -2 \)[/tex].
The focus is located at a distance of [tex]\( p \)[/tex] units from the vertex along the axis of symmetry of the parabola:
- Since [tex]\( p = -2 \)[/tex], and the parabola opens to the left, we move 2 units left from the vertex.
- Therefore, the focus is at [tex]\((1 - 2, 3) = (-1, 3)\)[/tex].
### Step 3: Determine the Directrix
The directrix of a parabola is a line perpendicular to the axis of symmetry and at the same distance [tex]\( p \)[/tex] from the vertex but in the opposite direction.
- Since [tex]\( p = -2 \)[/tex] and we went left for the focus, we go right for the directrix.
- The directrix is a vertical line at [tex]\( x = h - p = 1 - (-2) = 1 + 2 = 3 \)[/tex].
Thus, the directrix is the vertical line [tex]\( x = 3 \)[/tex].
### Graph it:
- Plot the vertex at [tex]\( (1, 3) \)[/tex].
- Plot the focus at [tex]\( (-1, 3) \)[/tex].
- Draw the directrix as the vertical line [tex]\( x = 3 \)[/tex].
Use the "Point" tool to plot the focus at [tex]\((-1, 3)\)[/tex]. Use the "Line" tool to draw the vertical line for the directrix at [tex]\(x = 3\)[/tex].
