Problem 24 - Algebra 2

Question

Identify the vertex, the focus, and the directrix of each graph. Then sketch the graph. $$ x=\frac{1}{24} y^{2} $$

Step-by-Step Solution

Answer

The vertex of the parabola is at the origin (0,0), the focus is at (6,0), and the directrix is the vertical line \(x=-6\). The parabola opens to the right.

Step 1: Identify the Form of the Parabola

The given equation is \(x=\frac{1}{24} y^{2}\). This is in the form of \(x=\frac{1}{4p} y^{2}\), which represents a parabola that opens to the right (if \(p>0\)) or to the left (if \(p<0\)).

Step 2: Determine the Value of 'p'

Here, we compare the given equation with the standard equation. The value of \(\frac{1}{4p}\) is equal to \(\frac{1}{24}\), thus we can solve for \(p\). After solving, we get that \(p=6\).

Step 3: Find the Vertex

Since the parabola is expressed in the standard form, its vertex is (0,0).

Step 4: Identify the Focus

The focus lies inside the parabola, and is \(p\) units away from the vertex. Since \(p=6\) and the parabola opens to the right, we get the focus at (6,0).

Step 5: Find the Directrix

The directrix is a vertical line \(p\) units to the left of the vertex, because the parabola opens to the right. Therefore, the equation of the directrix is \(x = -6\).

Step 6: Sketch the Graph

On a paper or graph paper, plot the vertex at the origin (0,0), the focus at (6,0), and draw the directrix at \(x=-6\). Then sketch the parabola, which should be symmetric with respect to the line \(y=0\) and should open to the right.

Key Concepts

Understanding the Vertex of a Parabola

The vertex of a parabola is its highest or lowest point, depending on how the parabola opens. It serves as the turning point and is an essential feature when describing the graph of a parabola.
For the equation given in the problem, which is in the form of \( x = \frac{1}{4p} y^2 \), the vertex lies at the origin, (0, 0).
This is because the equation doesn't include any additional terms that shift the vertex away from the origin. In general:

• For a vertical parabola, the vertex can be found at (h, k) from the standard equation \( (x-h)^2 = 4p(y-k) \)
• For a horizontal parabola, it's found at (h, k) from the equation \( (y-k)^2 = 4p(x-h) \)

The simple (0,0) structure here indicates there's been no horizontal or vertical movement from the standard starting position.

The Focus of a Parabola Explained

The focus is a crucial point that lies on the interior of the parabola. It helps define its shape and is used along with the directrix to depict the parabola accurately.
For our parabola, which opens to the right, this focus is located at the point (6, 0) calculated using \[ \frac{1}{4p} = \frac{1}{24} \] which solved gives us \(p = 6\).
Its placement can be understood by:

• Recognizing that the focus is always "p" units away from the vertex along the parabola's axis. Since \(p = 6\), the focus is moved 6 units horizontally from the origin (rightward as \(p > 0\)).
• In general, a parabola's focus can be calculated using its vertex and the distance \( p \). It will either shift horizontally or vertically, according to the parabola's orientation.

The focus is crucial for understanding the directness or spread of the parabola as every point on a parabola is equidistant from the focus and the directrix.

Describing the Directrix of a Parabola

The directrix is another essential component of a parabola. It's a fixed line outside the parabola that assists in defining its shape. For a right-opening parabola, like ours, the directrix will be a vertical line.
With our parabola having a vertex at (0, 0) and \(p = 6\), this directrix is found 6 units to the left of the vertex, resulting in the equation \(x = -6\).
In general:

• The directrix complements the focus in defining the parabola. Together, they maintain a balanced structure, ensuring each point on the parabola is equidistant between the focus and this line.
• For vertical parabolas, the directrix is a horizontal line, whereas, for horizontal parabolas, it forms a vertical line.

It's important to note the interplay between focus and directrix, as this distance from both defines the parabola's curve precisely, maintaining parabola's symmetrical shape.

Mastering Graph Sketching of Parabolas

Sketching a parabola involves translating its equation into a visual form. Here, we'll turn our mathematical findings into an understandable graph.
Start by marking the vertex. With the origin as (0,0), place this point first.

• Next, locate the focus (6, 0) on the graph. This confirms where the parabola will stretch towards.
• Draw the directrix. In this particular problem, the line is at \(x = -6\).
• Finally, sketch the curve. Parabolas are symmetric relative to their axis, which here lies along the y-axis. Make sure that both arms stretch evenly, passing through and slightly beyond the axis where the focus and directrix are set.

By using these calculated points - vertex, focus, and directrix - you'll create a clear visualization. The parabola will face rightwards, opening up away from the directrix, with the band of the graph flaring symmetrically around the center focus.