Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. \(y=36 x^{2}\)

Step-by-Step Solution

Answer

The parabola opens upwards with vertex \((0, 0)\), focus \((0, \frac{1}{144})\), and directrix \(y = -\frac{1}{144}\).

Step 1: Identify Parabola Type

The given equation is in the form of a parabola. Since it is given as \(y = 36x^2\), it indicates a vertical parabola which opens upwards.

Step 2: Rewrite in Standard Form

Rewrite the given equation \(y = 36x^2\) in the standard form of a parabola \((y - k) = a(x - h)^2\). Here, \(h = 0\), \(k = 0\), and \(a = 36\). So the equation is \(y = 36x^2\).

Step 3: Identify the Vertex

In the standard form equation \((y - k) = a(x - h)^2\), the vertex is \((h, k)\). For this equation, the vertex is \((0, 0)\).

Step 4: Determine the Focus

For a vertical parabola with equation \(y = ax^2\), the distance from the vertex to the focus is \(\frac{1}{4a}\). Here, \(a = 36\), so the distance is \(\frac{1}{4 \times 36} = \frac{1}{144}\). The focus is located at \((0, \frac{1}{144})\).

Step 5: Determine the Directrix

The directrix of a vertical parabola \(y = ax^2\) is the line \(y = k - \frac{1}{4a}\). Since \(h = 0\) and \(k = 0\), the directrix is \(y = -\frac{1}{144}\).

Step 6: Sketch the Parabola

Plot the vertex at \((0, 0)\). Draw the parabola opening upwards with the vertex at this point. Mark the focus at \((0, \frac{1}{144})\) and the directrix as a horizontal line at \(y = -\frac{1}{144}\). Annotate these key elements on the graph for reference.

Key Concepts

Vertex Form

The vertex form of a parabola is a way to express the quadratic equation so that the vertex of the parabola is easily identifiable. The general vertex form of a quadratic equation is given by \[ y = a(x-h)^2 + k \]where

• \( (h, k) \) represents the vertex of the parabola.
• \( a \) indicates the direction and the width of the parabola.

For the equation \( y = 36x^2 \), we can rewrite it as \( y = 36(x-0)^2 + 0 \), thus recognizing \( h = 0 \) and \( k = 0 \). This tells us that the vertex of the parabola is at the origin, \( (0, 0) \), and since \( a = 36 \), this parabola will open upwards and be narrower than a parabola with a smaller \( a \) value due to the larger coefficient.

Focus of a Parabola

The focus of a parabola is a point from which all points on the parabola are equidistant from a corresponding line called the directrix. To find the focus of a vertically oriented parabola in the form \( y = ax^2 \), such as \( y = 36x^2 \), we determine its distance from the vertex using the formula \( \frac{1}{4a} \).

• For \( a = 36 \), the distance to the focus is \( \frac{1}{4 \times 36} = \frac{1}{144} \).
• Therefore, the focus is located at \( (0, \frac{1}{144}) \).

This means the focus is just a tiny distance above the vertex, which is at the origin in this case. The focus plays a crucial role in determining the set of all points that form the parabola.

Directrix

The directrix of a parabola is a line such that every point on the parabola is equidistant from the focus and the directrix. For a parabola of the form \( y = ax^2 \), the directrix is found using the equation \[ y = k - \frac{1}{4a} \]For our specific parabola, since \( k = 0 \) and \( a = 36 \), we find the directrix at: \[ y = 0 - \frac{1}{4 \times 36} = -\frac{1}{144} \]Thus, the directrix is a horizontal line that sits just below the vertex, oppositely from the focus. This line is instrumental in defining the geometric properties of the parabola.

• The directrix doesn't actually pass through the parabola; instead, it's used to help ensure that the set of points that makes up the parabola is equidistant.

Graphing Parabolas

Graphing a parabola involves sketching its shape based on certain key features such as the vertex, focus, directrix, and axis of symmetry. It’s essential to first identify the vertex form or standard form of the equation. For the parabola \( y = 36x^2 \), follow these steps:

• Begin by plotting the vertex at the origin \( (0, 0) \).
• Identify that the parabola opens upwards due to the positive value of \( a \).
• Mark the focus at \( (0, \frac{1}{144}) \), just above the vertex.
• Draw the directrix as a horizontal line at \( y = -\frac{1}{144} \).

Once these elements are outlined, sketch the symmetric "U" shape of the parabola opening upwards. The focus and the directrix act as guides for the curve, ensuring accuracy and reflecting the parabola's unique properties. Understanding these elements makes graphing more intuitive and less reliant on plotting individual points.