Problem 60 - Algebra 2

Question

Identify the focus and directrix of each parabola. Then graph the parabola. $$ y=\frac{1}{16} x^{2} $$

Step-by-Step Solution

Answer

The focus of parabola \(y = \frac{1}{16}x^2\) is at (0,4) and the directrix is \(y = -4\).

Step 1: Identify the values of a, h and k

The given equation is in vertex form \(y=a(x-h)^{2}+k\), so a = 1/16, h = 0 and k = 0. This indicates that the vertex of the parabola is at the origin, (0,0).

Step 2: Calculate the Focus and Directrix

For a parabola of form \(y = ax^2\), the focus is at the point (0, 1/4a) and the directrix at \(y = -1/4a\). On substituting a = 1/16 into these, we find that the focus is (0,4) and the directrix is \(y = -4\).

Step 3: Graph the Parabola

Plot the vertex at (0,0) on the graph. Plot the focus at (0,4) and draw the directrix at \(y = -4\). This gives the parabola with focus and directrix.

Key Concepts

Vertex Form

The vertex form of a parabola is a way to express its equation which highlights the vertex of the parabola. A parabola can be represented in this form:

• Equation: \( y = a(x-h)^2 + k \)

In this equation:

• \( a \) determines the "width" and direction of the parabola (upward or downward).
• \( h, k \) are the coordinates of the vertex.

This form is especially useful since the vertex \((h, k)\) is directly identifiable from the equation. For instance, with the equation \( y = \frac{1}{16}x^2 \), which is simplified from \( y = \frac{1}{16}(x-0)^2 + 0 \):

• The vertex is \((0,0)\).
• The parabola opens upward because \( a = \frac{1}{16} > 0 \).

The vertex form thus sets the stage for understanding other characteristics like the focus and directrix.

Focus and Directrix

The focus and directrix of a parabola are essential parts of its structure, which help define and visualize its path. The focus is a point, and the directrix is a line:

• The focus of the parabola with equation \( y = ax^2 \) is located at \((0, \frac{1}{4a})\).
• The directrix is a horizontal line given by \( y = -\frac{1}{4a} \).

To find them for \( y = \frac{1}{16}x^2 \), substitute \( a = \frac{1}{16} \):

• The focus is \((0, 4)\), which means it lies 4 units above the vertex.
• The directrix is \( y = -4 \), a line 4 units below the vertex.

These components show how a parabola "bends" or "curves." The points on the parabola are equidistant from both the focus and the directrix line.

Graphing Parabolas

Once the vertex, focus, and directrix are known, graphing the parabola becomes a straightforward task. Here’s how you can plot it:

• Begin by plotting the vertex, which serves as the turning point; for \( y = \frac{1}{16}x^2 \), it’s at \((0,0)\).
• Plot the focus \((0, 4)\) above the vertex.
• Draw the directrix \( y = -4 \) as a horizontal line below the vertex.

Next, sketch the parabola:

• Ensure it opens upward since \( a = \frac{1}{16} \) is positive.
• The graph should be equidistant from the focus and directrix at all points, making a smooth curve upwards.

This process visually demonstrates how a parabola is shaped by these components, providing a clear geometric representation of its algebraic properties.