Problem 33 - College Algebra with Corequisite Support

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. \(y=\frac{1}{36} x^{2}\)

Step-by-Step Solution

Answer

Vertex at (0, 0); focus at (0, 9); directrix at y = -9.

Step 1: Identify Parabola Form

The given equation of the parabola is in the form of \(y = ax^2\), where \(a = \frac{1}{36}\). This is a vertically oriented parabola, opening upwards.

Step 2: Determine the Vertex

Since the equation is in the form \(y = ax^2\), the vertex of the parabola is at the origin, \((0, 0)\).

Step 3: Find the Focus

For a parabola \(y = ax^2\), the focus is given by \( (0, \frac{1}{4a}) \). Substitute \(a = \frac{1}{36}\):\[ \frac{1}{4a} = \frac{1}{4 \times \frac{1}{36}} = 9 \]Therefore, the focus is at \((0, 9)\).

Step 4: Determine the Directrix

The directrix of a parabola \(y = ax^2\) is given by the equation \(y = -\frac{1}{4a}\). Using \(a = \frac{1}{36}\), we find:\[ -\frac{1}{4a} = -9 \]Thus, the directrix is the horizontal line \(y = -9\).

Step 5: Graph the Parabola

Plot the vertex at \((0, 0)\), the focus at \((0, 9)\), and sketch the parabola opening upwards. Draw the directrix as the line \(y = -9\). Highlight the symmetry axis along the y-axis.

Key Concepts

Focus of a Parabola

The focus of a parabola is a special point that lies inside the curve. It plays a crucial role in the parabola's geometric property: every point on the parabola is equidistant from the focus and the directrix. Think of the focus as a source of rays that reflect off the parabola, such that they all travel parallel to the axis of symmetry. This makes the focus essential in the formation and orientation of the parabola.
For the given equation, which is in the form of \(y = ax^2\), the focus can be found using the formula:

• Focus Formula: \((0, \frac{1}{4a})\)

Substitute the value of \(a = \frac{1}{36}\) to determine the coordinate of the focus:
\[\frac{1}{4a} = \frac{1}{4 \times \frac{1}{36}} = 9\]Hence, the focus of the given parabola is at the point \((0, 9)\). This position is directly above the vertex at a distance of 9 units and serves as the 'focal point' from which the parabola seems to emanate.

Directrix of a Parabola

The directrix of a parabola is a line that acts as a reference for the parabola's measured curves. It complements the focus in defining the shape of the parabola by being equidistant to the points on the curve opposite the focus. The directrix is perpendicular to the axis of symmetry and lies outside the parabola's curve.
For a vertically oriented parabola of the form \(y = ax^2\), the equation for the directrix can be calculated by:

• Directrix Formula: \(y = -\frac{1}{4a}\)

Using \(a = \frac{1}{36}\), you can find:\[-\frac{1}{4a} = -9\]Thus, the directrix for the given parabola is the horizontal line \(y = -9\). It lies 9 units below the vertex and provides a reference line that, together with the focus, governs the parabola’s symmetry and shape.

Parabola Vertex

The vertex of a parabola is the most central point on the parabola and marks the position where the curve changes direction. It is considered the "turning point" of the parabola. In terms of mathematical characteristics, the vertex is the point where the parabola reaches its maximum or minimum value, depending on its orientation.
For parabolas in the format \(y = ax^2\), the vertex is always at the origin \((0, 0)\) if there are no added constants shifting the graph horizontally or vertically.
In the current exercise, the equation given is \(y = \frac{1}{36} x^{2}\), which inherently means there are no shifts, and the vertex remains firmly at \((0, 0)\). Here, the vertex serves as the tip of the parabola around which it opens upward, portraying perfect symmetry along the y-axis.