Problem 39 - Algebra 2

Question

Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola with the given equation. Then find the length of the latus rectum and graph the parabola. $$ x=-\frac{1}{3} y^{2}-12 y+15 $$

Step-by-Step Solution

Answer

Vertex: (123, -18); Focus: (122.25, -18); Axis: y = -18; Directrix: x = 123.75; Opens left; Latus Rectum: 3.

Step 1: Rewrite the Original Equation

Start by rearranging the given parabola equation to identify its structure. The equation is given as \(x = -\frac{1}{3} y^{2} - 12y + 15\). We'll rewrite it in the form of \(x = a(y-k)^2 + h\). This form allows us to easily identify important features of the parabola (vertex, focus, axis of symmetry, etc.).

Step 2: Complete the Square

To transform the equation into the vertex form, we need to complete the square for the \(y\) terms:1. Factor out \(-\frac{1}{3}\) from the \(y^2\) and \(y\) terms: \[x = -\frac{1}{3}(y^2 + 36y) + 15\]2. Complete the square inside the parentheses. Take half of 36, which is 18, square it to get 324: \[x = -\frac{1}{3}((y + 18)^2 - 324) + 15\]3. Distribute \(-\frac{1}{3}\) and simplify: \[x = -\frac{1}{3}(y + 18)^2 + 108 + 15\] \[x = -\frac{1}{3}(y + 18)^2 + 123\].

Step 3: Identify the Vertex

The equation is now in vertex form: \(x = a(y-k)^2 + h\), where \(a = -\frac{1}{3}\), \(h = 123\), and \(k = -18\). Therefore, the vertex of the parabola is at \((123, -18)\).

Step 4: Determine the Focus

The focus of a parabola \(x = a(y-k)^2 + h\) is at \((h + \frac{1}{4a}, k)\). Calculate \(\frac{1}{4a}\):\[\frac{1}{4a} = \frac{1}{4(-\frac{1}{3})} = -\frac{3}{4}\]Thus, the focus is at \((123 - \frac{3}{4}, -18)\) or \((122.25, -18)\).

Step 5: Axis of Symmetry

The axis of symmetry for a horizontally oriented parabola like \(x = a(y-k)^2 + h\) is the horizontal line \(y = k\). Therefore, the axis of symmetry is \(y = -18\).

Step 6: Equation of the Directrix

The directrix of the parabola is a vertical line at \(x = h - \frac{1}{4a}\). Calculate this as follows:\[x = 123 + \frac{3}{4} = 123.75\],so the directrix is the line \(x = 123.75\).

Step 7: Direction of Opening

Since \(a = -\frac{1}{3}\) is negative, the parabola opens to the left.

Step 8: Length of the Latus Rectum

The length of the latus rectum for a parabola \(x = a(y-k)^2 + h\) is \(\left|\frac{1}{a}\right|\). Calculate:\[\left|\frac{1}{-\frac{1}{3}}\right| = 3\].Thus, the length of the latus rectum is 3.

Step 9: Plotting the Parabola

Using the information gathered:- Vertex: \((123, -18)\)- Focus: \((122.25, -18)\)- Direction: left- Axes of Symmetry: \(y = -18\)- Directrix: \(x = 123.75\)- Length of Latus Rectum: 3Graph these features to plot the parabola accurately.

Key Concepts

Vertex Form

The vertex form of a parabola equation makes it easy to identify the vertex and other essential features of the parabola. A parabola in vertex form is given by the equation \(x = a(y-k)^2 + h\), where \(a\), \(h\), and \(k\) are constants. Here:

• \(h\) and \(k\) are the coordinates of the vertex, \((h, k)\).
• \(a\) determines the direction in which the parabola opens, with a negative \(a\) indicating the parabola opens to the left for a horizontally oriented parabola.

For the given equation, when rewritten in vertex form as \(x = -\frac{1}{3}(y + 18)^2 + 123\), the vertex is at \((123, -18)\). This simple transformation allows us to quickly identify the critical points of the parabola.

Axis of Symmetry

The axis of symmetry is a line that divides the parabola into two mirror-image halves. For a parabola expressed in the form \(x = a(y-k)^2 + h\), this line is horizontal and has the equation \(y = k\).
The axis of symmetry is important because it passes through both the vertex and the focus, serving as a reference point that ensures the parabola is uniformly balanced.
In the context of our example, the axis of symmetry is the line \(y = -18\), placing it right at the vertex \((123, -18)\). This line underscores the symmetrical structure of the parabola.

Latus Rectum

The latus rectum of a parabola is a line segment perpendicular to the axis of symmetry and passes through the focus, providing key insights into the parabola's width.
The length of the latus rectum can be calculated with \(\left|\frac{1}{a}\right|\), where \(a\) is the coefficient of the \((y-k)^2\) term in the vertex form equation.
For the equation \(x = -\frac{1}{3}(y + 18)^2 + 123\), \(a = -\frac{1}{3}\), which means the length of the latus rectum is \(3\). This segment helps in detailing the parabola's geometry, indicating how stretched or compressed the graph is from the focus.

Focus and Directrix

A parabola is defined as the set of points that are equidistant from a fixed point called the focus, and a line called the directrix. These two features demonstrate the parabola's geometric properties and how it relates to the coordinate plane.

• The focus for a parabola in vertex form \(x = a(y-k)^2 + h\) is found at \((h + \frac{1}{4a}, k)\).
• The directrix is the line \(x = h - \frac{1}{4a}\).

In our specific case, the focus is \((122.25, -18)\) and the directrix is the line \(x = 123.75\). The relationship ensures every point on the parabola is equidistant to both the focus and directrix, shaping the unique contour of the curve.

Parabola Graphing

Graphing a parabola involves plotting its vertex, focus, directrix, and observing its axis of symmetry. These components work in concert to depict the parabolic shape accurately.
Begin by plotting the vertex \((123, -18)\) and then draw the axis of symmetry, creating a guide through the vertex.

• Next, mark the focus \((122.25, -18)\) and draw the directrix line \(x = 123.75\).
• Indicate the direction of opening by noting the sign of \(a\). Since \(a = -\frac{1}{3}\), the parabola opens to the left.
• Finally, use the latus rectum's length to further refine the graph. This will help to plot the width of the parabola as it branches out from the focus.

By following these steps, one can produce a complete and accurate graph of the given parabola.