Problem 8 - Precalculus: Functions and Graphs

Question

Exer. \(1-12\) : Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$(y+1)^{2}=-12(x+2)$$

Step-by-Step Solution

Answer

Vertex: \((-2, -1)\), Focus: \((-5, -1)\), Directrix: \(x = 1\).

Step 1: Identify the Equation Form

The given equation is \((y + 1)^2 = -12(x + 2)\). This equation of the parabola is in the form \((y - k)^2 = 4p(x - h)\). From the given equation, \(h = -2\) and \(k = -1\). The value \(4p = -12\), so \(p = -3\).

Step 2: Determine the Vertex of the Parabola

The vertex of a parabola in the form \((y - k)^2 = 4p(x - h)\) is \((h, k)\). For our equation, the vertex is \((-2, -1)\).

Step 3: Find the Focus of the Parabola

Since \(p = -3\) and the vertex is \((-2, -1)\), the focus is at \((h + p, k)\). Substituting the values, the focus is \((-2 - 3, -1) = (-5, -1)\).

Step 4: Locate the Directrix of the Parabola

The directrix of the parabola is a vertical line \(x = h - p\). Using \(h = -2\) and \(p = -3\), the directrix is \(x = -2 + 3 = 1\).

Step 5: Graph the Parabola

On a coordinate plane, plot the vertex at \((-2, -1)\), the focus at \((-5, -1)\), and draw the line \(x = 1\) for the directrix. Sketch the parabola opening to the left because \(p\) is negative.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key point that indicates its maximum or minimum. It is considered the "turning point" where the parabola changes direction. In a standard equation of a parabola, \[(y - k)^2 = 4p(x - h),\]the vertex is represented by the point \((h, k)\). For the given equation, \((y + 1)^2 = -12(x + 2),\)the vertex can be found by identifying \(h = -2\) and \(k = -1\). Thus, the vertex is \((-2, -1)\). This vertex is crucial as it serves as the reference point for finding both the focus and the directrix.

• Vertex in this equation form can be seen as the midpoint around which the curve is symmetric.
• If the equation has the squared term in \((y - k)^2\), the parabola opens along the x-axis, either left or right.

The vertex not only defines the center of the parabola's U-shape but also helps determine the distance to the focus and directrix, making it foundational in graphing parabolas.

Focus of a Parabola

The focus of a parabola is a special point that demonstrates its reflective property, where all lines that are parallel to the parabola's axis of symmetry and reflective off the curve will pass through this point. The position of the focus relative to the vertex determines the parabola's shape and opening direction.

• For the equation \((y - k)^2 = 4p(x - h),\)the focus is located at \((h + p, k)\).
• In our example, with \((y + 1)^2 = -12(x + 2),\)
• \(p = -3\), thus making the focus \((-5, -1)\).

The negative value of \(p\) indicates that the parabola opens to the left. The concept of a parabola's focus is central because it ties directly to applications such as satellite dishes and flashlights where the focus helps direct waves and signals. Understanding the focus empowers students to grasp why parabolas behave in specific reflective ways, emphasizing the importance of the focus in determining the parabolic curve.

Directrix of a Parabola

The directrix of a parabola is a line that serves as a reference for defining the curve's steepness and direction in tandem with the focus. It is part of the geometric definition of a parabola: the set of all points equidistant from the focus and the directrix. In the equation form \((y - k)^2 = 4p(x - h),\)the directrix is given by the vertical line \(x = h - p\).

• For our specific equation, this results in the line \(x = 1\).
• The position of the directrix helps confirm the parabola's orientation; \(x = 1\) illustrates the line opposite to where the parabola opens. Because \(p\) is negative, the parabola opens left, the directrix being on the right.

The relationship between the focus, the vertex, and the directrix solidifies the understanding of a parabola's symmetrical properties. By visualizing the directrix and focus together, one can predict how the parabola will behave when graphing or interpreting its physical applications.