Problem 43 - College Algebra

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. $$ y^{2}+2 y-12 x+61=0 $$

Step-by-Step Solution

Answer

Vertex: (5, -1). Focus: (8, -1). Directrix: x = 2.

Step 1: Rearrange the Equation

First, rearrange the given equation \( y^2 + 2y - 12x + 61 = 0 \) to express it in a more familiar form. Isolate the terms involving \(y\) on one side: \( y^2 + 2y = 12x - 61 \).

Step 2: Complete the Square

To make the \( y \) terms a perfect square trinomial, complete the square: \( y^2 + 2y \). The expression needed is \( (y+1)^2 = y^2 + 2y + 1 \). Add and subtract 1 on the left side: \( (y^2 + 2y + 1) - 1 = 12x - 61 \). Simplifying gives \( (y+1)^2 = 12x - 60 \).

Step 3: Simplify and Express in Standard Form

Modify the equation to match the standard form of a parabola \((y-k)^2 = 4p(x-h)\). Factor \(12\) on the right: \( (y+1)^2 = 12(x-5) \). Here, the standard form is compared: \( (y-k)^2 = 4p(x-h) \), with \( h = 5, k = -1, \) and \( 4p = 12 \). Thus, \( p = 3 \).

Step 4: Identify the Vertex, Focus, and Directrix

The vertex of the parabola in the form \((y-k)^2 = 4p(x-h)\) is \((h,k) = (5,-1)\). The focus, one unit \( p \) from the vertex along the direction of the parabola's axis (positive x-direction since \( 4p = 12 \) which is positive), is \((h+p, k) = (8, -1)\). The directrix is a vertical line p units in the opposite direction: \(x = h-p = 2\).

Step 5: Graph the Parabola

Draw the parabola using the vertex \((5,-1)\). Plot the focus \((8, -1)\) and draw the directrix \(x = 2\). The parabola opens to the right since the \( x \)-terms are positive \((4p = 12 > 0)\). The parabola is wider than a standard parabola since \(4p = 12\), meaning it stretches more horizontally.

Key Concepts

Completing the Square

Completing the square is a technique used to transform quadratic equations into a form that makes them easier to analyze or graph. It involves creating a perfect square trinomial from a quadratic expression. For our given equation, we started with the expression \( y^2 + 2y \). To complete the square:

• Focus on the quadratic and linear terms \( y^2 + 2y \).
• Take half of the coefficient of \( y \), which is \( 2/2 = 1 \), and square it, resulting in \( 1^2 = 1 \).
• Add and subtract this squared value within the equation to keep the balance: \( (y^2 + 2y + 1) - 1 \).

This gives us a new equation \( (y+1)^2 = 12x - 60 \), where \( (y+1)^2 \) is the perfect square trinomial. By completing the square, equations become simpler to express in standard parabola form, which is crucial for graphing.

Standard Form of a Parabola

The standard form of a parabola is crucial for understanding its geometric properties. Particularly for parabolas that open horizontally, the standard form is \( (y-k)^2 = 4p(x-h) \). This equation allows easy identification of the vertex and indicates the parabola's direction of openness. For the equation \( (y+1)^2 = 12(x-5) \):

• The standard form comparison gives \( h = 5 \) and \( k = -1 \).
• The factor \( 12 \) corresponds to \( 4p \), leading to \( 4p = 12 \), thus \( p = 3 \).

The orientation is determined by the \( x \)-term's coefficient, which is positive, indicating that our parabola opens to the right. Understanding this form facilitates analysis of the parabola's shape and position on the coordinate plane.

Graphing Parabolas

Graphing a parabola involves illustrating its shape on the coordinate plane while marking key features such as the vertex, focus, and directrix. From our equation \( (y+1)^2 = 12(x-5) \), the steps are as follows:

• Identify the vertex, which is \((h, k) = (5, -1)\).
• Determine the focus by adding \( p \) units along the parabola’s axis direction, hence \( (8, -1) \) in our case.
• Locate the directrix, a vertical line, by going \( p \) units in the opposite direction, thus \( x = 2 \).
• Since the parabola opens to the right, draw a symmetrical curve starting from the vertex and moving towards the focus, showing it is wider due to a larger \( 4p \).

Mapping these elements accurately ensures a clear representation of the parabola's structure on the graph.

Vertex of a Parabola

The vertex of a parabola is a pivotal point that indicates the parabola's turning point. In its standard form, \( (y-k)^2 = 4p(x-h) \), the vertex is denoted as \((h, k)\). For the given equation \( (y+1)^2 = 12(x-5) \):

• The vertex can be read directly as \( (5, -1) \).
• This point serves as the starting reference for plotting the parabola.
• From the vertex, one can determine the direction (rightward opening for our equation) by checking the coefficient’s sign in front of \((x-h)\).

The vertex not only guides the graphing process but also helps identify other features like the focus and directrix, making it fundamental in understanding a parabola's orientation and position on the plane.