Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. \(3 x^{2}+30 x-4 y+95=0\)

Step-by-Step Solution

Answer

Vertex: (-5, 5), Focus: (-5, 16/3), Directrix: y = 14/3; parabola opens upwards.

Step 1: Rewrite the equation in a standard parabola form

Given the equation is \(3x^{2} + 30x - 4y + 95 = 0\). First, we need to rewrite it in the form \((x - h)^2 = 4p(y - k)\) which is the standard form of a parabola that opens vertically. To do this, first isolate the terms involving \(x\): \[3x^2 + 30x = 4y - 95\] Then factor out the 3 from the \(x\) terms:\[3(x^2 + 10x) = 4y - 95\]

Step 2: Complete the square

Complete the square for the \(x^2 + 10x\) expression inside the parenthesis. Take half of 10 (which is 5), square it to get 25, and then add and subtract it inside the parenthesis:\[x^2 + 10x + 25 - 25 = 4y - 95\] Then rewrite:\[3((x+5)^2 - 25) = 4y - 95\] Next, distribute the 3 and simplify:\[3(x+5)^2 - 75 = 4y - 95\]

Step 3: Rearrange to the parabola form

Move the constants to the other side of the equation:\[3(x+5)^2 = 4y - 20\]Now divide the entire equation by 4 to match the form:\[(x+5)^2 = \frac{4}{3}(y - 5)\]Thus, the equation is now in standard form where \((h,k) = (-5,5)\) and \(4p = \frac{4}{3}\), making \(p = \frac{1}{3}\).

Step 4: Identify focus and directrix

The vertex of the parabola is \((-5, 5)\). Since \(p = \frac{1}{3}\), the focus, \((h, k + p)\), is at \((-5, 5 + \frac{1}{3}) = (-5, \frac{16}{3})\). The directrix is \(y = k - p\) which means the equation for the directrix is:\[y = 5 - \frac{1}{3} = \frac{14}{3}\].

Step 5: Graph the parabola

Plot the vertex at \((-5, 5)\), the focus at \((-5, \frac{16}{3})\), and draw the directrix line at \(y = \frac{14}{3}\). Sketch the parabola with these points, noting it opens upwards as \(x\) terms are completed in the square and \((y - 5)\) is positive.

Key Concepts

Focus of a parabola

The focus of a parabola is a significant point that helps define the shape and direction of the parabola. It lies inside the curve, making it closer to the vertex of the parabola than any other point. To find the focus, one should identify the vertex of the parabola from the equation's standard form.

• The vertex form of a parabola is \((x - h)^2 = 4p(y - k)\) where \(h, k\) is the vertex and \(p\) determines the distance from the vertex to the focus.
• The focus is then located at \((h, k + p)\) for vertically opening parabolas.

In this problem, after writing the equation in standard form, the focus is positioned at \((-5, \,5 + \frac{1}{3}) = (-5, \frac{16}{3})\). The focus directs how the parabola opens, and any point on the parabola is equidistant to the focus and the directrix.

Directrix of a parabola

The directrix of a parabola is a line that helps to define the parabola's orientation and width. It is found opposite the focus on the other side of the vertex and is crucial in maintaining the equidistant property of points on the parabola.

• The vertex form equation \( (x - h)^2 = 4p(y - k) \) is used to identify the directrix.
• For a vertically oriented parabola, the directrix is the line \((y = k - p)\).

In this exercise, after setting the equation in standard form, the directrix is determined as \(y = \frac{14}{3}\). The focus and directrix are reflection partners, mirroring each other over the vertex, ensuring all parabola points maintain equal distances toward both.

Completing the square

Completing the square is a technique used in algebra to manipulate a quadratic equation into a format that makes it easier to identify key properties, such as the vertex in the case of a parabola.

• Begin by isolating the quadratic and linear terms on one side of the equation.
• Divide the coefficient of the linear term by 2 and square it, adding and subtracting this square inside the equation for continuity.
• This transforms the quadratic expression into a perfect square trinomial.

In this example, we started with \(3x^2 + 30x = 4y - 95\) and completed the square to convert it into \((x + 5)^2 \). This step is crucial for reconfiguring the quadratic equation into its standard form, allowing us to easily locate the vertex, focus, and directrix.

Standard form of a parabola

The standard form of a parabola is a rearranged version of the parabola's equation that makes it easier to analyze and plot its features, such as the vertex, focus, and directrix. For a parabola opening vertically (up or down), this standard form is given by:

• \((x - h)^2 = 4p(y - k)\)
• This provides immediate access to vital information about the parabola: \(h, k\) are the coordinates of the vertex, and \(p\) gives the directed distance between the vertex and the focus or directrix.

In this exercise, converting \(3x^2 + 30x - 4y + 95 = 0\) into the standard form \((x+5)^2 = \frac{4}{3}(y - 5)\) revealed a vertex at \((-5, 5)\) and allowed us to determine the values needed for graphing. Rewriting the equation this way simplifies finding the parabola’s orientation and its other characteristics, easing the graphing process.