Problem 25 - College Algebra with Corequisite Support

Question

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. \(5 x^{2}-50 x-4 y+113=0\)

Step-by-Step Solution

Answer

The vertex is \((5, -3)\), the focus is \((5, -\frac{43}{16})\), and the directrix is \(y = -\frac{53}{16}\).

Step 1: Rearrange the Equation

Start with the given equation: \[ 5x^2 - 50x - 4y + 113 = 0 \] Rearrange the terms to isolate the quadratic terms related to \(x\): \[ 5x^2 - 50x = 4y - 113 \]

Step 2: Complete the Square

Factor out the coefficient of \(x^2\) from the terms on the left side: \[ 5(x^2 - 10x) = 4y - 113 \] Complete the square inside the parentheses. Take half of the coefficient of \(x\), square it, and add and subtract that term inside the parentheses: \[ 5(x^2 - 10x + 25 - 25) = 4y - 113 \] Simplify this to: \[ 5((x-5)^2 - 25) = 4y - 113 \] Distribute the 5: \[ 5(x-5)^2 - 125 = 4y - 113 \]

Step 3: Solve for y

Isolate \(y\) to obtain the equation in the form \((x-h)^2 = 4p(y-k)\): \[ 5(x-5)^2 = 4y - 113 + 125 \] \[ 5(x-5)^2 = 4y + 12 \] Divide everything by 4 to solve for \(y\): \[ \frac{5}{4}(x-5)^2 = y + 3 \] This equation is now in vertex form \((x-h)^2 = 4p(y-k)\) where \(h = 5\), \(k = -3\), and \(4p = \frac{5}{4}\).

Step 4: Identify the Vertex, Focus, and Directrix

Identify the vertex from the equation \((h, k)\), which is \((5, -3)\). Calculate \(p\) by equating \(\frac{5}{4}\) to \(4p\): \[ 4p = \frac{5}{4} \Rightarrow p = \frac{5}{16} \] The focus \((h, k + p)\) is \((5, -3 + \frac{5}{16})\). The directrix \(y = k - p\) is \(y = -3 - \frac{5}{16}\).

Key Concepts

Standard Form

When dealing with parabolas, the standard form of a quadratic equation is an essential format because it allows easy extraction of important features of the parabola, like its vertex, focus, and directrix. The standard form of a parabola oriented vertically is \[ y = a(x-h)^2 + k \] where

• \( (h, k) \) is the vertex of the parabola
• \( a \) determines the direction and width of the parabola

Rewriting a parabolic equation into this form often involves completing the square, a crucial algebraic technique. This rearrangement lets us clearly identify the vertex, contributing to our understanding of the parabola's overall shape and orientation.

Vertex

The vertex of a parabola is a significant point that represents the peak (or trough) of the curve when the parabola opens upwards or downwards. By converting the equation \[ 5(x-5)^2 = 4y + 12 \] into the vertex form, we identify the vertex at \((x = 5, y = -3)\).For any parabola given in the form \((x-h)^2 = 4p(y-k)\), the vertex is easily obtained as \((h, k)\). This point provides a reliable reference for plotting the parabola accurately on the coordinate plane, defining it as the point where the parabola changes direction, whether it's opening upwards or downwards.

Focus

The focus of a parabola is a fixed point used, along with the directrix, to define the parabola. For a parabola in the form \((x-h)^2 = 4p(y-k)\), the focus is located at \((h, k + p)\). In our example, with the vertex at \((5, -3)\) and \(p = \frac{5}{16} \), the focus is calculated as \((5, -3 + \frac{5}{16}) = (5, -\frac{43}{16})\). This point is crucial for understanding the parabola's focal property. Any point on the parabola maintains an equal distance to the focus and the directrix, reinforcing the parabola's symmetrical properties and geometric shape.

Directrix

The directrix of a parabola is a fixed line that, together with the focus, defines the parabola. Unlike the focus, which is a point, the directrix is a straight line. For a vertical parabola described by \((x-h)^2 = 4p(y-k)\), the directrix can be determined by the equation \(y = k - p\).In the given exercise, with \(k = -3\) and \(p = \frac{5}{16}\), the directrix is located at \(y = -3 - \frac{5}{16} = -\frac{53}{16}\). This line and the focus work together to maintain the parabola's definition. Each point on the parabola is equidistant to the directrix and the focus, showing its reflective symmetry. Understanding the directrix helps in visualizing how the parabola will appear against the Cartesian plane.