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Chapter 8: Problem 25

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. $$ 5 x^{2}-50 x-4 y+113=0 $$

Short Answer

Expert verified
The vertex is \((5, -3)\), focus is \((5, -\frac{14}{5})\), and directrix is \(y = -\frac{16}{5}\).

Step by step solution

01

Rearrange the Equation

Start by rearranging the given equation to separate the quadratic term from the linear terms: \[5x^2 - 50x = 4y - 113\]
02

Factor and Complete the Square

Factor out the coefficient of \(x^2\), which is 5, from the quadratic terms on the left side: \[5(x^2 - 10x) = 4y - 113\] Next, complete the square for the quadratic expression inside the parentheses. Take half of the coefficient of \(x\), square it, and add it inside the parentheses: \[x^2 - 10x \to (x - 5)^2 - 25\] This changes the equation to: \[5((x-5)^2 - 25) = 4y - 113\] Distribute the factor and simplify: \[5(x-5)^2 - 125 = 4y - 113\]
03

Isolate the Quadratic Term

Isolate the quadratic term by moving the constant to the other side of the equation: \[5(x-5)^2 = 4y - 113 + 125\] Simplify the right side: \[5(x-5)^2 = 4y + 12\]
04

Convert to Vertex Form

Divide the entire equation by 4 to get the equation closer to the standard form of a parabola: \[\frac{5}{4}(x-5)^2 = y + 3\] Convert it to make \(y\) the subject: \[y = \frac{5}{4}(x-5)^2 - 3\] This equation is now in the vertex form \(y = a(x-h)^2 + k\).
05

Identify the Vertex

From the vertex form \(y = \frac{5}{4}(x-5)^2 - 3\), identify the coordinates of the vertex: The vertex \((V)\) is at \((h, k) = (5, -3)\).
06

Determine the Focus and Directrix

The parabola is vertical, so the formula for the focus and directrix is based on the vertex form with \(a = \frac{1}{4p}\): Given \(a = \frac{5}{4}\), solve for \(p\): \[\frac{5}{4} = \frac{1}{4p} \Rightarrow p = \frac{1}{5}\] The focus \((F)\) of a vertical parabola is \((h, k + p)\), giving us: \((5, -3 + \frac{1}{5}) = (5, -\frac{14}{5})\) The directrix \((d)\) is \(y = k - p\), so: \(y = -3 - \frac{1}{5} = -\frac{16}{5}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex Form
The vertex form of a parabola is a special way to write the equation that makes it very easy to identify the vertex of the parabola. The general formula to express a parabola in vertex form is:
  • \[ y = a(x - h)^2 + k \]
In this formula, \((h, k)\) represents the vertex of the parabola, and \(a\) shows the direction and "width" of the parabola:
  • If \(a > 0\), the parabola opens upwards.
  • If \(a < 0\), it opens downwards.
The vertex form is handy because it reveals the vertex directly, without additional calculations. For example, in the equation \(y = \frac{5}{4}(x-5)^2 - 3\), the vertex is at \((h, k) = (5, -3)\). A simple switch from standard to vertex form often involves completing the square.
Focus of Parabola
The focus of a parabola is a special point that helps define and understand the shape and orientation of the parabola. Every parabola is defined as the set of all points equidistant from the focus and a line called the directrix:
  • The focus is always located inside the "cup" of the parabola.
  • In a vertical parabola given by the equation \(y = a(x-h)^2 + k\), the focus is located at \((h, k + p)\) where \(p\) is calculated as follows: \[a = \frac{1}{4p} \]Thus, \(p = \frac{1}{4a}\).
For example, in our case with the equation \(y = \frac{5}{4}(x-5)^2 - 3\), \(p\) is calculated to be \(\frac{1}{5}\), meaning the focus is at \((5, -\frac{14}{5})\). Knowing the focus is crucial when drawing or analyzing the graph of the parabola.
Completing the Square
Completing the square is a mathematical technique used to convert a quadratic equation into vertex form. It involves manipulating a quadratic expression to make it a perfect square trinomial:
  • First, factor out any coefficient of \(x^2\) from the quadratic terms.
  • Take half of the \(x\)-term's coefficient, square it, and add it inside the parentheses.
  • Adjust the equation by adding and subtracting this squared term.
The technique allows us to rewrite the quadratic expression as a perfect square, leading to an easily recognizable vertex form. For example, starting from \(5(x^2 - 10x)\), we complete the square to achieve \((x-5)^2 - 25\). This crucial step simplifies identifying key features of the parabola, like the vertex as we convert it to \((x-5)^2\).
Directrix of Parabola
The directrix is a line that, along with the focus, describes the geometric properties of a parabola. It is always found on the side opposite the focus:
  • The directrix is parallel to the axis of symmetry of the parabola.
  • For a vertical parabola, the equation for the directrix is \(y = k - p\), where \(k\) is the \(y\)-coordinate of the vertex, and \(p\) is the distance from the vertex to the focus.
Finding the directrix in our example with vertex \((5, -3)\) and \(p = \frac{1}{5}\) is straightforward: the directrix is \(y = -\frac{16}{5}\). Understanding the directrix helps in exploring the parabola's symmetry and structure, making it an essential part of the parabola's definition.

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Most popular questions from this chapter

Express each equation in polar form with \(r\) as a function of \(\theta\). $$ 16 x^{2}+24 x y+9 y^{2}=4 $$

Graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r(3-4 \sin \theta)=9 $$

For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r=\frac{5}{2+\cos \theta} $$

Express each equation in polar form with \(r\) as a function of \(\theta\). $$ 2 x y+y=1 $$

Find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(y=2 ; e=2.5\)
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