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Chapter 9: Problem 79

Parabolic car headlights: The cross section of a typical car headlight can be modeled by an equation similar to \(25 x=16 y^{2},\) where \(x\) and \(y\) are in inches and \(x \in[0,4] .\) Use this information to graph the relation for the indicated domain.

Short Answer

Expert verified
Graph the parabola opening along the positive \(x\)-axis with vertex at \((0,0)\) and extending to \(x=4\).

Step by step solution

01

Understand the Equation

The given equation is \(25x = 16y^{2}\). It's essential to recognize that this represents a parabola, which is typically written as \(x = ay^2\) or \(y = ax^2\). In this form, \(x\) is expressed in terms of \(y^2\), indicating that the parabola opens along the \(x\)-axis.
02

Rewrite the Equation

To better analyze the equation, rewrite it in a more familiar form: \(x = \frac{16}{25}y^2\). This shows \(x\) as a function of \(y\), making it easier to plot.
03

Identify Key Features

The vertex of this parabola is at the origin \((0,0)\), and it opens along the positive \(x\)-axis since the coefficient \(\frac{16}{25}\) is positive. It will extend to \(x=4\) because of the domain \([0,4]\).
04

Solve for \(y\)

To find \(y\) in terms of \(x\), solve the equation \(x = \frac{16}{25}y^2\) for \(y\). Rearranging gives \(y^2 = \frac{25}{16}x\), and therefore, \(y = \pm\sqrt{\frac{25}{16}x}\). This will allow us to calculate corresponding \(y\) values for \(x\) values in the domain.
05

Calculate Key Points

For a few specific \(x\) values within the domain, calculate \(y\): - When \(x=0\), \(y=0\). - When \(x=4\), \(y = \pm\frac{5}{2}\). Calculate more points if needed for a smoother curve.
06

Graph the Curve

Plot the points calculated in the previous step on a graph. The \(x\)-axis will represent values from \([0,4]\), and the corresponding \(y\) values will be positive and negative square roots calculated earlier. Connect the points to form a parabola opening along the \(x\)-axis.

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Key Concepts

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

Parabolas
A parabola is one of the fundamental shapes in geometry, described as a curve where each point is equidistant from a fixed point, called the focus, and a fixed line, known as the directrix. Parabolas are commonly found in the quadratic equation form such as \[ y = ax^2 + bx + c \] or in their standard form \[ x = ay^2 + by + c \]. Parabolas have distinct characteristics:
  • They are symmetrical around a central axis, known as the axis of symmetry.
  • The highest or lowest point on the parabola, depending on its orientation, is called the vertex.
  • Parabolas can open upwards, downwards, or along the horizontal axis (like in our example) depending on their equation.
  • The orientation of the parabola greatly impacts its practical application, such as in satellite dishes or car headlights.
Understanding these features is essential, as they help in identifying the structure and behavior of parabolas in various scenarios.
Vertex of a Parabola
The vertex of a parabola is a crucial concept, representing the point where it changes direction. For parabolas described by equations such as \( ax^2 \) or \( ay^2 \), the vertex acts as a turning point. In mathematical terms, it is the minimum or maximum point of the curve.
  • For a parabola that opens vertically, the vertex is the lowest point if the parabola opens upwards, and the highest if it opens downwards.
  • In the equation \( x = ay^2 + by + c \), the vertex is at the origin \((0,0)\) if there are no horizontal or vertical shifts.
  • Determining the vertex is essential for graphing a parabola accurately.
  • When graphing, the vertex serves as a reference point to measure the parabola's stretch, orientation, and symmetry.
In the context of our problem, the vertex at \((0,0)\) indicates that the parabola starts at the origin and symmetrically extends along the positive \(x\)-axis.
Graphing Parabolas
Graphing a parabola involves plotting its path on a coordinate grid using specific characteristics such as the vertex, axis of symmetry, and key points along its curve. Here's a simple approach to graphing parabolas:
  • Identify the vertex: Start by locating the vertex, as it plays a pivotal role in shaping the parabola. For example, in \( x = ay^2 \), the vertex is the origin.
  • Calculate key points: Use the parabola's equation to find various points through which the curve passes. In the case of \( x = \frac{16}{25}y^2 \), solve for \(x\) to find bounds such as \( x = 0 \) or \( x = 4 \), and calculate corresponding \(y\) values.
  • Symmetry and Domain: Because parabolas are symmetrical, plotting points only on one side helps establish the curve's shape. Also, adhere to the given domain, in this case, \([0,4]\), to ensure accurate representation.
  • Draw the Curve: Connect the calculated points, smoothing out the transitions to depict the curve. Ensure the shape conveys the correct opening direction, whether vertical or horizontal.
By carefully following these steps, one can accurately draw a parabola, utilizing its unique characteristics and ensuring the graph reflects its mathematical equation correctly.

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Most popular questions from this chapter

The Perpendicular Distance from a Point to a Line: \(d=\left|\frac{A x_{1}+B y_{1}+C}{\sqrt{A^{2}+B^{2}}}\right|\) The perpendicular distance from a point \(\left(x_{1}, y_{1}\right)\) to a given line can be found using the formula shown, where \(A x+B y+C=0\) is the equation of the line in standard form \((A, B, \text { and } C\) are integers). Use the formula to verify that \(P(-6,2)\) and \(Q(6,4)\) are an equal distance from the line \(y=-\frac{1}{2} x+3\) .

Planetary motion: The perihelion, aphelion, and orbital period of the planets Jupiter, Saturn, Uranus, and Neptune are shown in the table. Use the information to answer or complete the following exercises. The formula $$L=2 \pi \sqrt{0.5\left(a^{2}+b^{2}\right)}$$ can be used to estimate the length of the orbital path. Recall for an cllipse, \(c^{2}=a^{2}-b^{2}\). $$\begin{array}{|l|c|c|c|} \hline \text { Planet } & \begin{array}{c} \text { Perihelion } \\ \left(10^{6} \mathrm{mi}\right) \end{array} & \begin{array}{c} \text { Aphelion } \\ \left(10^{6} \mathrm{mi}\right) \end{array} & \begin{array}{c} \text { Period } \\ (y r) \end{array} \\ \hline \text { Jupiter } & 460 & 507 & 11.9 \\ \hline \text { Saturn } & 840 & 941 & 29.5 \\ \hline \text { Uranus } & 1703 & 1866 & 84 \\ \hline \text { Neptune } & 2762 & 2824 & 164.8 \\ \hline \end{array}$$ Find the eccentricity of the planets Uranus and Neptune.

Sketch the graph of each ellipse. $$\frac{(x-2)^{2}}{25}+\frac{(y+3)^{2}}{4}=1$$

(a) graph the curves defined by the parametric equations using the specified interval and identify the graph (if possible) and (b) eliminate the parameter (Exercises \(7 \text { to } 16 \text { only })\) and write the corresponding rectangular form. $$\begin{aligned}&x=\frac{t^{3}}{10} ; t \in[-5,5]\\\&y=|t|\end{aligned}$$

Find the vertex, focus, and directrix for the parabolas defined by the equations given, then use this information to sketch a complete graph (illustrate and name these features). For Exercises 43 to 60 , also include the focal chord. $$y^{2}-2 y+8 x+9=0$$
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