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Chapter 7: Problem 51

Write an equation for each conic. Each parabola has vertex at the origin, and each ellipse or hyperbola is centered at the origin. $$\text { Focus }(3,0) ; e=\frac{1}{2}$$

Short Answer

Expert verified
The equation of the ellipse is \( \frac{x^2}{36} + \frac{y^2}{27} = 1 \).

Step by step solution

01

Determine the Conic Type with Given Data

The conic with the given information (a focus and eccentricity) can either be an ellipse or a hyperbola. Since the eccentricity \( e = \frac{1}{2} \) is less than 1, it is an ellipse.
02

Use the Definition of Eccentricity for an Ellipse

The eccentricity \( e \) of an ellipse is given by \( e = \frac{c}{a} \), where \( c \) is the distance from the center to a focus, and \( a \) is the semimajor axis. We know \( c = 3 \) from the focus \((3,0)\), and \( e = \frac{1}{2} \). Solve for \( a \): \[ \frac{3}{a} = \frac{1}{2} \] Clearing the fraction gives: \[ 3 = \frac{a}{2} \] Multiplying both sides by 2, we get \( a = 6 \).
03

Find the Length of the Minor Axis

For ellipses, the relationship between the axes is \( a^2 = b^2 + c^2 \). From Step 2, we know \( a = 6 \) and \( c = 3 \). Substituting into the equation gives: \[ 6^2 = b^2 + 3^2 \] \[ 36 = b^2 + 9 \] Solving for \( b^2 \), subtract 9 from both sides: \[ b^2 = 27 \] Thus, \( b = \sqrt{27} = 3\sqrt{3} \).
04

Write the Equation of the Ellipse

The standard form equation of an ellipse centered at the origin with the major axis along the x-axis is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] From previous steps, we have \( a = 6 \) and \( b^2 = 27 \). Substituting these values, the equation becomes: \[ \frac{x^2}{36} + \frac{y^2}{27} = 1 \]

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Key Concepts

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

Ellipse Equation
An ellipse is a type of conic section that can be understood as a flattened circle. To find the equation of an ellipse with its center at the origin, we use the standard form: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Here, \(a\) represents the semi-major axis, and \(b\) is the semi-minor axis. These are essentially the "radii" of the ellipse in the horizontal and vertical directions, respectively. If an ellipse is aligned along the x-axis, then \(a^2\) corresponds to the denominator of the \(x^2\) term and \(b^2\) to the \(y^2\) term, and vice versa if it is aligned along the y-axis.
  • Given a focus (3,0) and eccentricity \(e = \frac{1}{2}\), the ellipse has \(c = 3\).
  • With \(e = \frac{c}{a}\), solving for \(a\) confirms it as 6.
  • The relationship \(a^2 = b^2 + c^2\) helps find \(b^2 = 27\), leading to the ellipse equation: \(\frac{x^2}{36} + \frac{y^2}{27} = 1 \).
Eccentricity
Eccentricity is a measure that helps us understand how much a conic section deviates from being a circle. For different conics, eccentricities vary:
  • Circles have an eccentricity of zero.
  • Ellipses have eccentricities between 0 and 1.
  • Parabolas have an eccentricity of exactly 1.
  • Hyperbolas have eccentricities greater than 1.
For an ellipse, the eccentricity is calculated as \(e = \frac{c}{a}\), where:
  • \(c\) is the distance from the center to any of the foci.
  • \(a\) is the length of the semi-major axis.
  • An eccentricity of \(\frac{1}{2}\) indicates the ellipse is not too stretched, maintaining a certain roundness.
Foci of Conics
The foci (plural for focus) of a conic section are crucial in defining its shape and other properties. In an ellipse, which resembles an elongated circle:
  • The foci are two fixed points located symmetrically along the major axis.
  • The sum of the distances from any point on the ellipse to the two foci remains constant.
  • This unique property of ellipses ensures their shape, derived using the constant: \(2a\).
For the given problem, where one focus is \((3,0)\), these foci guide how the ellipse stretches along its main direction:
  • If the major axis is horizontal, the foci lie along the x-axis.
  • If the major axis is vertical, they are positioned along the y-axis.
  • The distance from the center to one focus is \(c = 3\), critical for calculating the ellipse equation.
Understanding foci not only enhances your grasp of the basic properties of ellipses but also aids in solving related geometric problems.

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Most popular questions from this chapter

Solve each problem. Structure of an Atom In \(1911,\) Ernest Rutherford discovered the basic structure of the atom by "shooting" positively charged alpha particles with a speed of \(10^{7}\) meters per second at a piece of gold foil \(6 \times 10^{-7}\) meter thick. Only a small percentage of the alpha particles struck a gold nucleus head-on and were deflected directly back toward their source. The rest of the particles often followed a hyperbolic trajectory because they were repelled by positively charged gold nuclei. Thus, Rutherford proposed that the atom was composed of mostly empty space and a small, dense nucleus. The figure shows an alpha particle \(A\) initially approaching a gold nucleus \(N\) and being deflected at an angle \(\theta=90^{\circ}\) \(N\) is located at a focus of the hyperbola, and the trajectory of \(A\) passes through a vertex of the hyperbola. (a) Determine the equation of the trajectory of the alpha particle if \(d=5 \times 10^{-14}\) meter. (b) Approximate the minimum distance between the centers of the alpha particle and the gold nucleus. (GRAPH CAN'T COPY)

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$2 x=y^{2}-4 y+6$$

Find an equation for each hyperbola. Center \((9,-7) \text { ; focus ( } 9,3 \text { ); vertex ( } 9,-1)\)

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$x=y^{2}+2 y-8$$

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$y=(x+3)^{2}-4$$
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