/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Find an equation of the ellipse ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 53

Find an equation of the ellipse with vertices (卤5,0) and eccentricity \(e=\frac{3}{5}.\)

Short Answer

Expert verified
The equation of the ellipse with vertices (卤5,0) and eccentricity \(e=\frac{3}{5}\) is \(\frac{x^2}{25} + \frac{y^2}{16} = 1\).

Step by step solution

01

Identify the given elements

From the problem, we identify that the vertices of the ellipse are given by (卤5,0), which means the length of the semi-major axis, \(a\), is 5. The eccentricity \(e\), is given as \(\frac{3}{5}\). Since the vertices are along the x-axis, the ellipse is horizontally oriented, and thus, the \(x\) component of the equation will contain \(a\).
02

Determine the value of b

Eccentricity is calculated by the formula \(e = \sqrt{1 - \frac{b^2}{a^2}}\). We can use this formula and given values to solve for \(b\). After substituting \(a = 5\) and \(e = \frac{3}{5}\) into this formula, we get: \(\frac{3}{5} = \sqrt{1 - \frac{b^2}{5^2}}\). Squaring both sides and rearranging gives us \(b^2 = 5^2 - (5* (\frac{3}{5}))^2 = 16\), which implies \(b = 4\).
03

Write the equation of an ellipse

The elliptical equation based on the given data would be: \(\frac{(x-0)^2}{5^2} + \frac{(y-0)^2}{4^2} = 1\), which can be simplified to: \(\frac{x^2}{25} + \frac{y^2}{16} = 1\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vertices of an Ellipse
In an ellipse, the vertices are crucial points that lie at the extremities of the major axis. The ellipse given in our exercise has its vertices at (卤5,0). This indicates that the ellipse is centered at the origin with a horizontal orientation, since the vertices are positioned along the x-axis.

The distance between each vertex and the center of the ellipse gives us the length of the semi-major axis, denoted by the letter "a". The full set of vertices can always be described as ( 卤a, 0) for horizontal ellipses or (0, 卤a) for vertical ones. Knowing the position of the vertices helps determine other properties of the ellipse, such as its overall shape and orientation.
Eccentricity of an Ellipse
Eccentricity in an ellipse measures how "stretched" or "elongated" the ellipse is. Essentially, it is a numerical value between 0 and 1. An eccentricity closer to 0 indicates an ellipse that is more circular, while a value closer to 1 represents a more elongated shape.

For our specific problem, the eccentricity is given as \( e = \frac{3}{5} \). This value provides information about the relationship between the ellipse's semi-major axis and semi-minor axis while allowing calculation of one if the other is known. It plays a vital role in the ellipse's formula, indicating the "flatness" of the ellipse by the formula:
  • \( e = \sqrt{1 - \frac{b^2}{a^2}} \)
Where \( b \) represents the semi-minor axis.
Semi-major Axis
The semi-major axis of an ellipse is one of its most defining characteristics. It represents one-half of the longest diameter of the ellipse, stretching from the center to the ellipse's edge. In our example, where the vertices are at (卤5,0), the length of the semi-major axis is 5.

The semi-major axis can often be found from the vertices of the ellipse. For horizontally oriented ellipses like ours, \( a \) can be derived directly from the x-coordinates of the vertices. This axis helps define the size and stretching of the ellipse, encapsulating the maximum reach of the curve.
Semi-minor Axis
The semi-minor axis of an ellipse is perpendicular to the semi-major axis and represents one-half of the shortest diameter. In our problem, we calculated this using the formula for eccentricity.

Given \( e = \frac{3}{5} \) and \( a = 5 \), solving the equation \( e = \sqrt{1 - \frac{b^2}{a^2}} \) allows us to find \( b \). By substituting these values and solving for \( b \), we discovered that \( b = 4 \).

This axis, combined with the semi-major axis, shapes the ellipse and dictates its overall width. Understanding both the major and minor axes is key to visualizing and working with ellipses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine whether the statement is true or false. Justify your answer. The graph of \(r=10 \sin 5 \theta\) is a rose curve with five petals.

The equation \(r=\frac{e p}{1 \pm e \sin \theta}\) is the equation of an ellipse with \(e<1 .\) What happens to the lengths of both the major axis and the minor axis when the value of \(e\) remains fixed and the value of \(p\) changes? Use an example to explain your reasoning.

Use a graphing utility to graph the polar equation. Identify the graph. $$r=\frac{12}{2-\cos \theta}$$

Use a graphing utility to graph the rotated conic. $$r=\frac{4}{4+\sin (\theta-\pi / 3)}$$

Use a graphing utility to graph the polar equation. Identify the graph. $$r=\frac{-1}{1-\sin \theta}$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.