/*! 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 20 Find the center, vertices, and f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 20

Find the center, vertices, and foci of each ellipse and graph it. $$x^{2}+\frac{y^{2}}{16}=1$$

Short Answer

Expert verified
Center: (0,0); Vertices: (卤1,0), (0,卤4); Foci: (0,卤鈭�15); see graph above.

Step by step solution

01

Identify the Equation Form

The given equation is \(x^{2} + \frac{y^{2}}{16} = 1\). Recognize that this is the standard form of an ellipse equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\).
02

Determine the Values of \(a\) and \(b\)

Compare the given equation to the standard form. We have \(a^{2} = 1\), so \(a = 1\), and \(b^{2} = 16\), so \(b = 4\).
03

Find the Center of the Ellipse

Since the ellipse is not shifted horizontally or vertically, its center is at the origin: \( (0, 0) \).
04

Locate the Vertices

For an ellipse in standard form \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), the vertices are at \((\text{卤}a, 0)\) and \((0, \text{卤}b)\). Here, they are at \((\text{卤}1, 0)\) and \((0, \text{卤}4)\).
05

Compute the Foci

The distance from the center to each focus is given by \(c = \sqrt{b^{2} - a^{2}}\). Calculate \(c = \sqrt{16 - 1} = \sqrt{15}\). Therefore, the foci are at \((0, \text{卤} \sqrt{15})\).
06

Graph the Ellipse

Plot the center at \(0,0)\), vertices at \(1,0\), \(-1,0\), \(0,4\), and \(0,-4\), and foci at \(0, \sqrt{15}\) and \(0, -\sqrt{15}\). Draw the ellipse through the vertices.

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.

ellipse equation
An ellipse is a curved shape, like an elongated circle, traced by a point moving such that the sum of its distances from two fixed points (foci) is constant. The general form of the equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
In this equation, \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively. If \(a > b\), the longer axis is horizontal, and if \(b > a\), it's vertical. The numbers underneath \(x^2\) and \(y^2\) tell you how much to stretch the ellipse along the x-axis and y-axis.
ellipse center
The center of an ellipse is the point around which the entire ellipse is symmetric. It鈥檚 the midpoint between the foci and also the intersection of the major and minor axes.
For the equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] the ellipse is centered at the origin unless there are horizontal or vertical shifts in the equation. For the given problem, the equation is: \( x^2 + \frac{y^2}{16} = 1 \). This indicates the ellipse is centered at (0,0), as there are no added or subtracted terms within the equation which would indicate shifts.
ellipse vertices
Vertices of an ellipse are the points where the ellipse intersects the major axis. There are two pairs of vertices for an ellipse: one pair along the x-axis and one along the y-axis.
For the standard form of the ellipse's equation, \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] the vertices are located at \((\pm a, 0)\) and \((0, \pm b)\). In this exercise, we have \(a=1\) and \(b=4\), so the vertices are at
  • \((\pm 1, 0)\)
  • \((0, \pm 4)\)
.
ellipse foci
The foci (plural of focus) are fixed points used in the geometric definition of the ellipse. They lie along the major axis, inside the ellipse, and help to define its shape. The distance from the center to each focus is given by \(c = \sqrt{b^2 - a^2}\).

For our given ellipse equation, we calculated \(c\) as: \[ c = \sqrt{16 - 1} = \sqrt{15} \] The foci, therefore, are located at
  • \((0, \pm\sqrt{15})\)
. Understanding the location of foci helps to get a complete picture of the ellipse鈥檚 geometry and how it relates to its axis and center.

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

The pedals of an elliptical exercise machine travel an elliptical path as the user is exercising. If the stride length (length of the major axis) for one machine is 20 inches and the maximum vertical pedal displacement (length of the minor axis) is 9 inches, find the equation of the pedal path, assuming it is centered at the origin.

Transform the polar equation \(r=6 \sin \theta\) to an equatiol in rectangular coordinates. Then identify and graph the equation.

Analyze each equation. \(\frac{(y+2)^{2}}{16}-\frac{(x-2)^{2}}{4}=1\)

Suppose that Karla hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of \(45^{\circ}\) to the horizontal on the Moon (gravity on the Moon is one-sixth of that on Earth). (a) Find parametric equations that model the position of the ball as a function of time. (b) How long is the ball in the air? (c) Determine the horizontal distance that the ball travels. (d) When is the ball at its maximum height? Determine the maximum height of the ball. (e) Using a graphing utility, simultaneously graph the equations found in part (a).

In Problems 43 and \(44,\) parametric equations of four plane curves are given. Graph each of them, indicating the orientation. \(\begin{array}{ll}C_{1}: & x(t)=t, \quad y(t)=t^{2} ; \quad-4 \leq t \leq 4 \\\ C_{2}: & x(t)=\cos t, \quad y(t)=1-\sin ^{2} t ; \quad 0 \leq t \leq \pi \\\ C_{3}: & x(t)=e^{t}, \quad y(t)=e^{2 t} ; \quad 0 \leq t \leq \ln 4 \\\ C_{4}: & x(t)=\sqrt{t}, \quad y(t)=t ; \quad 0 \leq t \leq 16\end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

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.