/*! 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 45 In Exercises 29-52, identify the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 45

In Exercises 29-52, identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(3x^2+y^2+18x-2y-8=0\)

Short Answer

Expert verified
The given equation represents an ellipse with a center at (-3,1), vertices at (-3,4) and (-3,-2), foci at (-3, 1+sqrt(6)) and (-3, 1-sqrt(6)), and eccentricity of sqrt(6)/3.

Step by step solution

01

Identify the Conic Shape

From the given equation \(3x^2+y^2+18x-2y-8=0\), we can see that the coefficients of \(x^2\) and \(y^2\) are unequal, which means the equation represents an ellipse.
02

Rewrite in Standard form

Rearrange the given equation in the form \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\) where \((h,k)\) are the coordinates of center. This can be achieved by completing the squares:\n\nGroup the \(x\)s and then the \(y\)s, move the constant to the right side: \(3(x^2+6x)+1(y^2-2y)=-8 \Rightarrow 3[(x+3)^{2}-9]+(y-1)^{2}=9\) => \(3(x+3)^{2}+(y-1)^{2}=9+9\) => \((x+3)^2/3 + (y-1)^2/9 =1 \).
03

Identify the Center

Looking at the equation \((x+3)^2/3 + (y-1)^2/9 =1 \), the ellipse's center, \((h,k)\), is the point that makes both \(x-h\) and \(y-k\) equal to zero. From this equation, \(h=-3\) and \(k=1\), so the center is at point \((-3,1).\)
04

Identify the Vertices

The vertices are situated a distance of \(a\) (semi-major axis) away from the center in both directions along the major axis. Since the denominator under \(y-1\)^2 is larger, the major axis is vertical. Starting from the center \((-3,1)\), move \(a=3\) units up and down to obtain the vertices. The vertices are \((-3,1+3)\) and \((-3,1-3)\) or points \((-3,4)\) and \((-3,-2)\)
05

Calculate the Length of Semi-minor Axis

The semi-minor axis, represented as \(b\), is the square root of the denominator under \(x+3\)^2 in the standard form of equation. So, \(b= \sqrt{3}\)
06

Calculate the Foci

The foci can be found using the relationship \(c=\sqrt{a^{2}-b^{2}}\), where \(c\) is the distance from either focus to the center. Substituting \(a=3\) and \(b= \sqrt{3}\), \(c=\sqrt{9-3}= \sqrt{6}\). The foci would be found on the major axis \(\sqrt{6}\) units away from the center. This gives coordinates \((-3, 1+\sqrt{6})\) and \((-3, 1-\sqrt{6})\)
07

Find Eccentricity of the Ellipse

The eccentricity of an ellipse, denoted as \(e\), is given by the formula \(e=c/a = \sqrt{6}/3 \
08

Sketch its graph

Plot the center, vertices, and foci on a graph and draw an ellipse through those points. The points \((-3, -2)\), \((-3, 4)\), \((-3, 1+\sqrt{6})\), and \((-3, 1-\sqrt{6})\) should all lie on the ellipse.

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!

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

In Exercises 13-18, test for symmetry with respect to \(\theta = \pi/2\), the polar axis, and the pole. \(r = \dfrac{2}{1\ +\ \sin\ \theta}\)

In Exercises 39-54, find a polar equation of the conic with its focus at the pole. \(\textit{Conic}\) Parabola \(\textit{Eccentricity}\) \(e=1\) \(\textit{Directrix}\) \(y=-4\)

Sketch the graph of \(r\ =\ 6\ \cos\ \theta\) over each interval.Describe the part of the graph obtained in each case. (a) \(0 \leq\ \theta\ \leq\ \dfrac{\pi}{2} \quad \quad\) (b) \(\dfrac{\pi}{2} \leq\ \theta\ \leq\ \pi\) (c) \(-\dfrac{\pi}{2} \leq\ \theta\ \dfrac{\pi}{2}\ \quad \quad\) (d) \(\dfrac{\pi}{4} \leq\ \theta\ \leq\ \dfrac{3\pi}{4}\)

In Exercises 29-34, use a graphing utility to graph the polar equation. Identify the graph. \(r=\dfrac{14}{14+17 \sin\ \theta}\)

In Exercises 59-64, use a graphing utility to graph the polar equation. Find an interval for \(\theta\) for which the graph is traced only once. \(r=3\ -\ 8\ \cos\ \theta\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.