/*! 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 Classify the graph of the equati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$9 x^{2}+4 y^{2}-18 x+16 y-119=0$$

Short Answer

Expert verified
The graph of the given equation is an ellipse.

Step by step solution

01

Group x and y Terms

First, arrange the equation by grouping x terms and y terms together: \(9x^{2}-18x+4y^{2}+16y-119=0\).
02

Complete the Square for x and y

To complete the square, take the coefficients of the x and y terms, divide by two and square it. For x terms, \((-18/2(9))^2=1\) and for y terms, \((16/2(4))^2=4\). Adding these into the equation gives us \(9(x^{2}-2x+1)+4(y^{2}+4y+4)=119+9+16\).
03

Simplify the Equation

This simplifies to: \(9(x-1)^{2}+4(y+2)^{2}=144\). Divide entire equation by 144 to get it in standard form. This gives us \(\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{36}=1\).
04

Identify the Conic Section

The standard form \(\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{36}=1\) is an equation of ellipse, because it is in the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).

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 91Ó°ÊÓ!

Key Concepts

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

Ellipse
An ellipse is one of the conic sections that represents an elongated circle, characterized by its distinct oval shape. It is defined mathematically as the set of all points such that the sum of the distances from two fixed points, called foci, is constant. Imagine stretching a circle; that's essentially how an ellipse forms.

Ellipses have two main axes: the major axis, which is the longest diameter running through the center and both foci, and the minor axis, which is the shortest diameter perpendicular to the major axis. Understanding an ellipse's axes helps in visualizing its shape and orientation.

Some practical examples of ellipses include the orbits of planets, which due to gravitational forces, often exhibit slight elliptical paths rather than perfect circles.
Completing the Square
Completing the square is an algebraic technique used to transform a quadratic equation into a perfect square trinomial. This method is particularly useful when dealing with conic sections, like ellipses, to help simplify and solve their equations.

The process involves taking the coefficient of the linear term (like \(x\) or \(y\) terms), dividing it by two, and then squaring the result. This squared value is then used to re-write the quadratic part of the equation as a perfect square.

For example, if you have \(9x^2 - 18x\), you'd focus on the \(-18x\). Half of \(-18/9\) is 1, and squaring it gives 1. Add this inside the equation to get a perfect square, so \((x-1)^2\).

Completing the square thus facilitates rewriting complex equations into simpler forms, making them easier to solve and analyze.
Equation of an Ellipse
An ellipse's equation in standard form is essential for identifying its properties and understanding its orientation. The standard form of an ellipse with a horizontal major axis is expressed as \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). Conversely, if the major axis is vertical, the form is \((x-h)^2/b^2 + (y-k)^2/a^2 = 1\).

In these equations, \(h, k\) represent the center of the ellipse, while \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. When \(a > b\), the major axis is horizontal, and when \(b > a\), it is vertical.

Understanding the equation of an ellipse allows us to determine key features like its center, foci, and axes. These insights are crucial for applications in physics and astronomy, where ellipses frequently model real-world phenomena.

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

Use a graphing utility to graph the polar equation. Identify the graph. $$r=\frac{4}{3-\cos \theta}$$

Consider the equation \(r=3 \sin k \theta\). (a) Use a graphing utility to graph the equation for \(k=1.5 .\) Find the interval for \(\theta\) over which the graph is traced only once. (b) Use the graphing utility to graph the equation for \(k=2.5 .\) Find the interval for \(\theta\) over which the graph is traced only once. (c) Is it possible to find an interval for \(\theta\) over which the graph is traced only once for any rational number \(k ?\) Explain.

Convert the polar equation to rectangular form. $$r^{2}=\cos 2 \theta$$

A satellite in a 100 -mile-high circular orbit around Earth has a velocity of approximately 17,500 miles per hour. If this velocity is multiplied by \(\sqrt{2},\) then the satellite will have the minimum velocity necessary to escape Earth's gravity and will follow a parabolic path with the center of Earth as the focus (see figure). (a) Find a polar equation of the parabolic path of the satellite (assume the radius of Earth is 4000 miles). (b) Use a graphing utility to graph the equation you found in part (a). (c) Find the distance between the surface of the Earth and the satellite when \(\theta=30^{\circ}\). (d) Find the distance between the surface of Earth and the satellite when \(\theta=60^{\circ}\).

Equation Show that the polar equation of the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad \text { is } \quad r^{2}=\frac{-b^{2}}{1-e^{2} \cos ^{2} \theta}$$.

See all solutions

Recommended explanations on Math Textbooks

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.