/*! 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 11 Find the vertices and foci of th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the vertices and foci of the ellipse and sketch its graph. \( \dfrac{x^2}{2} + \dfrac{y^2}{4} = 1 \)

Short Answer

Expert verified
Vertices: \((0, \pm 2), (\pm \sqrt{2}, 0)\); Foci: \((0, \pm \sqrt{2})\).

Step by step solution

01

Identify and Write the Equation in Standard Form

The given equation is already in the standard form of an ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with \( a^2 = 2 \) and \( b^2 = 4 \). We deduce that the major axis is vertical since \( b^2 > a^2 \). So \( a = \sqrt{2} \) and \( b = 2 \).
02

Determine the Center

Since the ellipse is expressed in its standard form centered at the origin, the center of the ellipse is \((0, 0)\).
03

Calculate the Vertices

The vertices align with the major and minor axes. The length of the semi-major axis is \( b = 2 \), giving vertices at \((0, \pm 2)\). The length of the semi-minor axis is \( a = \sqrt{2} \), giving vertices at \((\pm \sqrt{2}, 0)\).
04

Calculate the Foci

The distance of the foci from the center is \( c \), found using the formula \( c^2 = b^2 - a^2 \). Substituting, we get \( c^2 = 4 - 2 = 2 \), so \( c = \sqrt{2} \). Hence, the foci are located at \((0, \pm \sqrt{2})\).
05

Sketch the Ellipse

Draw the ellipse centered at the origin. The vertices on the vertical axis are at \((0, \pm 2)\) and on the horizontal axis at \((\pm \sqrt{2}, 0)\). The foci are points on the vertical axis at \((0, \pm \sqrt{2})\). Ensure the ellipse is elongated vertically.

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.

Standard Form of an Ellipse
An ellipse is a closed curve and a type of conic section that looks like an elongated circle. Its equation in standard form is structured as: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This formula allows us to understand and determine the geometric properties of the ellipse, including its orientation, center, and axes lengths. In this equation, \(a\) and \(b\) refer to the lengths of the semi-major and semi-minor axes, respectively. The larger of the two denominators \(a^2\) or \(b^2\) indicates whether the major axis is horizontal or vertical, which is crucial for sketching the ellipse accurately. Understanding this standard form helps in easily identifying these critical features of the ellipse.
Vertices of an Ellipse
The vertices of an ellipse are the points where the ellipse intersects its axes. They play a vital role in defining the shape and size of the ellipse. For a vertically oriented ellipse, the vertices on the major axis are found at \((0, \pm b)\), while for a horizontally oriented ellipse, they would be at \((\pm a, 0)\). In the given example, since \(b^2 > a^2\), the major axis is vertical, and therefore, the vertices are located at \((0, \pm 2)\), aligning with the semi-major axis. This understanding is essential for drawing the shape of the ellipse and determining its overall dimensions.
Foci of an Ellipse
The foci (singular: focus) of an ellipse are two special points located along the major axis. They have an intrinsic geometric property where the sum of the distances from any point on the ellipse to the two foci is constant. The foci help in generating the ellipse and are essential in applications involving orbits and optics. To find them, use the formula \( c^2 = b^2 - a^2 \) to calculate the distance \(c\) from the center to each focus. From the example equation, we found \(c = \sqrt{2}\), meaning the foci are at \((0, \pm \sqrt{2})\) since the ellipse is vertically oriented. This location of the foci influences the shape and extent of the ellipse.
Semi-Major Axis
The semi-major axis of an ellipse is the longest radius, extending from the center to a vertex along the major axis. It is represented in the standard form equation as either \(a\) or \(b\), depending on the ellipse's orientation. In the given ellipse, it’s identified by \(b\) because \(b^2 > a^2\). Thus, \(b = 2\) represents the semi-major axis. The length of this axis determines the height of the ellipse when vertically shaped and the width when horizontally oriented. Understanding the semi-major axis helps in determining the focal length and in calculating the eccentricity, which describes how much the ellipse deviates from being a circle.
Semi-Minor Axis
Conversely, the semi-minor axis of an ellipse is the shortest radius, stretching from the center to the ellipse edge along the minor axis. It is a crucial dimension for understanding the extent of the ellipse perpendicularly to the major axis. In this exercise, the semi-minor axis is denoted by \(a = \sqrt{2}\), representing the half-length of the smaller width of our vertically elongated ellipse. The interplay of the semi-major and semi-minor axes defines the overall shape and configuration of the ellipse, allowing for clear distinction from a circle, where both axes would be equal.

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

Describe the motion of a particle with position \( (x, y) \) as \( t \) varies in a given interval. \( x = 5 + 2\cos \pi t \), \( \; y = 3 + 2 \sin \pi t \), \( \; 1 \leqslant t \leqslant 2 \)

The LORAN (LOng RAnge Navigation) radio navigation system was widely used until the 1990s when it was superseded by the GPS system. In the LORAN system, two radio stations located at \( A \) and \( B \) transmit simultaneous signals to a ship or an aircraft located at \( P \). The onboard computer converts the time difference in receiving these signals into a distance difference \( | PA | - | PB | \), and this, according to the definition of a hyperbola, locates the ship or aircraft on one branch of a hyperbola (see the figure). Suppose that station \( B \) is located \( 400\;mi \) due east of station \( A \) on a coastline. A ship received the signal from \( B \) \( 1200 \) microseconds \( (\mu s) \) before it received the signal from \( A \). (a) Assuming that radio signals travel at a speed of \( 980\; ft/\mu s \), find an equation of the hyperbola on which the ship lies. (b) If the ship is due north of \( B \). how far off the coastline is the ship?

Find an equation for the conic that satisfies the given conditions. Ellipse, center \( (-1, 4) \), vertex \( (-1, 0) \), focus \( (-1, 6) \)

The orbit of Mars around the sun is an ellipse with eccentricity 0.093 and semi major axis \(2.28 \times 10^{8} \mathrm{km} .\) Find a polar equation for the orbit.

(a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as parameter increases. \( x = \dfrac{1}{2}\cos\theta \), \( \quad y = 2\sin\theta \), \( \quad 0 \leqslant \theta \leqslant \pi \)

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.