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Chapter 9: Problem 21

Find the center, vertices, and foci of the ellipse that satisfies the given equation, and sketch the ellipse. $$\frac{(x-1)^{2}}{100}+\frac{(y+1)^{2}}{36}=1$$

Short Answer

Expert verified
The center is at \((1, -1)\), the vertices are at \((-9, -1)\), \((11, -1)\), \((1, -7)\), and \((1, 5)\), and the foci are at \((-7, -1)\) and \((9, -1)\).

Step by step solution

01

Find the center

The center of the ellipse \((h, k)\) can be determined by equating the given equation with the general equation of the ellipse. Here, \(h\) corresponds to 1 and \(k\) corresponds to -1. Therefore, the center of the ellipse is at \((1, -1)\).
02

Find the vertices

The values of \(a^2\) and \(b^2\) determine the vertices of the ellipse. For this problem, \(a^2 = 100\) and \(b^2 = 36\). The square roots of these values gives \(a = 10\) and \(b = 6\). The vertices of the ellipse along the x-axis are at points \((1-a, -1)\) and \((1+a, -1)\) which are \((-9, -1)\) and \((11, -1)\). The vertices along the y-axis are at points \((1, -1-b)\) and \((1, -1+b)\) which are \((1, -7)\) and \((1, 5)\).
03

Find the foci

The distance from the center to each focus is given by \(\sqrt{|a^2-b^2|} = \sqrt{|100-36|} = 8\). Therefore, the foci are at \((1-8, -1)\) and \((1+8, -1)\), which are \((-7, -1)\) and \((9, -1)\).
04

Sketch the ellipse

Plot the center, vertices, and foci on a graph. Draw an ellipse by connecting these points. The center is at \((1, -1)\), the vertices are at \((-9, -1)\), \((11, -1)\), \((1, -7)\), and \((1, 5)\), and the foci are at \((-7, -1)\) and \((9, -1)\). The longer axis passes through the vertices \((-9, -1)\) and \((11, -1)\) while the shorter axis passes through the vertices \((1, -7)\) and \((1, 5)\).

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Key Concepts

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

Conic Sections
Conic sections are curves generated by the intersection of a plane and a double-napped cone. These curves are fundamental in mathematics and include circles, ellipses, parabolas, and hyperbolas.
Each shape has unique properties and equations that define their placements on the coordinate plane.
  • **Circle**: A special ellipse where the foci are at the same point, characterized by the equation \(x^2 + y^2 = r^2\).
  • **Ellipse**: Oval-shaped curves with two focal points, described by \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
  • **Parabola**: A curve with a single focus and directrix, with the equation \(y = ax^2 + bx + c\).
  • **Hyperbola**: Two separate curves with two foci, given by \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).
Understanding these shapes helps in various applications such as physics, engineering, and architecture.
Ellipse Center
The center of an ellipse is the midpoint between its foci. In a standard ellipse equation like \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the center is represented by \(h, k\).
For the equation given in the exercise, \(\frac{(x-1)^{2}}{100} + \frac{(y+1)^{2}}{36} = 1\), the center is at \((1, -1)\).
This center serves as the reference point for finding the other components of the ellipse.
All measurements such as vertices and foci are taken from this central location. Identifying the center helps simplify sketching and understanding the symmetry of the ellipse.
Ellipse Vertices
Vertices are the points on an ellipse that are furthest from the center in either the horizontal or vertical direction. These points lie along the main axes of the ellipse.
For an ellipse centered at \(h, k\), the vertices are determined by the lengths of the semi-major and semi-minor axes.
In the exercise, \(a^2 = 100\) and \(b^2 = 36\), giving \(a = 10\) and \(b = 6\).
  • Horizontal vertices are at \(h-a, k\) and \(h+a, k\), which results in \((-9, -1)\) and \(11, -1)\).
  • Vertical vertices are at \(h, k-b\) and \(h, k+b\), leading to \(1, -7\) and \(1, 5\).
Recognizing these points helps understand the ellipse's orientation and proportions.
Ellipse Foci
Foci are two special points inside the ellipse from which the sum of the distances to any point on the ellipse is constant.
They play a crucial role in defining the shape and position of an ellipse.
To find the foci positions, use the formula \(\sqrt{|a^2-b^2|}\), which provides the distance from the center to each focus.
In the exercise, this distance is \(8\), resulting in foci at \((-7, -1)\) and \(9, -1)\).
These points illustrate how "stretched" the ellipse is along its major axis.
  • If the foci are far apart, the ellipse is more elongated.
  • If they are closer, the ellipse is closer to a circular shape.
Understanding the role of foci aids in comprehending the elliptical geometry.

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Most popular questions from this chapter

This set of exercises will draw on the ideas presented in this section and your general math background. What are the slopes of the asymptotes of a hyperbola that satisfics an cquation of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) if \(a=b>0 ?\) At what angle do the asymptotes intersect?

The eccentricity of an ellipse is defined as \(e=\frac{c}{a}\) \(\left(=\frac{\sqrt{a^{2}-b^{2}}}{a}\right),\) where \(a, b,\) and \(c\) are as defined in this section. since \(0 < c < a,\) the value of \(e\) lies between 0 and 1 In ellipses that are long and thin, \(b\) is small compared to \(a,\) so the eccentricity is close to \(1 .\) In ellipses that are nearly circular, \(b\) is almost as large as \(a,\) so the eccentricity is close to \(0 .\) What is the eccentricity of the ellipse with equation \(\frac{x^{2}}{9}+\frac{y^{2}}{25}=1 ?\) Does this ellipse have a greater or lesser eccentricity than the ellipse with equation \(\frac{x^{2}}{16}+\frac{y^{2}}{25}=1 ?\)

In this set of exercises, you will use hyperbolas to study real-world problems. Physics Because positively-charged particles repel each other, there is a limit on how close a small, positivelycharged particle can get to the nucleus of a heavy atom. (A nucleus is positively charged.) As a result, the smaller particle follows a hyperbolic path in the neighborhood of the nucleus. If the asymptotes of the hyperbola have slopcs of \(\pm 1,\) what is the overall change in the direction of the path of the smaller particle as it first approaches the nucleus of the heavy atom and ultimately recedes from it?

Determine the equation in standard form of the hyperbola that satisfies the given conditions. Transverse axis of length \(6 ;\) one vertex at (7,8)\(;\) one focus at (7,5)

Graph the parametric equations \(x=\frac{1}{t}, y=\frac{1}{t+1}\) for \(0 \leq t \leq 10,\) indicating direction of increasing \(t .\) Will (0,0) ever be reached? Explain.
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