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Chapter 12: Problem 3

The graph of the equation \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) with \(a>b>0\) is an ellipse with vertices (_, _) and (_, _) and foci \((0, \pm c),\) where \(c=\) __________ \((0, \pm c),\) where \(c=\) is an ellipse with vertices (_, _) and (_, _) and foci (_,_) and (_, _)

Short Answer

Expert verified
Vertices: (0, -a) and (0, a); Foci: (0, -c) and (0, c), where c = \(\sqrt{a^2 - b^2}\).

Step by step solution

01

Identify the Ellipse

The given equation is of the standard form for an ellipse: \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\), where \(a > b > 0\). This indicates a vertically oriented ellipse.
02

Determine the Vertices

For a vertically oriented ellipse, the vertices are located at \((0, \pm a)\). This comes from the larger value being under \(y\), signifying vertical orientation.
03

Calculate the Foci

The foci of the ellipse are found using the relation \(c = \sqrt{a^2 - b^2}\). The foci are located at \((0, \pm c)\).
04

Compile the Results

From the calculations, the vertices of the ellipse are \((0, -a)\) and \((0, a)\). The foci are \((0, -c)\) and \((0, c)\), where \(c = \sqrt{a^2 - b^2}\).

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Key Concepts

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

Vertices of an Ellipse
In the world of ellipses, vertices hold crucial importance as they define the extent of the ellipse along its major axis. The major axis of an ellipse is the longest diameter that determines its primary direction. When dealing with a vertically oriented ellipse, like in our original exercise, the vertices are positioned along the vertical line, at
  • \((0, -a)\)
  • \((0, a)\)
This implies that the ellipse stretches from the point \((0, -a)\) to \((0, a)\) vertically. The essential observation here is that the value of \(a\), which is greater than \(b\), influences this vertical spread. Each vertex is a center of elliptical arc, one at each end of the major axis.
Foci of an Ellipse
Foci are two distinctive points inside an ellipse that help in describing its specific shape. In our example, the foci are placed along the major axis, and they determine how "stretched" or "elongated" the ellipse appears. Calculated using the formula
  • \(c = \sqrt{a^2 - b^2}\)
we can understand that \(c\) gives the distance from the center to each focus. This ellipse, being vertically oriented, has its foci at
  • \((0, -c)\)
  • \((0, c)\)
Understanding the position of the foci helps one gauge the eccentricity, or the degree of "ovality," of the ellipse.
Ellipse Equation
An ellipse's equation is fundamental in defining its geometry. The given equation of our ellipse is
  • \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)
This formulation indicates several things:
- The ellipse is centered at the origin point (0,0).
- It is vertically oriented, as the coefficient of
  • \(y^2\)
is greater than that of
  • \(x^2\).
- The lengths of semi-major and semi-minor axes are determined by \(a\) and \(b\) respectively. This equation anchors the ellipse to the coordinate grid, detailing its proportional dimensions.
Standard Form of Ellipse
The standard form of an ellipse is a neatly organized equation that lays out its key characteristics. Our standard form, repeatedly referenced, is:
  • \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)
In this equation, \(a\) represents the semi-major axis (length from center to vertex), while \(b\) represents the semi-minor axis. For vertically oriented ellipses, the larger value under
  • \(y^2\)
confirms this orientation. By analyzing this standard form, one can instantly see whether the ellipse is elongated horizontally or vertically. Displaying an ellipse in standard form simplifies the process of identifying its properties, easing calculations, and enhancing understanding.

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Most popular questions from this chapter

A polar equation of a conic is given. (a) Find the eccentricity and the directrix of the conic. (b) If this conic is rotated about the origin through the given angle , write the resulting equation. (c) Draw graphs of the original conic and the rotated conic on the same screen. $$ r=\frac{2}{1+\sin \theta} ; \quad \theta=-\frac{\pi}{4} $$

(a) For the hyperbola $$ \frac{x^{2}}{9}-\frac{y^{2}}{16}=1 $$ determine the values of \(a, b,\) and \(c,\) and find the coordinates of the foci \(F_{1}\) and \(F_{2}\) . (b) Show that the point \(P\left(5, \frac{16}{3}\right)\) lies on this hyperbola. (c) Find \(d\left(P, F_{1}\right)\) and \(d\left(P, F_{2}\right) .\) (d) Verify that the difference between \(d\left(P, F_{1}\right)\) and \(d\left(P, F_{2}\right)\) is \(2 a .\)

\(29-32\) . (a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$ 9 x^{2}-6 x y+y^{2}+6 x-2 y=0 $$

Graph the conics \(r=e /(1-e \cos \theta)\) with \(e=0.4,0.6,0.8\) and 1.0 on a common screen. How does the value of \(e\) affect the shape of the curve?

\(23-34\) Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$ x^{2}-4 y^{2}-2 x+16 y=20 $$
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