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Chapter 7: Problem 42

Graph each ellipse and give the location of its foci. $$\frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1$$

Short Answer

Expert verified
The ellipse's foci are approximately located at (3, -3.6458) and (3, 1.6458).

Step by step solution

01

Identify the centers

The standard form of the equation of an ellipse is \(\frac{(x-h)^{2}}{a^2}+\frac{(y-k)^{2}}{b^2}=1\), where (h, k) is the center of the ellipse. In this case, our equation is \(\frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1\), so the centers are at (h, k) = (3, -1).
02

Identify a and b

The values of 'a' and 'b' are the square roots of the denominators in our equation. Hence, a=sqrt(9)=3, and b=sqrt(16)=4.
03

Determine the vertices

The vertices of the ellipse along the x-axis are given by (h卤a, k), and the vertices along the y-axis are given by (h, k卤b). Hence, for our equation, the vertices are (3卤3, -1) and (3, -1卤4), or in coordinates: (0, -1), (6, -1), (3, -5), and (3, 3).
04

Determine the foci

The foci are located at (h, k卤c), where \(c=\sqrt{b^{2}-a^{2}}\). Thus, \(c=\sqrt{16-9}=\sqrt{7}\). So the foci are at (3, -1卤sqrt(7)), or in decimals, approximately at (3, -3.6458) and (3, 1.6458).
05

Sketch the ellipse

The ellipse can now be sketched with the given vertices and foci. It is wider along the y-axis because b>a, with a horizontal spread of 6 (from 0 to 6) and a vertical spread of 8 (from -5 to 3). The foci bisect the ellipse along the y-axis.

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Most popular questions from this chapter

Consult the research department of your library or the Internet to find an example of architecture that incorporates one or more conic sections in its design. Share this example with other group members. Explain precisely how conic sections are used. Do conic sections enhance the appeal of the architecture? In what ways?

Describe one similarity and one difference between the graphs of \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) and \(\frac{(x-1)^{2}}{25}+\frac{(y-1)^{2}}{16}=1\)

Isolate the terms involving \(y\) on the left side of the equation: $$ y^{2}+2 y+12 x-23-0 $$ Then write the equation in an equivalent form by completing the square on the left side.

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y=-x^{2}+4 x-3$$

Will help you prepare for the material covered in the next section. Consider the equation \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) a. Find the \(y\) -intercepts. b. Explain why there are no \(x\) -intercepts.
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