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Chapter 7: Problem 45

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

Short Answer

Expert verified
The ellipse is centered at (-3,2), with lengths along the major and minor axes of 6 units and 2 units, respectively. The foci are located at (-3-sqrt(8),2) and (-3+sqrt(8),2)

Step by step solution

01

Determine the Center of the Ellipse

The center of the ellipse is given by the values of h and k in the equation. For this problem, the equation can be seen as \(\frac{(x - (-3))^{2}}{9} + \frac{(y - 2)^{2}}{1} = 1\). Thus, the center of the ellipse is at (-3,2).
02

Determine the Lengths of the Major and Minor Axes

The lengths of the major and minor axes are determined by the values of a^2 and b^2 in the equation. For this problem, we have a^2 = 9 and b^2 = 1. This means that the ellipse has a length of 2*sqrt(9) = 6 units along the major axis, and a length of 2*sqrt(1) = 2 units along the minor axis.
03

Locate the Foci

The foci of the ellipse are located on the major axis, a distance of c units from the center, where c^2 = a^2 - b^2. In this problem, a^2 - b^2 = 9 - 1 = 8, which implies that the distance is sqrt(8) units from the center. From the center (-3,2), we go sqrt(8) units left and right since the x term has the larger denominator in the equation, so the major axis is horizontal. This gives us the foci at points (-3-sqrt(8),2) and (-3+sqrt(8),2).
04

Sketch the Ellipse

Using the points gathered, an ellipse can be sketched. The major axis is along the x-axis and measures 2*sqrt(9)=6 units. The minor axis is along the y-axis and measures 2*sqrt(1)=2 units. The foci are at points (-3-sqrt(8),2) and (-3+sqrt(8),2)

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