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Chapter 7: Problem 46

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

Short Answer

Expert verified
The foci of the ellipse are at points (-2 - \(\sqrt{15}\), 3) and (-2 + \(\sqrt{15}\), 3).

Step by step solution

01

Identify the Center (h, k)

In the give ellipse equation, the center can be found by locating the terms associated with x and y. The center (h, k) of this ellipse is (-2, 3).
02

Identify Constants a and b

We need to determine the values of a and b. The value under the x-term is \(a^2\), which is 16 in this equation. Thus, a = \(\sqrt{16}\) = 4. Since the value under the y-term is \(b^2\) and equals to 1, b is thus also equal to 1.
03

Calculate c for the Foci

Use the formula for c to find the focus points. \(c=\sqrt{a^2 - b^2}\) = \(\sqrt{4^2-1^2}\) = \(\sqrt{15}\).
04

Determine the Foci

Because a>b, the foci are located at (h卤c, k), which are (-2 - \(\sqrt{15}\), 3) and (-2 + \(\sqrt{15}\), 3).
05

Graph the Ellipse

Plot the center point on the graph. Then move a = 4 units left and right from this point, b = 1 unit up and down marking those points. Draw a rough oval shape to form the ellipse. Lastly, mark the foci.

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