Problem 46 - College Algebra

Question

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

Step-by-Step Solution

Answer

The center of the ellipse is at (-2, 3), with major semi-axis of 4 units along x-direction and minor semi-axis of 1 unit along y-direction. Foci are located at (-2 - \sqrt{15}, 3) and (-2 + \sqrt{15}, 3)

Step 1: Determine the center, semi-axis lengths and orientation

By comparing the given equation \((x+2)^2/16 + (y-3)^2 = 1\) with the standard form, we can find that the center of the ellipse is at \((-2, 3)\). Since 16 is larger than 1, the larger semi-axis is along the x-direction and it has a length of \(\sqrt{16} = 4\). The minor semi-axis is along the y-direction and it has a length of \(\sqrt{1} = 1\). So this is a horizontal ellipse.

Step 2: Calculate the foci

The foci of an ellipse are given by the equation \(c=\sqrt{a^2 - b^2}\) where \(c\) is the distance from the center to each focus, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axis respectively. In this case, applying the foci formula, we get \(c = \sqrt{4^2 - 1^2} = \sqrt{15}\). The foci will be located at \((-2 - \sqrt{15}, 3)\) and \((-2 + \sqrt{15}, 3)\) respectively.

Step 3: Graph the ellipse and mark the foci

Plot the center \((-2, 3)\). Draw the major axis with length of 8 units (\(2 * 4\)) and minor axis with length of 2 units (\(2 * 1\)). Then, plot the foci at \((-2 - \sqrt{15}, 3)\) and \((-2 + \sqrt{15}, 3)\).