Problem 39 - College Algebra

Question

Graph each ellipse and give the location of its foci. $$(x+3)^{2}+4(y-2)^{2}=16$$

Step-by-Step Solution

Answer

The ellipse is centered at (-3,2), with a major axis of 4 units and minor axis of 2 units. The foci are at (-3 ± \(2\sqrt{3}\),2)

Step 1: Identifying Center and Axes

Rewrite equation in standard form to visualize the center and the lengths of major and minor axes. The standard form is \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). So the given equation \((x+3)^{2}+4(y-2)^{2}=16\) simplifies to \((x+3)^2/16 + (y-2)^2/4 = 1\. The center is at (-3,2). The length of major axis is sqrt(16) = 4 units (it lies along the x-axis) and minor axis is sqrt(4) = 2 units (it lies along the y-axis).

Step 2: Plotting the Ellipse

Use this information to sketch the ellipse by first drawing a rectangle with the center at (-3,2), using the lengths of major axis (4 units) and minor axis (2 units) as lengths of sides. An ellipse will touch the rectangle at the midpoint of each of its sides.

Step 3: Determining Foci

To find the foci, use the relationship \(c^2 = a^2 - b^2\), where c is the distance from the center to each focus. After substituting \(a^2=16\) and \(b^2=4\), it follows that \(c^2 = 16 - 4 = 12\), so \(c = \sqrt{12} = 2\sqrt{3}\). Since the major axis is along the x-axis, the foci will be along the x-axis as well, located at (-3 ± \(2\sqrt{3}\),2)