Problem 42 - College Algebra

Question

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

Step-by-Step Solution

Answer

The ellipse has its center at (3, -1), semi-major axis of length 4, and semi-minor axis of length 3. The foci of the ellipse are at (3, -1±sqrt(7)).

Step 1: Identify the Center, and Semi-major and Semi-minor Axes

The center of the ellipse is the point (h, k), which here is (3, -1). Since the denominator of the x-term is less than that of the y-term, the semi-major axis is along the y-axis and has length \(a=\sqrt{16}=4\) and the semi-minor axis is along the x-axis and has length \(b=\sqrt{9}=3\).

Step 2: Draw the Ellipse

Start by plotting the center of the ellipse at (3, -1). Sketch the ellipse by plotting points a units above and below the center, and b units to the right and left of the center. This gives points at (3, 3), (3, -5), (0, -1), and (6, -1). Draw a smooth curve through these four points to finish the ellipse.

Step 3: Find the Foci

The foci of an ellipse are located on the major axis, c units from the center, where \(c^2=a^2-b^2\). Here, \(c=\sqrt{16-9}=\sqrt{7}\). Therefore, the foci are at (3, -1±sqrt(7)).