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's foci are approximately located at (3, -3.6458) and (3, 1.6458).

Step 1: Identify the centers

The standard form of the equation of an ellipse is \(\frac{(x-h)^{2}}{a^2}+\frac{(y-k)^{2}}{b^2}=1\), where (h, k) is the center of the ellipse. In this case, our equation is \(\frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1\), so the centers are at (h, k) = (3, -1).

Step 2: Identify a and b

The values of 'a' and 'b' are the square roots of the denominators in our equation. Hence, a=sqrt(9)=3, and b=sqrt(16)=4.

Step 3: Determine the vertices

The vertices of the ellipse along the x-axis are given by (h±a, k), and the vertices along the y-axis are given by (h, k±b). Hence, for our equation, the vertices are (3±3, -1) and (3, -1±4), or in coordinates: (0, -1), (6, -1), (3, -5), and (3, 3).

Step 4: Determine the foci

The foci are located at (h, k±c), where \(c=\sqrt{b^{2}-a^{2}}\). Thus, \(c=\sqrt{16-9}=\sqrt{7}\). So the foci are at (3, -1±sqrt(7)), or in decimals, approximately at (3, -3.6458) and (3, 1.6458).

Step 5: Sketch the ellipse

The ellipse can now be sketched with the given vertices and foci. It is wider along the y-axis because b>a, with a horizontal spread of 6 (from 0 to 6) and a vertical spread of 8 (from -5 to 3). The foci bisect the ellipse along the y-axis.