Problem 37 - College Algebra

Question

Graph each ellipse and give the location of its foci. $$\frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{4}=1$$

Step-by-Step Solution

Answer

The ellipse is centered at the point (2, 1) with a semi-major axis length of 3 and a semi-minor axis length of 2. The foci are approximately located at (4.24, 1) and (-0.24, 1).

Step 1: Identify the Center of the Ellipse

From the given equation, 'h' and 'k' can be identified as 2 and 1 respectively. Therefore, the center of the ellipse is at the point (2, 1).

Step 2: Determine the Lengths of the Semi-Major and Semi-Minor Axes

In the given equation, 'a^2' and 'b^2' are 9 and 4 respectively. This means that 'a' is 3 and 'b' is 2, showing that the semi-major axis is 3 units long and the semi-minor axis is 2 units long.

Step 3: Calculate the Distances from the Center to Each Focus

Use the formula \(c = \sqrt{a^2 - b^2}\) to calculate 'c'. Substituting 'a' and 'b' into the formula gives \(c = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5}\). Therefore, each focus is \(\sqrt{5}\) units from the center.

Step 4: Identify the Location of the Foci

Because the 'a^2' term is under the 'x' term in the equation, we know that the foci are along the x-axis. They will be at the positions (h+c, k) and (h-c, k). Substituting the values h = 2, k = 1, c = \(\sqrt{5}\) yields the foci at (2+\(\sqrt{5}\), 1) and (2-\(\sqrt{5}\), 1). It means that the foci are approximately located at (4.24, 1) and (-0.24, 1).