Problem 40 - College Algebra

Question

Graph each ellipse and give the location of its foci. $$(x-3)^{2}+9(y+2)^{2}=18$$

Step-by-Step Solution

Answer

The locations of the foci are (-1, -2) and (7, -2). The graph will be an ellipse centered at (3, -2) with a vertical minor axis of length \(\sqrt{2}\) and a horizontal major axis of length \(\sqrt{18}\).

Step 1: Standardize the equation

Firstly standardize the given equation to the form of an ellipse equation. In our case divide each side of the given equation by 18 to get \((x-3)^2/18 + (y+2)^2/2 = 1\)

Step 2: Identify the center, semi-major and semi-minor axes

Comparing with standard ellipse equation, it's clear the center of the ellipse is at point (3, -2). Also, \(a^2 = 18\) and \(b^2 = 2\), so, the semi-major axis (a) is \(\sqrt{18}\) and the semi-minor axis (b) is \(\sqrt{2}\).

Step 3: Graph the ellipse

Plot the center at (3, -2) on a graph. Draw the major axes as a horizontal line through the center having a length of 2a = 2 * \(\sqrt{18}\), and minor axes as a vertical line through the center having a length of 2b = 2 * \(\sqrt{2}\). Then draw the ellipse around these axes.

Step 4: Find the foci

The foci of an ellipse are located on the major axis at a distance of \(\sqrt{a^2 - b^2}\) from the center. Substituting the values for a and b gives \(\sqrt{18 - 2} = 4\). So, the foci are located at (3-4, -2) and (3+4, -2), or (-1, -2) and (7, -2). You can denote these points on your graph to complete it.