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 foci of the ellipse are located at (-1, -2) and (7, -2).

Step 1: Rearrange the given equation to standard form

Standard form of an ellipse equation is \((x-h)^{2}/a^{2}+(y-k)^{2}/b^{2}=1\), the given equation is \((x-3)^{2}+9(y+2)^{2}=18\). To rewrite, divide each term by 18. This yields \((x-3)^{2}/18+(y+2)^{2}/2=1\). Therefore, a^2 = 18, b^2 = 2, h = 3 and k = -2.

Step 2: Identify semi-major and semi-minor axes

The major axis is always the larger of the two, meaning a = \(\sqrt{18}\) and b = \(\sqrt{2}\). Hence, the semi-major axis is along the x-direction and semi-minor axis is along the y-direction.

Step 3: Find the location of the foci

The distance from the center to each focus (c) is found by \(c=\sqrt{a^{2}-b^{2}}\). Substituting a^2=18 and b^2=2, we get \(c=\sqrt{16}=4\). The foci will be 4 units from the center (h, k) along the major axis, which is the x-axis. Hence, foci will be at (h-c, k) and (h+c, k), or (3-4, -2) and (3+4, -2), which simplifies to (-1, -2) and (7, -2)

Step 4: Graph the Ellipse

Sketch an x-y coordinate system. Plot the center of the ellipse at point (3, -2). Draw the major axis 2a = 2(\(\sqrt{18}\)) units long and minor axis 2b = 2(\(\sqrt{2}\)) units long. The ellipse will be elongated along the x-axis. Then, mark the two foci on the major axis at points (-1, -2) and (7, -2).