Problem 45 - College Algebra

Question

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

Step-by-Step Solution

Answer

The foci of the ellipse given by \(\frac{(x+3)^{2}}{9}+(y-2)^{2}=1\) are located at (-3 - sqrt(8), 2) and (-3 + sqrt(8), 2).

Step 1: Identify the center

The ellipse equation is in the form \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). From the given equation we see that h=-3 and k=2. Therefore, the center of the ellipse is (-3,2).

Step 2: Identify a and b

The lengths of the semi-major axis 'a' and semi-minor axis 'b' are the square roots of the denominators in the equation. Therefore, we have: a=3, and b=1.

Step 3: Calculate c (distance to foci)

The distance from the center to each focus 'c' can be found using the equation \(c=\sqrt{a^{2}-b^{2}}\). Substituting a=3 and b=1 into the equation, we get: \(c=\sqrt{9-1}=\sqrt{8}\)

Step 4: Identify the foci

The foci of the ellipse are given by (h-c,k) and (h+c,k). Hence, the foci are located at (-3 - sqrt(8), 2) and (-3 + sqrt(8), 2)

Step 5: Graph the ellipse

With the center at (-3,2), draw an ellipse with semi-major axis of length 3 and semi-minor axis of length 1. Mark the foci at (-3 - sqrt(8), 2) and (-3 + sqrt(8), 2)