Problem 43 - College Algebra

Question

Graph each ellipse and give the location of its foci. $$\frac{x^{2}}{25}+\frac{(y-2)^{2}}{36}=1$$

Step-by-Step Solution

Answer

The foci of the ellipse are at (0, 2 - √11) and (0, 2 + √11).

Step 1: Identify the center, semi-major axis, and semi-minor axis

The equation is in the form \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\). Compare this with the given equation to find that the center of the ellipse is at (h, k) = (0, 2), the length of the semi-major axis a is \(\sqrt{36}=6\), and the length of the semi-minor axis b is \(\sqrt{25}=5\). The major axis is vertical because a > b.

Step 2: Find the distance from the center to a focus

The distance from the center to a focus is \(c=\sqrt{a^{2}-b^{2}}=\sqrt{6^{2}-5^{2}}=\sqrt{11}\). So, the foci are at (0, 2 - √11) and (0, 2+ √11).

Step 3: Sketch the ellipse

Placed the center at (0, 2) on the graph. The major axis is vertical, and has length 2a = 12, and the minor axis has length 2b = 10. Sketch the ellipse based on these axes. Place the foci at (0, 2 - √11) and (0, 2 + √11).