Problem 44 - College Algebra

Question

Graph each ellipse and give the location of its foci. $$\frac{(x-4)^{2}}{4}+\frac{y^{2}}{25}=1$$

Step-by-Step Solution

Answer

The ellipse is centered at (4, 0) with a semi-major axis of length 5 and a semi-minor axis of length 2. The foci are located at \((4+\sqrt{21}, 0)\) and \((4-\sqrt{21}, 0)\)

Step 1: Identify the center

In \((x-4)^{2}/4 + y^{2}/25 = 1\), compare this with the standard equation of an ellipse to get the center, \((h,k)\). \(h\) corresponds to 4 and \(k\) corresponds to 0 since there's no explicit translation of the y-term. So, the center of the ellipse is at (4, 0).

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

\(a^2\) and \(b^2\) are given by the terms below \(x^2\) and \(y^2\) respectively. So, the lengths of the semi-major axis \(a\) and the semi-minor axis \(b\) are 5 and 2, respectively.

Step 3: Sketch the ellipse

The ellipse is centered at (4, 0) and extends 2 units in the x-direction and 5 units in the y-direction. Sketch the ellipse based on these observations.

Step 4: Find the foci

For any ellipse, the foci are located c units from the center, where \(c^2 = a^2 - b^2\). In this case, \(c^2 = 5^2 - 2^2 = 21\), giving \(c = \sqrt{21}\). The foci are therefore located at \((4+\sqrt{21}, 0)\) and \((4-\sqrt{21}, 0)\)