Problem 16 - College Algebra

Question

Graph each ellipse and locate the foci. $$4 x^{2}+25 y^{2}=100$$

Step-by-Step Solution

Answer

The values of \(a=5\) and \(b=2\) were obtained from given equation. The c-value was calculated to be \(c=\sqrt{21}\). The foci were located at (-\sqrt{21}, 0) and (+\sqrt{21}, 0). The ellipse was drawn by plotting the center, vertices and co-vertices then sketching the ellipse passing through these points.

Step 1: Express the given equation in standard form

We notice the given equation can be rewritten as \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), by dividing all terms by 100. Doing this, our equation becomes: \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\) which is the standard form of an ellipse. Here, it can be noticed that \(a^{2}=25\) and \(b^{2}=4\), which means that \(a=5\) and \(b=2\).

Step 2: Determine the value of c

To find c, which is the distance from the center to a focus, we use the formula \(c=\sqrt{a^{2}-b^{2}}\). We substitute \(a=5\) and \(b=2\) to the equation to get \(c=\sqrt{25-4}=\sqrt{21}.\)

Step 3: Determine the coordinates of the foci

Since \(a > b\), the foci will be located at \((h - c, k)\) and \((h + c, k)\). Given that the center (h, k) is at (0, 0), the foci will be at (-\sqrt{21}, 0) and (+\sqrt{21}, 0).

Step 4: Graph the ellipse and locate the foci

To graph, first draw the major axis horizontal line of length 2a=10 units and minor axis vertical line of length 2b=4 units. Hence, the vertices are at (-5,0), (5,0) and co-vertices are at (0,-2), (0,2). Then, plot the center, vertices and co-vertices, and draw an ellipse passing through these points which will be the graph of the given equation. The foci will be located at (-\sqrt{21}, 0) and (+\sqrt{21}, 0) on the major axis.