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Chapter 7: Problem 16

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

Short Answer

Expert verified
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 by step solution

01

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 \(b=2\).
02

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 \(c=\sqrt{25-4}=\sqrt{21}.\)
03

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).
04

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.

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Most popular questions from this chapter

Determine whether each statement is true or false If the statement is false, make the necessary change(s) to produce a true statement. The parabola whose equation is \(x=2 y-y^{2}+5\) opens to the right.

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