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Chapter 7: Problem 44

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

Short Answer

Expert verified
The foci of the ellipse are located at (4, \(\sqrt{21}\)) and (4, -\(\sqrt{21}\)). The semi-minor axis is 2 and the semi-major axis is 5 with the center of the ellipse at (4, 0).

Step by step solution

01

Identify the Ellipse Parameters

In the given equation \((x-4)^{2}/4 + y^{2}/25 = 1\), the coordinates of the center (h, k) can be found as h = 4 and k = 0 because the equation is in the form \((x-h)^2/a^2 + (y-k)^{2}/b^2 = 1\), here a^2 = 4 so a = 2, and b^2 = 25 so b = 5. Thus, the center is (4, 0), the semi-minor axis is 2 and the semi-major axis is 5.
02

Find the Foci

Using the relationship for the foci \(c= \sqrt{|a^{2}-b^{2} |}\), we find \(c = \sqrt{|2^{2} - 5^{2}|} = \sqrt{21}\) since \(-21\) under a sqrt sign yields an imaginary result.
03

Plot the Ellipse

Using the center (4, 0), points along the semi-major axis at (4, 5) and (4, -5), and points along the semi-minor axis at (2, 0) and (6, 0), we can sketch the ellipse. The foci are at \((4, \sqrt{21})\) and \((4, -\sqrt{21})\)

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Most popular questions from this chapter

What is an ellipse?

Wre a graphing utility to graph \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=0 .\) Is the graph a hyperbola? In general, what is the graph of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0 ?\)

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Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. What happens to the shape of the graph of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) as \(\frac{c}{a} \rightarrow \infty,\) where \(c^{2}=a^{2}+b^{2} ?\)
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