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Chapter 7: Problem 37

Graph each ellipse and give the location of its foci. $$\frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{4}=1$$

Short Answer

Expert verified
The ellipse is centered at the point (2, 1) with a semi-major axis length of 3 and a semi-minor axis length of 2. The foci are approximately located at (4.24, 1) and (-0.24, 1).

Step by step solution

01

Identify the Center of the Ellipse

From the given equation, 'h' and 'k' can be identified as 2 and 1 respectively. Therefore, the center of the ellipse is at the point (2, 1).
02

Determine the Lengths of the Semi-Major and Semi-Minor Axes

In the given equation, 'a^2' and 'b^2' are 9 and 4 respectively. This means that 'a' is 3 and 'b' is 2, showing that the semi-major axis is 3 units long and the semi-minor axis is 2 units long.
03

Calculate the Distances from the Center to Each Focus

Use the formula \(c = \sqrt{a^2 - b^2}\) to calculate 'c'. Substituting 'a' and 'b' into the formula gives \(c = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5}\). Therefore, each focus is \(\sqrt{5}\) units from the center.
04

Identify the Location of the Foci

Because the 'a^2' term is under the 'x' term in the equation, we know that the foci are along the x-axis. They will be at the positions (h+c, k) and (h-c, k). Substituting the values h = 2, k = 1, c = \(\sqrt{5}\) yields the foci at (2+\(\sqrt{5}\), 1) and (2-\(\sqrt{5}\), 1). It means that the foci are approximately located at (4.24, 1) and (-0.24, 1).

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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. 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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