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Chapter 7: Problem 38

Graph each ellipse and give the location of its foci. $$\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1$$

Short Answer

Expert verified
The center of the ellipse is at (1, -2), the lengths of the major and minor axes are 8 and 6 respectively, and the foci are located at (1 卤 \(\sqrt{7}\), -2). The major axis is oriented along the x-axis.

Step by step solution

01

Identify the center, and lengths of the major and minor axis

Looking at the given equation, we see that it is already in standard form. From the equation, we can see that (h, k) = (1, -2), a = 4 and b = 3. Since \(a^{2} > b^{2}\), the major axis is along the x axis and its length is 2a = 8. The length of the minor axis is 2b = 6.
02

Identify the orientation of the ellipse

Since \(a^{2} > b^{2}\), the major axis is along the x-axis.
03

Calculate the distance to the foci from the center

We can calculate the distance to the foci from the center with the formula for c which is \(c = \sqrt{a^2 - b^2}\). Substituting the appropriate values, we have \(c = \sqrt{4^2 - 3^2} = \sqrt{7}\). So the foci are at a distance of \(\sqrt{7}\) along the major axis from the center.
04

Graph the ellipse

Start at the center (1, -2). Draw the major axis along the x-axis with a length of 2a = 8 and the minor axis along the y-axis with a length of 2b = 6. Place the foci at a distance of \(\sqrt{7}\) along the major axis from the center. This gives the graph of the ellipse.
05

Identify the location of the foci

The foci are located at a distance of \(\sqrt{7}\) from the center along the major (x) axis. Since the center of the ellipse is at (1, -2), the foci are located at (1 卤 \(\sqrt{7}\), -2).

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Most popular questions from this chapter

An architect designs two houses that are shaped and positioned like a part of the branches of the hyperbola whose equation is \(625 y^{2}-400 x^{2}=250,000,\) where \(x\) and \(y\) are in yards. How far apart are the houses at their closest point?

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$4 x^{2}-y^{2}+32 x+6 y+39=0$$

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