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Chapter 7: Problem 48

Graph each ellipse and give the location of its foci. $$\frac{(x+1)^{2}}{2}+\frac{(y-3)^{2}}{5}=1$$

Short Answer

Expert verified
The ellipse is centered at (-1, 3) with semi-major axis length \( \sqrt{5} \) and semi-minor axis length \( \sqrt{2} \). The coordinates of the foci are \((-1, 3+ \sqrt{3}) \) and \((-1, 3 - \sqrt{3}) \).

Step by step solution

01

Identify the Center

To find the center of the ellipse, we equate the expression inside the parentheses to zero. Hence the center of the ellipse is at \((-1, 3)\).
02

Identify the semi-major and semi-minor axes

The values under x and y are the squares of the semi-major axis and semi-minor axis. The larger denominator is under y, so we have semi-major axis a as \( \sqrt{5} \) and semi-minor axis b as \( \sqrt{2} \).
03

Calculate the Foci

Now we calculate the foci. The distance from the center to each focus is given by \(c = \sqrt{a^2 - b^2}\). Plugging in the values we have, \(c = \sqrt{5-2} = \sqrt{3}\). The foci are \(c\) units above and below the center along the major axis (vertical axis) since we have a > b. Therefore, the foci are at \((-1, 3+ \sqrt{3}) \) and \((-1, 3 - \sqrt{3})\)

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Most popular questions from this chapter

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{25}+\frac{y^{2}}{64}=1$$

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$9 x^{2}+25 y^{2}-36 x+50 y-164=0$$

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In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{9}+\frac{y^{2}}{36}=1$$

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