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Chapter 7: Problem 45

Graph each ellipse and give the location of its foci. $$\frac{(x+3)^{2}}{9}+(y-2)^{2}=1$$

Short Answer

Expert verified
The foci of the ellipse given by \(\frac{(x+3)^{2}}{9}+(y-2)^{2}=1\) are located at (-3 - sqrt(8), 2) and (-3 + sqrt(8), 2).

Step by step solution

01

Identify the center

The ellipse equation is in the form \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). From the given equation we see that h=-3 and k=2. Therefore, the center of the ellipse is (-3,2).
02

Identify a and b

The lengths of the semi-major axis 'a' and semi-minor axis 'b' are the square roots of the denominators in the equation. Therefore, we have: a=3, and b=1.
03

Calculate c (distance to foci)

The distance from the center to each focus 'c' can be found using the equation \(c=\sqrt{a^{2}-b^{2}}\). Substituting a=3 and b=1 into the equation, we get: \(c=\sqrt{9-1}=\sqrt{8}\)
04

Identify the foci

The foci of the ellipse are given by (h-c,k) and (h+c,k). Hence, the foci are located at (-3 - sqrt(8), 2) and (-3 + sqrt(8), 2)
05

Graph the ellipse

With the center at (-3,2), draw an ellipse with semi-major axis of length 3 and semi-minor axis of length 1. Mark the foci at (-3 - sqrt(8), 2) and (-3 + sqrt(8), 2)

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Most popular questions from this chapter

Convert each equation to standard form by completing the square on \(x\) or \(y .\) Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. $$ x^{2}-2 x-4 y+9=0 $$

Graph each ellipse and give the location of its foci. $$(x-3)^{2}+9(y+2)^{2}=18$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$4 x^{2}+16 y^{2}=64$$

Find the standard form of the equation of each ellipse satisfying the given conditions. Endpoints of major axis: \((7,9)\) and \((7,3)\) Endpoints of minor axis: \((5,6)\) and \((9,6)\)

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