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Q 59.

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Precalculus Enhanced with Graphing Utilities
Found in: Page 655
Precalculus Enhanced with Graphing Utilities

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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Short Answer

Find an equation for each ellipse. Graph the equation by hand.

Foci at (5,1) and (-1,1): length of the major axis is 8.

The equation of the ellipse is (x-2)216+(y-1)27=1 and graph of the ellipse is

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Step by Step Solution

Step 1. Given information. 

Foci at (5,1) and (-1,1):length of the major axis is 8.

The foci lie on the liney=1, so the major axis is parallel to the x-axis.

The two foci (5,1) and (-1,1), we can find the center of the ellipse which is the midpoint of these two points.

Thus, the center of the ellipse is (2,1).

Step 2. The equation of ellipse.  

The length of the major axis 2a is given as 8. So, we have

2a=8a=4

The distance from the center of the ellipse (2,1) to the focus (5,1) is c=3

Substitute c=3 and a=4 in

role="math" localid="1646719438001" b2=a2-c2b2=42-32b2=16-9b2=7

The equation of ellipse is (x-2)216+(y-1)27=1

Step 3 .Graph of the ellipse  

The major axis parallel to the x- axis. So the vertices of the ellipse are a=4units left and right of the center (2,1).

Thus, the vertices are v1=(-2,1) and v2=(6,1).

We use b=7 to find the two points above and below the center. The two points are (2,1+7)and (2,1-7).

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