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Q 63.

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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.

Center at (1,2): vertex at (4,2):contains the point(1,5)

The equation of the ellipse is (x-1)29+(y-2)29=1and graph of the ellipse is

See the step by step solution

Step by Step Solution

Step 1 . Given information. 

Center at (1,2): vertex at (4,2):contains the point (1,5).

On comparing with the standard equation of the ellipse (x-h)2a2+(y-k)2b2=1

Here,x=1, y=5, h=1, k=2 and h+a=4a=4-1a=3

Now put these values in the equation of the ellipse.

(1-1)232+(5-2)2b2=109+9b2=19b2=1b2=9

Step 2. The equation of ellipse.   

Substitute the value of a2=9 and b2=9 we get

(x-1)29+(y-2)29=1

Step 3 .Graph of the ellipse .  

The major axis is parallel to the x-axis.So the vertices are a=3units left and right of the center (1,2).Therefore the vertices are localid="1646732539408" v1=(-2,2) and localid="1646732554304" v2=(4,2)

Since,

localid="1646732713719" c2=a2-b2c2=9-9c2=0

Major axis is parallel to the x-axis, foci are localid="1646732722940" (1-0,2) and localid="1646732737337" (1,2).

We use the value localid="1646732752572" b=3 to find the two points above and below the center as localid="1646732764537" (1,-1) and localid="1646732779266" (1,5)

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