SOLUTION: Identify the vertices and foci.
(x+3)^2/16+(y-1)^2/25=1
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Question 467420: Identify the vertices and foci.
(x+3)^2/16+(y-1)^2/25=1
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Identify the vertices and foci.
(x+3)^2/16+(y-1)^2/25=1
...
(x+3)^2/16+(y-1)^2/25=1
This is an ellipse with a vertical major axis.
Center:(-3,1)
a^2=25
a=5
length of major axis=2a=10
vertices at (-3,1±5) or (-3,6) and (-3,-4)
..
b^2=16
b=4
length of minor axis=2b=8
..
c^2=a^2-b^2=25-16=9
c=√9=3
foci:(-3±3) or (-3,0) and (-3,-6)
See graph below as a visual check on answers
..
y=(25-25(x+3)^2/16)^.5+1
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