Question 467169: what is the foci, center, and vertices of (x+3)^(2)/(16)+(y-1)^(2)/(25)=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! what is the foci, center, and vertices of (x+3)^(2)/(16)+(y-1)^(2)/(25)=1
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Standard form for ellipses:
(x-h)^2/a^2+(y-k)^2/b^2=1 (horizontal major axis),a>b, with(h,k) being the (x,y) coordinates of the center.
(y-k)^2/a^2+(x-h)^2/b^2=1 (vertical major axis),a>b,with(h,k) being the (x,y) coordinates of the center.
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(x+3)^(2)/(16)+(y-1)^(2)/(25)=1
Given function is an ellipse with a vertical major axis
Center:(-3,1)
a^2=25
a=5
length of major axis=2a=10
vertices: (-3,1±5) or (-3,6) and (-3,4)
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b^2=16
b=4
length of minor axis=2b=8
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c^2=a^2-b^2=25-16=9
c=√9=3
foci: (-3,1±3) or (-3,4) and (-3,-2)
see graph below as a visual check on the answers.
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y=(25-25(x+3)^2/16)^.5+1
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