SOLUTION: Find center, vertices, and foci of the ellipse: (x+1)^2/16+(y-3)^2/25=1

Question 563053: Find center, vertices, and foci of the ellipse: (x+1)^2/16+(y-3)^2/25=1
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find center, vertices, and foci of the ellipse: (x+1)^2/16+(y-3)^2/25=1
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Standard form of equation for an ellipse with vertical major axis:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, with (h,k) being the (x,y) coordinates of the center.
For given elllipse: (x+1)^2/16+(y-3)^2/25=1
center: (-1,3)
a^2=25
a=√25=5
Vertices: (-1,3�a)=(-1,3�5),=(-1,-2) and (-1, 8)
b^2=16
b=√16=4
Foci:
c^2=a^2-b^2=25-16=9
c=√9=3
Foci: (-1,3�c)=(-1,3�3),=(-1,0) and (-1, 6)
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