SOLUTION: (x-2)^2+4y^2=16 I need to get the center, the vertices, and foci.

Question 971219: (x-2)^2+4y^2=16
I need to get the center, the vertices, and foci.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
(x-2)^2+4y^2=16
divide by 16
x^2/16 +4y^2/16 =1= x^2/16 + y^2/4
This is an ellipse centered on the origin.
The x-axis goes from (-4,0) to (0,4).
The y-axis goes from (0-2) to (0,2) Those points are the vertices.
The foci are
f=(sqrt (16-4))=2 sqrt (3)
They are along the x-axis at (-2sqrt(3),0) and (2 sqrt (3),0)
2sqrt(3)=3.464

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