SOLUTION: Find the center,foci,and vertices of the ellipse and determine the lenghts of the major and minor axis. Then sketch the graph. (x+2)^2/4+y^2=1 I thought both numbers in the equa

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Find the center,foci,and vertices of the ellipse and determine the lenghts of the major and minor axis. Then sketch the graph. (x+2)^2/4+y^2=1 I thought both numbers in the equa      Log On

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Question 70091: Find the center,foci,and vertices of the ellipse and determine the lenghts of the major and minor axis. Then sketch the graph.
(x+2)^2/4+y^2=1 I thought both numbers in the equation needed a denominator. I put a 1 under the y^2. Is this correct?
The center: (2,0)
a^2=4 so a=+or-2
Vertices: (-2,0+or-2)-->(-2,2),(-2,-2)
Major Axis: 2a=2*2=4
Minor Axis: 2b=1
b^2=1-->b=+or-1
Foci: (-4,0) (0,0)
Is this correct? I don't have a problem with the sketching. I just do not understand how to do it without the denominator under y^2.
Answer by stanbon(48569) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center,foci,and vertices of the ellipse and determine the lenghts of the major and minor axis. Then sketch the graph.
(x+2)^2/4+y^2=1
I thought both numbers in the equation needed a denominator. I put a 1 under the y^2. Is this correct?
COMMENT: Yes
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The center: (-2,0)
a^2=4 so a=2 (a is always positive)
Vertices: (0,0) and (-4,0)
Major Axis: 2a=2*2=4
b^2=1 so b=1
Minor Axis: 2b=2*1=2
Foci: To find the foci you need to find "c" where c=sqrt(a^2-b^2)
c=Sqrt(4-1)=sqrt3
Then Foci are (-2+sqrt3,0) and (-2-sqrt3,0)
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Cheers,
Stan H.