SOLUTION: Problem: 3x^2+12y^2=12 How do I first write a standard equation of the ellipse and find the coordinates of the center, vertices, co-vertices, and foci?

Question 541721: Problem: 3x^2+12y^2=12
How do I first write a standard equation of the ellipse and find the coordinates of the center, vertices, co-vertices, and foci?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Problem: 3x^2+12y^2=12
How do I first write a standard equation of the ellipse and find the coordinates of the center, vertices, co-vertices, and foci?
**
Standard form of equation for ellipse with horizontal major axis:
(x-h)^2/a^2+(y-k)^2/b^2=1,a>b, (h,k)=(x,y) coordinates of center
..
For given problem:
3x^2+12y^2=12
divide by 12
x^2/4+y^2/1=1
x^2/4+y^2=1 (standard form of equation for given ellipse)
..
Center: (0,0)
a^2=4
a=2
vertices: (0�a,0)=(0�2,0)=(-2,0) and (2,0)
..
b^2=1
b=1
co-vertices: (0,0�b)=(0,0�1)=(0,-1) and (0,1)
..
c^2=a^2-b^2=4-1=3
c=√3
Foci: (0�c,0)=(0�√3,0)=(-√3,0) and (√3,0)
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