Question 786519: How do i find the center, vertices, co-vertices and foci of these equations?
1. 25x2 + 9y2 = 225
2. 49x2 + y2 = 49
3. x2 - 2x + 9y2 - 8 = 0
4. x2 + 4y2 - 18x -8y + 81 = 0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! How do i find the center, vertices, co-vertices and foci of these equations?
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Standard form of equation for ellipses:
1)With horizontal major axis:  , a>b, (h,k)=(x,y) coordinates of the center.
2)With vertical major axis:  , a>b, (h,k)=(x,y) coordinates of the center.
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1. 25x2 + 9y2 = 225
divide by 225
x^2/9+y^2/25=1
ellipse has vertical major axis
center: (0,0)
a^2=25
a=√25=5
vertices: (0,±5)
b^2=9
b=√9=3
co-vertices:(±3,0)
c^2=a^2-b^2=25-9=16
c=√16=4
foci:(0,±4)
..
2. 49x2 + y2 = 49
divide by 49
x^2+y^2/49=1
ellipse has vertical major axis
center: (0,0)
a^2=49
a=√49=7
vertices: (0,±7)
b^2=1
b=1
co-vertices:(±1,0)
c^2=a^2-b^2=49-1=48
c=√48
foci:(0,±√48)
..
3. x2 - 2x + 9y2 - 8 = 0
complete the square
(x^2-2x+1)+9y^2=8+1
(x-1)^2+9y^2=9
divide by 9
x^2/9+y^2=1
ellipse has horizontal major axis
center: (1,0)
a^2=9
a=√9=3
vertices: (-2,0),(4,0)
b^2=1
b=1
co-vertices:(0,±1)
c^2=a^2-b^2=9-1=8
c=√8
foci:(1-√8,0),(1+√8,0)
..
4. x2 + 4y2 - 18x -8y + 81 = 0
x^2-18x+4y^2-8y=-81
complete the square:
(x^2-18x+81)+4(y^2-2y+1)=-81+81+4
(x-9)^2+4(y-1)^2=4
divide by 4
(x-9)^2/4+(y-1)^2=1
ellipse has horizontal major axis
center: (9,1)
a^2=4
a=√4=2
vertices: (7,1),(11,1)
b^2=1
b=1
co-vertices:(9,0),(9,2)
c^2=a^2-b^2=4-1=3
c=√3
foci:(9-√3,1),(9+√3,1)
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