Question 491835: find the center,vertices and foci of the ellipse:
16x^2+4y^2+64x-24y+36=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the center,vertices and foci of the ellipse:
16x^2+4y^2+64x-24y+36=0
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16x^2+4y^2+64x-24y+36=0
complete the squares:
16(x^2+4x+4)+4(y^2-6y+9)=-36+64+36=64
16(x+2)^2+4(y-3)^2=64
divide by 64
(x+2)^2/4+(y-3)^2/16=1
This is an ellipse with a vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, with (h,k) being the (x,y) coordinates of the center.
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For given problem:
Center: (-2,3)
a^2=16
a=√16=4
length of vertical major axis=2a=8
vertices: (end-points of major axis)=(-2, 3±a)=(-2, 3±4)=(-2,7) and (-2,-1)
Vertices: (-2,7) and (-2,-1)
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b^2=4
b=√4=2
..
c^2=a^2-b^2=16-4=12
c=√12=3.46
Foci: (-2,3±c)=(-2,3±3.46)=(-2, 6.46) and (-2, -.46)
Ans:
Center: (-2,3)
Vertices: (-2,7) and (-2,-1)
Foci:(-2, 6.46) and (-2, -.46)
See graph below as a visual check on answers:
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y=±((16-4(x+2)^2)^.5)+3
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