SOLUTION: Find the center, vertices and foci of the ellipse 32x^2+4y^2+192x-24y=-196

Question 461591: Find the center, vertices and foci of the ellipse 32x^2+4y^2+192x-24y=-196
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the center, vertices and foci of the ellipse 32x^2+4y^2+192x-24y=-196
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Standard form of ellipse with horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), with (h,k) being the (x,y) coordinates of the center.
Standard form of ellipse with vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), with (h,k) being the (x,y) coordinates of the center.
The difference between the two forms is the interchange of a^2 and b^2.
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32x^2+4y^2+192x-24y=-196
completing the square
32(x^2+6x+9)+4(y^2-6y+9)=-196+288+36=128
32(x+3)^2+4(y-3)^2=128
divide by 128
(x+3)^2/4+(y-3)^2/32=1
Because the y-term has the larger denominator, this ellipse has a vertical major axis, second form listed above.
center: (-3,3)
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a^2=32
a=√32=5.66
length of major axis=2a=2√32
vertices: (3,3�a)=(3,3�√32)
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b^2=4
b=2
length of minor axis=2b=4
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c^2=a^2-b^2=32-4=28
c=√28=5.29..
Foci:(3,3�c)=(3,3�√28)
See the graph below as a visual check on the answers
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y= (32-8(x+3)^2)^.5+3

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