SOLUTION: Write an equation for each ellipse described below: The foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5) and (0,-5).

Question 454543: Write an equation for each ellipse described below:
The foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5) and (0,-5).
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Write an equation for each ellipse described below:
The foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5) and (0,-5).
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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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Given coordinates of the foci shows that this ellipse has a horizontal major axis.(foci and major axis are on the same line, y=0, or x-axis.) This also tells you that the y-coordinate of the center=0. Given coordinates of the minor axis show that the x-coordinate of the center=0, and the length of the minor axis=10. This also shows c=12.
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Center (0,0)
length of minor axis=10=2b
b=5
b^2=25
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c^2=a^2-b^2
a^2=c^2+b^2=12^2+25=144+25=169
a^2=169
a=√169=13
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Equation of ellipse:
(x-0)^2/169+(y-0)^2/25=1
x^2/169+y^2/25=1
See the graph as a visual check on the answer
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y=(25-25x^2/169)^.5

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