SOLUTION: Write the equation of the Ellipse, in Standard Form: Foci: (-1, -2), (-1, 6) Vertices: (-1, -6), (-1, 10)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Write the equation of the Ellipse, in Standard Form: Foci: (-1, -2), (-1, 6) Vertices: (-1, -6), (-1, 10)      Log On


   



Question 428941: Write the equation of the Ellipse, in Standard Form: Foci: (-1, -2), (-1, 6) Vertices: (-1, -6), (-1, 10)
Answer by lwsshak3(11628) About Me  (Show Source):
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Write the equation of the Ellipse, in Standard Form: Foci: (-1, -2), (-1, 6) Vertices: (-1, -6), (-1, 10)
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Standard form of ellipse for horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1 (a>b)
Standard form of ellipse for vertical major axis: (y-k)^2/a^2+(x-h)^2/b^2=1 (a>b)
In both forms, (h,k) represent the (x,y) coordinates of the center.
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Given data shows this ellipse has a vertical major axis on line x=-1, so, it is of the second form:
Center:(-1,2) (half way between foci or vertices on the major axis)
Length of major axis=16 (between vertices on the major axis)=2a
2a=16
a=8
a^2=64
c=4(from ctr to either foci on major axis)
c^2=a^2-b^2
b^2=a^2-c^2=64-16=48
b=sqrt(48)
b^2=48
We now have the information we need to write the equation of this ellipse:
(y-2)^2/64+(x+1)^2/48=1
see the graph below as a visual check on the answers above
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y=(64(1-(x+1)^2/48))^.5+2