SOLUTION: What is the "equation for an ellipse if the endpoints of the major axis are at (1,6) and (1,-6) and the endpoints of the minor axis are at (5,0) and (-3,0"

Question 446589: What is the "equation for an ellipse if the endpoints of the major axis are at (1,6) and (1,-6) and the endpoints of the minor axis are at (5,0) and (-3,0"
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
What is the "equation for an ellipse if the endpoints of the major axis are at (1,6) and (1,-6) and the endpoints of the minor axis are at (5,0) and (-3,0"
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Standard forms of an ellipse:
With horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1,with (h,k) as (x,y) coordinates of the center (a>b)
With vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1, with (h,k) as (x,y) coordinates of the center (a>b)
Note that if a, the larger number, is under x^2, the ellipse will have a horizontal major axis.
If the larger number, is under y^2, the ellipse will have a vertical major axis.
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The best way to start solving given problem to draw a sketch of the ellipse with the points given.
You will notice right away that the major axis is vertical.
Length of major axis=12=2a
a=6
a^2=36
length of minor axis=8=2b
b=4
b^2=16
The center is somewhere on the major axis which lies on the line x=1
The center is also somewhere on the minor axis which lies on the line y=0 or the x-axis
Where these two lines cross is the center at (1,0)
Now, we have the information to write the equation of the given ellipse as follows:
(x+1)^2/16+y^2/36=1
see the graph below
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y=(36(1-((x-1)^2/16)))^.5

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