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Question 449299: We are working on conic sections. I need help finding the equation of an ellipse with these given points: vartices at (-1,2) and (7,2)and a focus at (5,2).
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! We are working on conic sections. I need help finding the equation of an ellipse with these given points: vartices at (-1,2) and (7,2)and a focus at (5,2).
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Standard form for an 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 for an 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. Note that the only difference between these two forms is that the a^2 and b^2 terms are interchanged.
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For given ellipse:
Since the vertices are horizontal, this ellipse is of the first form listed above.
x-coordinate of center =(-1+7)/2=3
y-coordinate of center=2
Center is at (3,2)
Length of major axis=distance between end points of vertices=8=2a
a=4
a^2=16
c=distance from center to focal point on major axis=5-3=2
c^2=4
c^2=a^2-b^2
b^2=a^2-c^2=16-4=12
b=√12
you now have all the information you need to write the equation of given ellipse as follows:
(x-3)^2/16+(y-2)^2/12=1
See the graph below as visual evidence of this equation
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y=(12(1-(x-3)^2/16))^.5+2
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