Question 598910: trying to graph the following, but I've never seen something like this before:
(x-2)^2+(y-3)^2/64=1
help please
Found 3 solutions by solver91311, stanbon, lwsshak3:
Answer by solver91311(24713)   (Show Source):
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! trying to graph the following, but I've never seen something like this before:
(x-2)^2+(y-3)^2/64=1
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Draw an ellipse with
center at (2,3)
horizontal axis has length = 2
vertical axis has length = 16
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Cheers,
Stan H.
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Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! trying to graph the following, but I've never seen something like this before:
(x-2)^2+(y-3)^2/64=1
This is an equation of an ellipse with vertical major axis.
Its standard form: (x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center.
For given equation: (x-2)^2+(y-3)^2/64=1
center: (2,3)
a^2=64
a=√64=8
vertices: (2,3±a)=(2,3±8)=(2,-5) and (2,11)
..
b^2=1
b=1
end-points of minor axis: (2±b,3)=(2±1,3)=(1,3) and (3,3)
..
Foci:
c^2=a^2-b^2=64-1=63
c=√63≈7.9
Foci:(2,3±c)=(2,3±7.9)=(2,-4.9) and (2,10.9)
see graph below:
y=(64-64(x-2)^2)^.5+3
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