How do you write an equation of a ellipse with foci (0,0), (0,8) and major axis of length 16?

1 Answer
seph
Dec 30, 2016

#x^2/48 + (y - 4)^2/64 = 1#

Explanation:

The midpoint between the foci is the center

#C: ((0 + 0)/2, (0 + 8)/2)#

#=> C: (0, 4)#

The distance between the foci is equal to #2c#

#2c = sqrt((0 - 0)^2 + (0 - 8)^2)#

#=> 2c = sqrt(0 + 64)#

#=> 2c = 8#

#=> c = 4#

The major axis length is equal to #2a#

#2a = 16#

#=> a = 8#

#c^2 = a^2 - b^2#

#=> b^2 = a^2 - c^2#

#=> b^2 = 8^2 - 4^2#

#=> b^2 = 64 - 16#

#=> b^2 = 48#

Between the coordinates of the foci, only the y-coordinate changes, this means the major axis is vertical. The standard equation of an ellipse with a vertical major axis is

#(x - h)^2/b^2 + (y - k)^2/a^2 = 1#

#=> (x - 0)^2/48 + (y - 4)^2/8^2 = 1#

#=> x^2/48 + (y - 4)^2/64 = 1#

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