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

1 Answer
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