Foci: (0,0), (0,8), major axis of length 16

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SoosteethicU · Accepted Answer

The foci are in the vertical direction so the ellipse is vertical. Its equation must then be of the form $(\frac{{(x - h)}^{2}}{b^{2}}) + (\frac{{(y - k)}^{2}}{a^{2}}) = 1 .$
The center of the ellipse is the midpoint between the foci. Since the foci are (0,0) and (0,8), the center is $(h, k) = (\frac{0 + 0}{2}, \frac{0 + 8}{2}) = (\frac{0}{2}, \frac{8}{2}) = (0, 4)$ The distance from (0,4) to (0,0) is 4 so c=4
The length of the major axis of an ellipse is 2a so the length of the major axis is 16, then 2a=16. Dividing both sides by 2 then gives a=8.
The values of aa, bb, and cc are related by the equation $c_{2} = a_{2} - b_{2}$ . Substitute the values of a=8 and c=4 into this equation to find $b_{2}$ :
$c^{2} = a^{2} - b^{2}$
$4^{2} = 8^{2} - b^{2}$
$16 = 64 - b^{2}$
$- 48 = - b^{2}$
$48 = b^{2}$
Substituting in h=0, k=4, a=8 and $b_{2} = 48$ then gives:
$(\frac{{(x - h)}^{2}}{b^{2}}) + \frac{{(y - k)}^{2}}{a^{2}} = 1$
$(\frac{{(x - 0)}^{2}}{48}) + \frac{{(y - 4)}^{2}}{8^{2}} = 1$
$(x^{2} + 48) + \frac{{(y - 4)}^{2}}{64} = 1$

Foci: (0,0), (0,8), major axis of length 16

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