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

Dec 30, 2016

${x}^{2} / 48 + {\left(y - 4\right)}^{2} / 64 = 1$

#### Explanation:

The midpoint between the foci is the center

$C : \left(\frac{0 + 0}{2} , \frac{0 + 8}{2}\right)$

$\implies C : \left(0 , 4\right)$

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

$2 c = \sqrt{{\left(0 - 0\right)}^{2} + {\left(0 - 8\right)}^{2}}$

$\implies 2 c = \sqrt{0 + 64}$

$\implies 2 c = 8$

$\implies c = 4$

The major axis length is equal to $2 a$

$2 a = 16$

$\implies a = 8$

${c}^{2} = {a}^{2} - {b}^{2}$

$\implies {b}^{2} = {a}^{2} - {c}^{2}$

$\implies {b}^{2} = {8}^{2} - {4}^{2}$

$\implies {b}^{2} = 64 - 16$

$\implies {b}^{2} = 48$

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

${\left(x - h\right)}^{2} / {b}^{2} + {\left(y - k\right)}^{2} / {a}^{2} = 1$

$\implies {\left(x - 0\right)}^{2} / 48 + {\left(y - 4\right)}^{2} / {8}^{2} = 1$

$\implies {x}^{2} / 48 + {\left(y - 4\right)}^{2} / 64 = 1$