# How do you write an equation of an ellipse for the given Foci (0,±8) Co-Vertices (±8,0)?

${x}^{2} / 64 + {y}^{2} / 128 = 1$

#### Explanation:

From the given:
Foci: ${F}_{1} \left(0 , 8\right) , {F}_{2} \left(0 , - 8\right)$
Co-vertices at $\left(8 , 0\right)$ and (-8, 0)

by inspection, the center of the ellipse is at Origin $\left(0 , 0\right)$
that means

Center $\left(h , k\right) = \left(0 , 0\right)$

Also, by inspection, $c = 8$ the distance from the center to a focus.
Also, $b = 8$ the distance from the center to one end point of the minor axis also called semi-minor axis length.

Compute for semi-major axis length $a$:

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

$a = \sqrt{{b}^{2} + {c}^{2}}$

a=sqrt(8^2+8^2)=sqrt(64+64#

$a = \sqrt{2 \cdot 64}$

$a = 8 \sqrt{2}$

Use now the standard Form of the equation of ellipse for Vertical Major Axis.

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

Let us put on the values of $h = 0 , k = 0 , a = 8 \sqrt{2} , b = 8$

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

${x}^{2} / 64 + {y}^{2} / 128 = 1$

Kindly check the graph ....
graph{(x^2/64+y^2/128-1)=0[-25,25,-15,15]}

Have a nice day !!! from the Philippines...