# How do you write an equation of an ellipse in standard form given foci (+/-4,0) and co-vertices at (0,+/-2)?

${x}^{2} / 20 + {y}^{2} / 4 = 1$

#### Explanation:

From the given data:

Foci(4, 0) and (-4, 0)
Co-vertices (0, 2) and (0, -2) are the endpoints of the minor axis.

The Center is at the Origin $\left(0 , 0\right)$ by inspection.

The standard form equation for Horizontal major axis Ellipse

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

$b = 2$ the length of 1/2 of the minor axis called semi-minor axis.
$c = 4$ the distance from center to a focus.

$a$ is the length of 1/2 of the major axis also called semi-major axis.

Compute for $a$

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

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

$a = \sqrt{{4}^{2} + {2}^{2}}$

$a = \sqrt{16 + 4}$

$a = \sqrt{20} = 2 \sqrt{5}$

Use now the equation

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ with Center $\left(h , k\right) = \left(0 , 0\right)$

${\left(x - 0\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} + {\left(y - 0\right)}^{2} / {2}^{2} = 1$

which simplifies to

${x}^{2} / 20 + {y}^{2} / 4 = 1$

Kindly check the graph...

graph{(x^2/20+y^2/4-1)=0[-10,10,-5,5]}

Have a nice day !!! from the Philippines .