# How do you find the coordinates of the center, foci, lengths of the major and minor axes given 4x^2+8y^2=32?

Oct 29, 2016

The center is $\left(0 , 0\right)$
The foci are $\left(2 , 0\right)$ and $\left(- 2 , 0\right)$
Length of major axis is $= 2 \sqrt{8}$
Length of minor axis is$= 4$

#### Explanation:

$4 {x}^{2} + 8 {y}^{2} = 32$
Dividing by 32, we get
${x}^{2} / 8 + {y}^{2} / 4 = 1$
Comparing this to the equation of the ellipse
${\left(x - {x}_{0}\right)}^{2} / {a}^{2} + \frac{y - {y}_{0}^{2}}{b} ^ 2 = 1$

The center is $\left({x}_{0} , {y}_{0}\right) = \left(0 , 0\right)$
To determine the foci, we calculate $c = \sqrt{{a}^{2} - {b}^{2}} = \sqrt{8 - 4} = \sqrt{4} = 2$
Therefore the foci are $\left(2 , 0\right)$ and $\left(- 2 , 0\right)$
Length of half the major axis is $a = \sqrt{8}$
and the length of half the minor axis is $b = 2$
graph{4x^2+8y^2=32 [-6.244, 6.243, -3.12, 3.123]}