# How do you find the coordinates of the center, foci, the length of the major and minor axis given 3x^2+9y^2=27?

Dec 29, 2016

The center is $\left(0 , 0\right)$
The foci are $F = \left(\sqrt{6} , 0\right)$ and $F ' = \left(- \sqrt{6} , 0\right)$
The length of the major axis is $= 6$
The length of the minor axis is $= 2 \sqrt{3}$

#### Explanation:

Let 's rearrange the equation

$3 {x}^{2} + 9 {y}^{2} = 27$

Dividing by $27$

$\frac{3 {x}^{2}}{27} + \frac{9 {y}^{2}}{27} = 1$

${x}^{2} / 9 + {y}^{2} / 3 = 1$

${x}^{2} / {3}^{2} + {y}^{2} / {\left(\sqrt{3}\right)}^{2} = 1$

We compare this to the general equation of the ellipse

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

$a = 3$ and $b = \sqrt{3}$

The center is $\left(h . k\right) = \left(0 , 0\right)$

To calculate the foci, we need

$c = \sqrt{{a}^{2} - {b}^{2}} = \sqrt{9 - 3} = \sqrt{6}$

The foci are F$= \left(h + c , k\right) = \left(\sqrt{6} , 0\right)$

and F'$= \left(h - c , k\right) = \left(- \sqrt{6} , 0\right)$

The length of the major axis is $= 2 a = 6$

The length of the minor axis is $= 2 b = 2 \sqrt{3}$

graph{3x^2+9y^2=27 [-5.546, 5.55, -2.773, 2.774]}