# How do you find the foci and sketch the ellipse 9x^2+4y^2=16?

May 12, 2017

The foci are $F = \left(0 , \frac{\sqrt{20}}{3}\right)$ and $F ' = \left(0 , - \frac{\sqrt{20}}{3}\right)$

#### Explanation:

The general equation of the ellipse is

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

We compare this equation to

$9 {x}^{2} + 4 {y}^{2} = 16$

$\frac{9}{16} {x}^{2} + \frac{4}{16} {y}^{2} = 1$

${x}^{2} / {\left(\frac{4}{3}\right)}^{2} + {y}^{2} / \left({2}^{2}\right) = 1$

The center of the ellipse is $C = \left(h , k\right) = \left(0 , 0\right)$ that is, the origin.

The major axis is vertical.

$A = \left(0 , 2\right)$ and $A ' = \left(0 , - 2\right)$

The other points are

$B = \left(\frac{4}{3} , 0\right)$ and $B ' = \left(- \frac{4}{3} , 0\right)$

We calculate

${c}^{2} = {b}^{2} - {a}^{2} = 4 - \frac{16}{9} = \frac{20}{9}$

$c = \pm \frac{\sqrt{20}}{3}$

The foci are

$F = \left(0 , \frac{\sqrt{20}}{3}\right)$ and $F ' = \left(0 , - \frac{\sqrt{20}}{3}\right)$

With the points $A$, $A '$, $B$ and $B '$, you can sketch the ellipse

graph{(9x^2+4y^2-16)((x)^2+(y-(sqrt20)/3)^2-0.001)((x)^2+(y-sqrt20/3)^2-0.001)=0 [-4.382, 4.386, -2.19, 2.197]}