# How do you graph the ellipse (x+2)^2/16+(y-5)^2/20=1 and find the center, the major and minor axis, vertices, foci and eccentricity?

Jun 4, 2017

#### Explanation:

The centre of the ellipse ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ is $\left(h , k\right)$.

The two axes are $2 a$ and $2 b$ and major axis is the greater of the two.

Vertices are $\left(h \pm a , k\right)$ and $\left(h , k \pm b\right)$. Here those along major axis are called vertices and those along minor axis are called covertices.

${\left(x + 2\right)}^{2} / 16 + {\left(y - 5\right)}^{2} / 20 = 1$ can be written as ${\left(x + 2\right)}^{2} / {4}^{2} + {\left(y - 5\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} = 1$

hence centre is $\left(- 2 , 5\right)$, maor axis is $2 \times 2 \sqrt{5} = 4 \sqrt{5}$ and minor axis is $2 \times 4 = 8$

Vertices are $\left(- 2 , 5 \pm 2 \sqrt{5}\right)$ i.e. $\left(- 2 , 5 \pm 2 \sqrt{5}\right)$ and $\left(- 2 , 5 + 2 \sqrt{5}\right)$and covertices are $\left(- 2 \pm 4 , 5\right)$ i.e. $\left(2 , 5\right)$ and $\left(- 6 , 5\right)$

Here we have a vertical major axis and as such eccentricity $e$ and focii are $\left(h , k \pm b e\right)$. $e$is given by the relation ${a}^{2} = {b}^{2} \left(1 - {e}^{2}\right)$ i.e. $e = \sqrt{1 - {a}^{2} / {b}^{2}}$ Here ${a}^{2} = 16$ and ${b}^{2} = 20$ and hence $e = \sqrt{1 - \frac{16}{20}} = \frac{1}{\sqrt{5}}$

and focii are $\left(- 2 , 5 \pm \left(2 \sqrt{5} \times \frac{1}{\sqrt{5}}\right)\right)$ or $\left(- 2 , 5 \pm 2\right)$ i.e. $\left(- 2 , 7\right)$ and $\left(- 2 , 3\right)$.

graph{(5(x+2)^2+4(y-5)^2-80)((x+2)^2+(y-7)^2-0.02)((x+2)^2+(y-3)^2-0.02)=0 [-12.33, 7.67, 0.1, 10.1]}