# How do you graph the ellipse (x+5)^2/16+(y-4)^2/1=1 and find the center, the major and minor axis, vertices, foci and eccentricity?

Sep 16, 2017

#### Explanation:

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

whose center is $\left(h , k\right)$, $2 a$ and $2 b$ are major axis and minor axis (if $a > b$, but if $a < b$, it is reverse). In the former case major axis is parallel to $x$-axis and in latter case major axis is parallel to minor axis.

vertices are $\left(h , k \pm b\right)$ and $\left(h \pm a , k\right)$

Eccentricity $e$ can be obtained using $e = \sqrt{1 - {\left(\text{minor axis")^2/("major axis}\right)}^{2}}$

and focii are at a distance of $a e$ on either side of center along major axis (or $b e$ if major axis is parallel to $y$-axis).

Hence, in ${\left(x + 5\right)}^{2} / 16 + {\left(y - 4\right)}^{2} / 1 = 1$

center is $\left(- 5 , 4\right)$, major axis is $2 \sqrt{16} = 8$, which is parallel to $x$-axis and minor axis is $2 \sqrt{1} = 2$

Vertices are $\left(- 5 , 4 \pm 1\right)$ and $\left(- 5 \pm 4 , 4\right)$ i.e. $\left(- 5 , 5\right)$, $\left(- 5 , 3\right)$, $\left(- 1 , 4\right)$ and $\left(- 9 , 4\right)$.

Eccentricity is $e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ and $a e = \sqrt{3}$

and focii are $\left(- 5 \pm \sqrt{3} , 4\right)$ i.e. $\left(- 5 - \sqrt{3} , 4\right)$and $\left(- 5 + \sqrt{3} , 4\right)$.

They appear as follows:

graph{((x+5)^2+16(y-4)^2-16)((x+5)^2+(y-4)^2-0.01)((x+5)^2+(y-5)^2-0.01)((x+5)^2+(y-3)^2-0.01)((x+1)^2+(y-4)^2-0.01)((x+9)^2+(y-4)^2-0.01)((x+5+sqrt3)^2+(y-4)^2-0.01)((x+5-sqrt3)^2+(y-4)^2-0.01)=0 [-10.12, -0.12, 1.46, 6.46]}