# How do you find the coordinates of the center, foci, the length of the major and minor axis given (x+11)^2/144+(y-5)^2/121=1?

Apr 23, 2017

Centre is $\left(- 11 , 5\right)$, foci are $\left(- 11 - \sqrt{23} , 5\right)$ and $\left(- 11 + \sqrt{23} , 5\right)$ and major axis is $24$ and minor axis is $22$.

#### Explanation:

As the equation ${\left(x + 11\right)}^{2} / 144 + {\left(y - 5\right)}^{2} / 121 = 1$ is of form

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$, whose centre is $\left(h , k\right)$ and major and minor axis are $2 a$ and $2 b$ respectively.

As such here centre is $\left(- 11 , 5\right)$ and major axis is $24$ and minor axis is $22$.

For focus we should calculate eccentricity of ellipse and as ${b}^{2} = {a}^{2} \left(1 - {e}^{2}\right)$, $e = \sqrt{1 - {b}^{2} / {a}^{2}} = \sqrt{1 - \frac{121}{144}} = \frac{\sqrt{23}}{12}$

and foci are given by $\left(h \pm a e , k\right)$ i.e. they are

$\left(- 11 \pm \sqrt{23} , 5\right)$ i.e. $\left(- 11 - \sqrt{23} , 5\right)$ and $\left(- 11 + \sqrt{23} , 5\right)$

graph{(x+11)^2/144+(y-5)^2/121=1 [-30, 10, -8, 18]}

Note - Here we had $a > b$, hence major axis is parallel to $x$-axis. But if we have $b > a$, then major axis is parallel to $y$-axis and $e = \sqrt{1 - {a}^{2} / {b}^{2}}$ and then foci are $\left(h , k \pm a e\right)$