# How do you find the center, vertices, and foci of an ellipse (x+5)^2/144 + (y+1)^2/225 = 1?

Aug 24, 2017

#### Explanation:

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ represents an ellipse with center $\left(h , k\right)$. Its vertices are $\left(h \pm a , k\right)$ and $\left(h , k \pm b\right)$.

If $b > a$, then it is an ellipse with major axis parallel to $y$-axis and eccentricity is $e = \sqrt{1 - {a}^{2} / {b}^{2}}$ and focii are $\left(h , k \pm b e\right)$. If $a > b$, then it is an ellipse with major axis parallel to $x$-axis and eccentricity is $e = \sqrt{1 - {b}^{2} / {a}^{2}}$ and focii are $\left(h \pm a e , k\right)$.

Hence for ${\left(x + 5\right)}^{2} / 144 + {\left(y + 1\right)}^{2} / 225 = 1$ or ${\left(x + 5\right)}^{2} / {12}^{2} + {\left(y + 1\right)}^{2} / {15}^{2} = 1$ and hence, we have ellipse with center as $\left(- 5 , - 1\right)$ and major axis parallel to $y$-axis.

Its vertices are $\left(- 5 \pm 12 , - 1\right)$ and $\left(- 5 , - 1 \pm 15\right)$ i.e. $\left(7 , - 1\right)$, $\left(- 17 , - 1\right)$, $\left(- 5 , 14\right)$ and $\left(- 5 , - 16\right)$.

Eccentricity is $\sqrt{1 - \frac{144}{225}} = \sqrt{\frac{81}{225}} = \frac{9}{15} = \frac{3}{5}$ and $b e = 15 \times \frac{3}{5} = 9$

and focii are $\left(- 5 , - 1 \pm 9\right)$ i.e. $\left(- 5 , 8\right)$ and $\left(- 5 , - 10\right)$.

Ellipse appears as follows:

graph{((x+5)^2/144+(y+1)^2/225-1)((x+5)^2+(y+1)^2-0.09)((x-7)^2+(y+1)^2-0.09)((x+17)^2+(y+1)^2-0.09)((x+5)^2+(y-14)^2-0.09)((x+5)^2+(y+16)^2-0.09)((x+5)^2+(y-8)^2-0.09)((x+5)^2+(y+10)^2-0.09)=0 [-40, 40, -20, 20]}