# How do you find the center, foci and vertices of (x-6)^2/4+(y+7)^2/16=1?

Dec 20, 2016

The center is $= \left(6 , - 7\right)$
The foci are $= \left(6 , - 7 + 2 \sqrt{3}\right)$ and $= \left(6 , - 7 - 2 \sqrt{3}\right)$
The vertices are $\left(6 , - 3\right)$, $\left(6 , - 11\right)$, $\left(8 , - 7\right)$ and $\left(4 , - 7\right)$

#### Explanation:

The equation represents an ellipse.

${\left(x - 6\right)}^{2} / 4 + {\left(y + 7\right)}^{2} / 16 = 1$

with the major axis vertical

The general equation is

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

The center of the ellipse is $\left(h , k\right) = \left(6 , - 7\right)$

Let's calculate $c = \sqrt{{b}^{2} - {a}^{2}} = \sqrt{16 - 4} = \sqrt{12} = 2 \sqrt{3}$

The foci are F$= \left(h , k + c\right) = \left(6 , - 7 + 2 \sqrt{3}\right)$

and F'$= \left(h , k - c\right) = \left(6 , - 7 - 2 \sqrt{3}\right)$

The vertices are

A$= \left(h , k + a\right) = \left(6 , - 7 + 4\right) = \left(6 , - 3\right)$

A'$= \left(h , k - a\right) = \left(6 , - 7 - 4\right) = \left(6 , - 11\right)$

B$= \left(h + b , k\right) = \left(6 + 2 , - 7\right) = \left(8 , - 7\right)$

and B'$= \left(h - b , k ,\right) = \left(6 - 2 , - 7\right) = \left(4 , - 7\right)$

graph{(x-6)^2/4+(y+7)^2/16=1 [-10.13, 15.18, -11.41, 1.25]}