# How do you identify the vertices, foci, and direction of (x-3)^2/4-(y+1)^2/9=1?

Oct 8, 2016

The direction is horizontal transverse
The vertices are at $\left(1 , - 1\right)$ and $\left(5 , - 1\right)$
The foci are $\left(3 - \sqrt{13} , - 1\right)$ and $\left(3 + \sqrt{13} , - 1\right)$

#### Explanation:

Please notice that the center of the hyperbola is $\left(3 , - 1\right)$.

To find the vertices, force the negative term to become zero by setting $y = - 1$:

((x - 3)²)/4 = 1

(x - 3)² = 4

We are about to take the square root of both sides so we need to write two equation one where the right side is 2 and the other where the right side is -2:

$x - 3 = 2$ and $x - 3 = - 2$

$x = 5 , \mathmr{and} x = 1$

The vertices are at $\left(1 , - 1\right)$ and $\left(5 , - 1\right)$

We know that the y coordinate of both foci is -1 but the x coordinate is plus or minus the distance c from the x coordinate of the center, 3 The formula for c is:

c² = a² + b² where a² and b² are the denominators in the equation.

c² = 4 + 9

$c = \sqrt{13}$

The foci are $\left(3 - \sqrt{13} , - 1\right)$ and $\left(3 + \sqrt{13} , - 1\right)$