How do you identify the vertices, foci, and direction of (x-3)^2/4-(y+1)^2/9=1(x3)24(y+1)29=1?

1 Answer
Oct 8, 2016

The direction is horizontal transverse
The vertices are at (1, -1)(1,1) and (5, -1)(5,1)
The foci are (3 - sqrt(13), -1)(313,1) and (3 + sqrt(13), -1)(3+13,1)

Explanation:

Please notice that the center of the hyperbola is (3, -1)(3,1).

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

The vertices are at (1, -1) and (5, -1)

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 and are the denominators in the equation.

c² = 4 + 9

c = sqrt(13)

The foci are (3 - sqrt(13), -1) and (3 + sqrt(13), -1)