Using the information given below write the equation of the hyperbola:

Question

a). The hyperbola has vertices (–2, 9) and (–2, 3) and foci (–2, 13) and (–2, –1).

b). The transverse axis of the hyperbola lies on the line y=–3 and has length 6; the conjugate axis lies on the line x=2 and has length 8.

PLEASE HELP AND THANK YOU IF YOU DO!!

Answer 1

a) Using the vertices, we can find the center of the hyperbola. It is half way between them. It would be (-2,6). We can also use the vertices to get the length of the traverse axis, it would be 9-3 = 6/2 = a=3

Using the foci, we can find the length of the c term and that will allow us to calculate the b² term based on this formula b²=c²-a². 13-(-1) = 14/2 = 7 = c

b²=7²- 3² = 40

A hyperbola will always use subtraction. It will also always =1. The traverse axis is vertical, so the y term will be the first term.

The result:

(y-6)²/9 - (x+2)²/40 = 1

b) By giving the traverse and conjugate axis lines, you are able to figure out the center of the hyperbola. (2,-3) would be the center.

The length of the axis gives you the value for the denominators. They length needs to be divided by 2 and then squared. Denominator for the traverse axis is 6/2 = 3² and the conjugate axis is 8/2 = 4²

A hyperbola will always use subtraction. It will also always =1. The traverse axis is horizontal, so the x term will be the first term.

The result:

(x-2)²/16 - (y+3)²/9 = 1