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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!!
Solution
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