Question

What are the vertices of the hyperbola given by the equation [(y-4)^2/121]-[(x+9)^2/144] = 1?

Answer 1

Vertices are a distance, 'a' away from the centre. Remember the std eqn: \[(y-k)^2/a^2-(x-h)^2/b^2=1\]

Answer 2

ok the distance A away so 121 is the vertice?

Answer 3

no i only have coordinates in the answers.

Answer 4

121=a^2. You have to take the sqrt. You know the hyperbola goes up & down because its y-x. you also know the center, (h,k). The vertices are 'a' away from the centre. Like, (h,k) +- (0,a)

Answer 5

k so the center is (9,4) and the vertice is 11 away from the center. but which way?

Answer 6

here is everything you need

Answer 7

http://www.wolframalpha.com/input/?i=%5B%28y-4%29%5E2%2F121%5D-%5B%28x%2B9%29%5E2%2F144%5D+%3D+1

Answer 8

These are the answers i have . Nothing anyone's told me looks like these :(
(-9, 16) and (-9, -8)

(-9, 15) and (-9, -7)

(2, 4) and (-20, 4)

(3, 4) and (-21, 4)

Answer 9

Do you know if the hyperbola is going up or down?

Answer 10

no I'm sorry. I may have posted the question wierd. here's just copy and paste
What are the vertices of the hyperbola given by the equation open parentheses y minus 4 close parentheses squared over 121 minus open parentheses x plus 9 close parentheses squared over 144 = 1?

Answer 11

Just give me a moment......

Answer 12

Sorry, but is this algebra 2, no?

Answer 13

Yup

Answer 14

Ohhhh.... ok. At first I thought it was Algebra 1, and right now, I'm completing geometry, so I won't be able to help with Algebra 2 until later this month. I'm so so sorry.

Answer 15

It's ok. do you know anyone else online that could help though?

Answer 16

Umm... Hero, Calcmathlete, uri, Loujoelou, ParthKohli.... AccessDenied, and a couple of others....

Answer 17

u still need help?

Answer 18

yes please!

Answer 19

first establish which variable can be zero for the equation to still make sense

Answer 20

if y=0; -x^2=1 makes no "real" sense; so the hyper opens along parallel to the y axis .... up and down

Answer 21

the vertex parts can then be evaluated by ignoring the x part and solving for y

Answer 22

does x still =0 or was that just for the beggining?

Answer 23

(y-4)^2/121 = 1

(y-4)^2 = 121

y-4 = +- 11

y = 4 +- 11

Answer 24

the vertexes are only position along the axis assosiated with the y in this case; so the x part is pointless; zero out the x part and we can determine the verts

Answer 25

oh ok

Answer 26

4+11 = 15
4-11 = -7

only one option has those as y part values

Answer 27

oh so the answer is b!

Answer 28

correct

Answer 29

what was RKfitz talking about before?

Answer 30

prolly something more technical ....

Answer 31

Ok thank you SO much. <3

Answer 32

youre welcome, and good luck