Question

please help...
Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).

Answer 1

@openstudier101x

Answer 2

somebody... anybody please....

Answer 3

@amistre64 pleeeease...

Answer 4

@jim_thompson5910

Answer 5

@Loser66

Answer 6

its a hyperbola, whats so tough about it?

Answer 7

i just don't get precalc...
.-.

Answer 8

give me something to work with, what would you start with?

Answer 9

i don't even know how to begin this problem....

Answer 10

don't hate me for being annoying...

Answer 11

well we know its a hyperbola, so lets start with the equation for one ....

Answer 12

(x-h)^2/a^2 - (y-k)^2/b^2=1

Answer 13

good, at worst we might have to swap x and y parts ...

now tell me what you think h and k represent on our construction

Answer 14

h is +/- 2 and k is +/-11??

Answer 15

not quite, h and k are generally used for center point of a function, it where we grab hold of it to move it about the grid.

what is the center of our hyperbola? what point is the center between the verts?

Answer 16

(0,0)?

Answer 17

i'm telling you i don't know...

Answer 18

yep, seems fair, so h and k are 0 so we can ignore them

Answer 19

okay

Answer 20

x^2/a^2 - y^2/b^2 = 1

now plug in a vertex, its an x and a y value so lets use it to start adjusting this thing.

Answer 21

x^2/+/-2 -y^2/ +/-11+1

Answer 22

opps =1

Answer 23

just one point (0,2) is fine

0^2/a^2 - 2^2 /b^2 = 1

this simplifies to:

- 2^2/b^2 = 1

what do you spose b will be?

Answer 24

4

Answer 25

meh, b^2 = 4, so b=2

now we have something to adjust

- 2^2/2^2 = 1 gives us: -1=1 does this make any sense? or should y be positive instead?

Answer 26

idk man....
is this the final answer??
y^2/121 - x^2/4 =1

Answer 27

no, its not

Answer 28

okay is it y^2/4 - x^2/121 =1??

Answer 29

im not playing a guessing game, im taking my time trying to help you learn it .. if you dont want to take the time ill just go do something else.

Answer 30

okay :/

Answer 31

should i stay or go?

Answer 32

stay

Answer 33

ok, then we still have a question

- 4/4 = 1 is telling us that -1=1 which is not right. would you agree that this means that the y parts need to be positive instead?

Answer 34

yes

Answer 35

then lets adjust

instead of: x^2 /a^2 - y^2 /4 = 1 we need to swap the x and y parts

y^2/4 - x^2/a^2 = 1 now we have one part left to determine, and we can do that with the pythag thrm

the distance a focus is from the center is a hypotenuse, a and b are legs.

whats the distance to our focus?

Answer 36

ummm.... 11^2 so 121..??

Answer 37

well 11, but yeah

a^2 + b^2 = 11^2

a^2 = 121 - 4

Answer 38

so the final answer is Y^2/4-x^2/117=1?

Answer 39

thatll do it

Answer 40

thank you! :)

Answer 41

youre welcome