Question

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

Answer 1

alright then first thing to take note of is the vertices are at (0, 2) and 0, -2)

Answer 2

Okay, got that. totally lost after that

Answer 3

so
\[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]

Answer 4

okay whats the next step?? sorry im so confused with this

Answer 5

our "a" = 2

because the major axis is the one with the vertices.

Answer 6

so the b will be -2

Answer 7

dunno that yet

Answer 8

from point (0,2) to point (0,-2) = 2a

Answer 9

the total length is 4, so a =2

Answer 10

okay got it

Answer 11

the foci are at (0,c) and (0,-c)

Answer 12

the foci and the vertices share the same x value in the (x,y) because the hyperbola is vertical

Answer 13

so c = 11 or (0,c)

Answer 14

\[c^2=a^2+b^2\]

Answer 15

\[11^2=2^2+b^2\]

Answer 16

ohhhh

Answer 17

i think im getting this now

Answer 18

hold on though, gonna check this make sure im right first

Answer 19

okay

Answer 20

ya

Answer 21

ok i think im getting the hang of it, im gonna show you the answer choices i have

Answer 22

y squared over 4 minus x squared over 117 equals 1

y squared over 4 minus x squared over 121 equals 1

y squared over 121 minus x squared over 4 equals 1

y squared over 117 minus x squared over 4 equals 1

Answer 23

\[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]

plug the values in for a and b

Answer 24

\[\frac{ y^2 }{ 4 } - \frac{ x^2 }{ ? } = 1\]

Answer 25

\[11^2=2^2+b^2\]

Answer 26

121?

Answer 27

\[121 = 4 + b^2\]

Answer 28

subtract the 4 first to isolate b^2

Answer 29

117

Answer 30

yep

Answer 31

so

y squared over 4 minus x squared over 117 equals 1

Answer 32

but remember in this case, the major axis was the vertical, if its the horizontal its done differently.

Answer 33

but ya thats right

Answer 34

thanks man! I think i gave you a medal, im a newbie to this site

Answer 35

yep, good luck