Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

Answer 1

do you recall the general equation for a hyperbola ?

Answer 2

\[\frac{(x-h)^{2} }{ a ^{2} }-\frac{(y-k)^{2} }{ b ^{2} }\]

Answer 3

... yeah, and =1 , can you tell me from the given information where the center of this will be?

Answer 4

its at origin, so itll be, x^2/a^2 - y^2/b^2 =1

Answer 5

sorry its easy to type than use the equation tool

Answer 6

good, and one more bit of information from the givens .... will x or y be after the subtraction sign?

Answer 7

if its in standard for, x should be before the subtraction sign

Answer 8

standard for a hyperbola can be defined 2 different ways due to the subtraction operation:

we are given that vertexes that when x=0, y is some value:

(0/a)^2 - (y/b)^2 = 1

- (y/b)^2 = 1 is not possible among the real numbers, so we need to subtract the x parts instead of the y parts

(y/b)^2 - (x/a)^2 = 1; now this is doable

Answer 9

using the vertex again; we are able to define the b value .... can you tell me what b has to be?

Answer 10

um, well i know that c^2 = a^2 + b^2
so i can use that to find b
b^2=c^2=a^2

and c is the distance that the foci are from the center, which in this case is origin

Answer 11

sorry one sec

Answer 12

b^2=c^2-a^2

Answer 13

that was going to be my next step :) but we can determine the value for b before we go thru that; and will use it in the process to find a

Answer 14

so c=9

b^2= 81-a^2

Answer 15

ok

Answer 16

the vertex tells us the when x=0, y = +-6

( |6|/b)^2 - (0/a)^2 = 1

(6/b)^2 = 1 ; what is b now?

Answer 17

(6/b)^2 = 1
6/b = sqrt(1)
6/b = 1
6 = b

Answer 18

good, now we can use that to determine "a" :)

Answer 19

okay so,

81 = a^2 + 36
a^2 = 81-36
a^2 = 45
a= sqrt(45)

Answer 20

3sqrt(5)

Answer 21

we can usually stop at a^2 = 45 since its asking for the standard equation and not some distances: this gives us the completion as:\[\frac{y^2}{36}-\frac{x^2}{45}=1\]

Answer 22

thank you for all the help :D

Answer 23

youre welcome :)