Question

PLEASE MAJOR HELP
Find the standard equation of the hyperbola given vertices of(0.+/-6) and asymptotes at y=+/-(3/5)x

Answer 1

the center of the vertex is the center of the hyper ... so we should be able to see that this is at least centered at the origin; and since the y parts have value they need to be the positive portion of the setup:\[\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\]

Answer 2

when x=0, y=+-6; so we can solve for b:\[\frac{6^2}{b^2}-\frac{0^2}{a^2}=1\]
\[\frac{6^2}{b^2}=1~:~b=6\]

Answer 3

the asymptote gives us a ratio of b to a; its the slope is defined as b'/a' for a given b and a

Answer 4

this gives us the follwing relationship
\[\frac ba:~\frac 35=\frac 6a\]

Answer 5

I'm confused..

Answer 6

youll have to show me where your confused then ...

Answer 7

haha everything! I'm not sure how to start with the..6^2/b^2 thing

Answer 8

lets start this over again then, tell me what you know of as the general setup for a basic hyperB equation

Answer 9

well that depends on y-axis or x-axis. im guessing this is y-axis so it's y^2/a^2-x^2/b^2

Answer 10

thats a good thought, but lets keep this easier on us: do you recall that the slope of a line is defined as a relationship of: y/x ?

Answer 11

if we standardize the b and a to a particular x or y part, then we dont have to play the mental juggling games with this

Answer 12

the general setup for a hyperbola can be expressed as:
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]
and the x and y parts are positioned according to the vertex values: lets use (0,6) to adjust it
\[\frac{0^2}{a^2}-\frac{6^2}{b^2}=1\]
\[-\frac{6^2}{b^2}=1\]
this is not doable so that tells us to swap the x and y parts and try again
\[\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\]
\[\frac{6^2}{b^2}-\frac{0^2}{a^2}=1\]
\[\frac{6^2}{b^2}=1\]
this IS doable, so we can solve for b

Answer 13

how do i solve for b..?

Answer 14

do you remember learning that anything divided by itself is equal to 1?

Answer 15

3/3 = 1

16/16 = 1

-35/-35 = 1

etc ...

Answer 16

0/0 is not 1 so thats the only exception :)

Answer 17

so that's the case for 6? then where do I go from there

Answer 18

plug it into the setup: b=6\[\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\]
\[\frac{y^2}{6^2}-\frac{x^2}{a^2}=1\]

now all thats left is to determine the value of "a"

Answer 19

they give us lines for the asymptote .... how do we define the slope of a line between 2 points?

Answer 20

oh god..i have no idea hah i haven't done that in so long

Answer 21

lets make it simple and say that one of the points is the origin, and the other is some generic (x,y)

the slope is defined as the change in y, divided by the change in x:\[slope=\frac{y-0}{x-0}=\frac yx\]

Answer 22

okauy but how does that help find a?

Answer 23

notice the equation we have setup:
\[\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\]
b relates to y; as a relates to x; so would you agree that the slope (y/x) of the asymptote line is related to: b/a ??

Answer 24

yes

Answer 25

and can you tell me what the slope of our asymptote line is given to be?

Answer 26

6^2/a^2?

Answer 27

... and asymptotes at y=+-(3/5)x

lets say that the slope of the asymptote line is: 3/5

Answer 28

\[\frac{3}{5}=\frac ba\]
we know b=6
\[\frac{3}{5}=\frac 6a\]
solve it for a