Question

Write an equation for the hyperbola that satisfies the given set of conditions.

Vertices (16,0) and (-16,0) conjugate axis of length 16 units.

Answer 1

@Hero

Answer 2

@amistre64

Answer 3

what are your thoughts so far?

Answer 4

honestly I don't know where to start

Answer 5

start with graphing the verts

Answer 6

also, start with the basic hyperbola equation to fill in

Answer 7

I don't know what the hyperbola equation is :/

Answer 8

then you should look in your notes, or course material since they tell you these things before attempting to give you problems to solve.

Answer 9

you finding it, and typing it out does more for your learning then me just giving it over to you. which is why i need you to find it for us

Answer 10

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

Answer 11

thats fine, but we will have to adjust it for the information

personal preference is i like to keep as and xs lined up; and bs and ys lined since they are related by the graph

Answer 12

now, plug in one of your vertexes to see if we need to swap x and y signs

Answer 13

so 0/a^2-16^2/b^2=1?

Answer 14

good

does -16^2/b^2 = 1 have any real solutions?

Answer 15

b=16?

Answer 16

not what im asking,

im asking for some basic math knowledge here.

-16^2 = b^2 .... but any real number that is ^2 is positive.

can a positive number be a negative number at the same time?

Answer 17

I'm really confused right now

Answer 18

about what?

Answer 19

everything

Answer 20

what are a and b?

Answer 21

we havent determined that yet. we are trying to determine if the setup is valid:

\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]
if its not, then we need to use the other setup
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

which one should we use? well, we are testing your fist
idea; test the point (16,0)
\[\cancel{\frac{0^2}{a^2}}^{=0}-\frac{16^2}{b^2}=1\color{red}{\implies}-\frac{16^2}{b^2}=1\]
can b^2 = -16^2 ? can a positive value ever be a negative value at the same time?

Answer 22

yes

Answer 23

youll have to show me how ....

Answer 24

in that equation?

Answer 25

its the equation we need to use, so yeah

Answer 26

logically,

you cant have 4 apples and NOT have 4 apples
you cant have 2 birds and NOT have 2 apples
you cant NOT have 10 dollars and have 10 dollars

but you say we can :)

Answer 27

2nd line should be all birds :/

Answer 28

so the equation is invalid?

Answer 29

the setup is invalid so we use the alternative

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]
when y=0, x = 16 ... 16^2 = a^2, what is a?

Answer 30

16

Answer 31

yep

Answer 32

now w have to deal with what a conjugate axis is