Find the equation of the hyperbola with transverse axis on the y-axis, slope of asymptote numerically equal to nine-tenths of the conjugate axis, one vertex at (0,-9)?

- anonymous

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- anonymous

So let's try and sketch a picture first, would you like to try?
Just to help us get an idea what it looks like. So do you know what the transverse axis is?

- anonymous

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- anonymous

the ta is the x axis..

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- anonymous

no the y axis

- anonymous

That point is one of the vertices at 0,-9 and the transverse axis is the line that goes through the vertices. So it would be reasonable to assume that the other vertex is here. It looks like our hyperbola is centered at the origin. And then the slope it says the slope of an asymptote is 9/10 of the conjugate axis so lets see...

- anonymous

side to side axis..

- anonymous

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- anonymous

it will look something like that, so let me put the general formula on the board just a moment.

- anonymous

yeah thats what i did last night.. and stuck up in the slope..

- anonymous

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- anonymous

So that is the general equation for a hyperbola on that lies along the y azix for us, the center is the origin. So now we just need to find what a and b are right?
B is just the distance from the center to a vertex so that shouldn't be hard to figure out, right?

- anonymous

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- anonymous

i think -9 is a.. yes. b.

- anonymous

What I like to do is draw a box the slope of the red asymptote that is just b/a rise/run does that make sense?

- anonymous

Also, wouldn't b=9?

- anonymous

a=b.

- anonymous

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- anonymous

Well, we don't know yet that a and b are equal we need to use this piece of information now though slope of asymptote numerically equal to nine-tenths of the conjugate axis..

- anonymous

So how can we express that statement using an equation?
Any idea? [:

- anonymous

y=mx + b???

- anonymous

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- anonymous

Well, y=mx+b is the equation of a line, right?
So we need to solve this equation

- anonymous

b= 9

- anonymous

i sub the vertex (0, -9), m = 9/10 ..

- anonymous

b= -9

- anonymous

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- anonymous

Slope of asymptote is numerically equal to nine-tenths of the conjugate axis, that is what that statement means mathematically do you see?
Yes! B is 9

- anonymous

Now we are trying to find a now the conjugate axis has length 2a so we plug that in and now we have an equation we can solve for a can you help me solve that equation for a?

- anonymous

b is not -9?

- anonymous

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- anonymous

Well, a and b are always positive since they represent distances but in the end we are squaring them so b^2 and a^2 are always postive

- anonymous

Do you understand it now? [:

- anonymous

stupid me. hahaha yes

- anonymous

The key thing was the conjugate axis length is 2a so we are plugging in for it and now trying to solve for a would you like to help me solve for a there?

- anonymous

lol :P

- anonymous

where am i going to plug 2a?

- anonymous

well, we already did right here watch!

- anonymous

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- anonymous

Slope of asymptote numerically equal to nine-tenths of the conjugate axis that's the equation for that sentence the slope of the asymptote was 9/a so we are setting that equal to nine-tenths of the conjugate axis and plugging in 2a for the conjugate axis. Does that make sense?

- anonymous

?

- anonymous

2.23

- anonymous

ok let me check

- anonymous

Yes, that is correct however, it may be best to keep it in the exact form a is just square root of 5 right?

- anonymous

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- anonymous

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- anonymous

So that would be our equation for the hyperbola it all came down to finding a and b; that was the tricky part. Does this problem make sense to you now?

- anonymous

yeheeey!! thank you :) what if the problem was slope of asymptote equal to square toor 26 / 500 of distance of directrices, transverse axis = 20..

- anonymous

Your welcome! (:

- anonymous

well, is that another problem?

- anonymous

i mean square root of 26 over 500 (only the 26 is squared)

- anonymous

yes yes.. :(

- anonymous

Alright! [":

- anonymous

all i know is that a = 10

- anonymous

ok so the problem said a=10?

- anonymous

This one might be a little different, so if you don't mind could you type the exact words of the equation for me? [:

- anonymous

no no.. yes sure..

- anonymous

conjugate axis on the y-axis, slope of asymptote numerically equal to square root of 26 / 500 of the distance between the directices, transverse axis = 20.. (the squared only is 26, 500 is not included.. hehe)

- anonymous

Ok thanks, I think that helps!

- anonymous

So maybe first we should just try and sketch it to get a rough idea how it looks, would you like to give it a try?

- anonymous

I can do it myself if you want?
Its just the matter of you understanding the problem, since its yourz not mine. [:

- anonymous

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- anonymous

I think for this problem we will have to assume the hyperbola is centered at the origin and the conjugate axis is on the y axis, so that means the transverse axis is on what axis it is something like that the transverse axis is the x axis.

- anonymous

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- anonymous

so that would be our basic shape just a rough sketch do you agree?

- anonymous

so then we make our box

- anonymous

yes!!!!

- anonymous

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- anonymous

OK!

- anonymous

So the slope of the cross line is just b/a do you agree?
So i agree with you that a=10 that is correct

- anonymous

well, we need to find a formula for the distance between the directices but before we can do that let's sketch them in so we can see where they are!

- anonymous

any idea where they are located on the graph?

- anonymous

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- anonymous

they are 2 lines, like you have got above. 2 vertical lines in this case and they have a formula a/c and -a/c so the distance between those 2 is what?
it is just 2a/x so that is our equation and we are trying to solve for b so there is a formula for c do that look familiar?

- anonymous

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- anonymous

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- anonymous

does that Pythagoras theorem look familiar?

- anonymous

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- anonymous

So now we have an equation just in terms of b we have to try and solve that now, any ideas how to do that?

- anonymous

@silverxx ????

- anonymous

yes yes. sorry, i went to the kitchen. hehe

- anonymous

So how would you solve that?

- anonymous

solving for b here; it involves a bit of algebra but we could begin by squaring both sides to get rid of the square roots, did you want to try writing that for me?

- anonymous

no.. i'll do it. ;)

- anonymous

can you show me, then?

- anonymous

we just need to square each side, see if you can help me do that!

- anonymous

Are you fine with this problem now, do you think you can handle the rest?
If yes- then good luck!
Bye for now!

- anonymous

b is not a whole number..

- anonymous

that's right!
it doesn't work out to be a nice integer unfortunately, did you solve for b already?

- anonymous

not yet..

- anonymous

its 0.0416??

- anonymous

did you square both sides?

- anonymous

Yes, for b^2 that is what I got
Good job! [:

- anonymous

whats next?

- anonymous

yes!!! i can do it now. ;) thanks!!

- anonymous

last last question....... ??

- anonymous

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- anonymous

So we know b^2 and a^2 and we know the center is at the origin so we should be in good shape now, right. ? So lets write the equation can you help me do that?

- anonymous

x^2 - 25y^2 = 100

- anonymous

conjugate axis on the y-axis,, latus rectum 18/5 distance between directrices 25/17 square root 34..

- anonymous

Sorry for the inconvenience, but I won't have time for your very last question. Since I have other work to do at the moment, and I've already spent enough time with you.
Maybe we can work on this a bit later?

- anonymous

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