anonymous
  • anonymous
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)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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
  • anonymous
|dw:1378615378731:dw|
anonymous
  • anonymous
the ta is the x axis..

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anonymous
  • anonymous
no the y axis
anonymous
  • 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
  • anonymous
side to side axis..
anonymous
  • anonymous
|dw:1378615543018:dw|
anonymous
  • anonymous
it will look something like that, so let me put the general formula on the board just a moment.
anonymous
  • anonymous
yeah thats what i did last night.. and stuck up in the slope..
anonymous
  • anonymous
|dw:1378615676603:dw|
anonymous
  • 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
  • anonymous
|dw:1378615820443:dw|
anonymous
  • anonymous
i think -9 is a.. yes. b.
anonymous
  • 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
  • anonymous
Also, wouldn't b=9?
anonymous
  • anonymous
a=b.
anonymous
  • anonymous
|dw:1378615988198:dw|
anonymous
  • 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
  • anonymous
So how can we express that statement using an equation? Any idea? [:
anonymous
  • anonymous
y=mx + b???
anonymous
  • anonymous
|dw:1378616221733:dw|
anonymous
  • anonymous
Well, y=mx+b is the equation of a line, right? So we need to solve this equation
anonymous
  • anonymous
b= 9
anonymous
  • anonymous
i sub the vertex (0, -9), m = 9/10 ..
anonymous
  • anonymous
b= -9
anonymous
  • anonymous
|dw:1378616293129:dw|
anonymous
  • 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
  • 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
  • anonymous
b is not -9?
anonymous
  • anonymous
|dw:1378616481969:dw|
anonymous
  • 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
  • anonymous
Do you understand it now? [:
anonymous
  • anonymous
stupid me. hahaha yes
anonymous
  • 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
  • anonymous
lol :P
anonymous
  • anonymous
where am i going to plug 2a?
anonymous
  • anonymous
well, we already did right here watch!
anonymous
  • anonymous
|dw:1378616757532:dw|
anonymous
  • 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
?
anonymous
  • anonymous
2.23
anonymous
  • anonymous
ok let me check
anonymous
  • 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
  • anonymous
|dw:1378617035849:dw|
anonymous
  • anonymous
|dw:1378617068701:dw|
anonymous
  • 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
  • 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
  • anonymous
Your welcome! (:
anonymous
  • anonymous
well, is that another problem?
anonymous
  • anonymous
i mean square root of 26 over 500 (only the 26 is squared)
anonymous
  • anonymous
yes yes.. :(
anonymous
  • anonymous
Alright! [":
anonymous
  • anonymous
all i know is that a = 10
anonymous
  • anonymous
ok so the problem said a=10?
anonymous
  • 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
  • anonymous
no no.. yes sure..
anonymous
  • 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
  • anonymous
Ok thanks, I think that helps!
anonymous
  • 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
  • anonymous
I can do it myself if you want? Its just the matter of you understanding the problem, since its yourz not mine. [:
anonymous
  • anonymous
|dw:1378714600249:dw|
anonymous
  • 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
  • anonymous
|dw:1378617844006:dw|
anonymous
  • anonymous
so that would be our basic shape just a rough sketch do you agree?
anonymous
  • anonymous
so then we make our box
anonymous
  • anonymous
yes!!!!
anonymous
  • anonymous
|dw:1378617943603:dw|
anonymous
  • anonymous
OK!
anonymous
  • 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
  • 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
  • anonymous
any idea where they are located on the graph?
anonymous
  • anonymous
|dw:1378715183660:dw|
anonymous
  • 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
  • anonymous
|dw:1378618618414:dw|
anonymous
  • anonymous
|dw:1378618725220:dw|
anonymous
  • anonymous
does that Pythagoras theorem look familiar?
anonymous
  • anonymous
|dw:1378618774215:dw|
anonymous
  • 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
  • anonymous
@silverxx ????
anonymous
  • anonymous
yes yes. sorry, i went to the kitchen. hehe
anonymous
  • anonymous
So how would you solve that?
anonymous
  • 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
  • anonymous
no.. i'll do it. ;)
anonymous
  • anonymous
can you show me, then?
anonymous
  • anonymous
we just need to square each side, see if you can help me do that!
anonymous
  • 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
  • anonymous
b is not a whole number..
anonymous
  • anonymous
that's right! it doesn't work out to be a nice integer unfortunately, did you solve for b already?
anonymous
  • anonymous
not yet..
anonymous
  • anonymous
its 0.0416??
anonymous
  • anonymous
did you square both sides?
anonymous
  • anonymous
Yes, for b^2 that is what I got Good job! [:
anonymous
  • anonymous
whats next?
anonymous
  • anonymous
yes!!! i can do it now. ;) thanks!!
anonymous
  • anonymous
last last question....... ??
anonymous
  • anonymous
|dw:1378619464324:dw|
anonymous
  • 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
  • anonymous
x^2 - 25y^2 = 100
anonymous
  • anonymous
conjugate axis on the y-axis,, latus rectum 18/5 distance between directrices 25/17 square root 34..
anonymous
  • 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
  • anonymous
|dw:1378619656140:dw|