# Hyperbolic functionsWatch

Thread starter 1 week ago
#1
How can I show that the hyperbola can be represented in the form x=acoshu , y=bsinhu ?
Last edited by joyoustele; 1 week ago
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1 week ago
#2
(Original post by joyoustele)
How can I show that the hyperbola can be represented in the form x=acoshu , y=bsinhu ?
That's an ellipse.

You need .

Probably just a typo but if not that could be your problem.
1
1 week ago
#3
(Original post by joyoustele)
How can I show that the hyperbola can be represented in the form x=acoshu , y=bsinhu ?
Substitute the x,y and show that the Cartesian eq. is satisfied for all . Do note the comment above, though.
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Thread starter 1 week ago
#4
(Original post by BuryMathsTutor)
That's an ellipse.

You need .

Probably just a typo but if not that could be your problem.
Sorry, I typed the question in wrong.
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Thread starter 1 week ago
#5
(Original post by RDKGames)
Substitute the x,y and show that the Cartesian eq. is satisfied for all . Do note the comment above, though.
I dont understand what you mean by substitute the x,y ?
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1 week ago
#6
(Original post by joyoustele)
I dont understand what you mean by substitute the x,y ?
You have and ... so sub them in, and give a reason why the resulting equation holds for all .
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Thread starter 1 week ago
#7
(Original post by RDKGames)
You have and ... so sub them in, and give a reason why the resulting equation holds for all .
Oh, ops, I thought I had to prove the x values would be in the form and from the equation. I suppose thats how im supposed to do it.

Thanks again RDK
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1 week ago
#8
(Original post by joyoustele)
Oh, ops, I thought I had to prove the x values would be in the form and from the equation. I suppose thats how im supposed to do it.

Thanks again RDK
You can do that too, but (a) there are infinitely many ways in which you can parameterise this curve (*), and (b) it's much easier to just take the simple proposition like this and show that it holds in these cases.

(*) I.e. note that the parameterisation is valid as well. If for your parameterisation, then it doesn't matter what you put in the argument in this case (as long as its surjective on ) and we don't need to adjust the domain. If we had, say, for your parameterisation, then this parameterisation would hold if we define instead and produce the exact same thing.
Last edited by RDKGames; 1 week ago
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Thread starter 1 week ago
#9
(Original post by RDKGames)
You can do that too, but (a) there are infinitely many ways in which you can parameterise this curve (*), and (b) it's much easier to just take the simple proposition like this and show that it holds in these cases.

(*) I.e. note that the parameterisation is valid as well. If for your parameterisation, then it doesn't matter what you put in the argument in this case (as long as its surjective on ) and we don't need to adjust the domain. If we had, say, for your parameterisation, then this parameterisation would hold if we define instead and produce the exact same thing.
thanks so much.
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