Hyperbolic functions Watch

joyoustele
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How can I show that the hyperbola \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 can be represented in the form x=acoshu , y=bsinhu ?
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BuryMathsTutor
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(Original post by joyoustele)
How can I show that the hyperbola \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2  }=1 can be represented in the form x=acoshu , y=bsinhu ?
That's an ellipse.

You need \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.

Probably just a typo but if not that could be your problem.
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RDKGames
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(Original post by joyoustele)
How can I show that the hyperbola \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2  }=1 can be represented in the form x=acoshu , y=bsinhu ?
Substitute the x,y and show that the Cartesian eq. is satisfied for all u. Do note the comment above, though.
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joyoustele
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(Original post by BuryMathsTutor)
That's an ellipse.

You need \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.

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

Thanks again RDK
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RDKGames
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(Original post by joyoustele)
Oh, ops, I thought I had to prove the x values would be in the form x = a \cosh u and y = b \sinh u 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 x = a \cosh (2u), \quad y = b \sinh (2u) is valid as well. If u \in (-\infty, \infty) for your parameterisation, then it doesn't matter what you put in the argument in this case (as long as its surjective on (-\infty, \infty)) and we don't need to adjust the domain. If we had, say, u \in [-2,2] for your parameterisation, then this parameterisation would hold if we define u \in [-1,1] instead and produce the exact same thing.
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joyoustele
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(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 x = a \cosh (2u), \quad y = b \sinh (2u) is valid as well. If u \in (-\infty, \infty) for your parameterisation, then it doesn't matter what you put in the argument in this case (as long as its surjective on (-\infty, \infty)) and we don't need to adjust the domain. If we had, say, u \in [-2,2] for your parameterisation, then this parameterisation would hold if we define u \in [-1,1] instead and produce the exact same thing.
thanks so much.
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