Hyperbola conjugate,a>b questions

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Discussion Overview

The discussion revolves around the properties and equations of hyperbolas, particularly focusing on the relationship between the parameters \(a\), \(b\), and the eccentricity \(e\). Participants explore the implications of the relative sizes of \(a\) and \(b\), the equations for conjugate hyperbolas, and specific examples related to eccentricity calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the equation \(b^2 = a^2(e^2 - 1)\) holds when \(a\) is greater than \(b\).
  • Another participant asserts that the equation applies for any values of \(a\) and \(b\), suggesting that \(e^2 - 1\) can be any positive number.
  • A participant raises a concern about a specific example involving the hyperbola \(x^2/5 - y^2/3 = 1\) and its eccentricity, questioning the correctness of the eccentricity calculation using the formula \(b^2 = a^2(e^2 - 1)\).
  • There is a correction regarding the formula for eccentricity, with one participant indicating that the correct form should be \(e^2 = 1 + b^2/a^2\).
  • Another participant confirms their understanding of the conjugate hyperbola, stating that it can be derived by changing the sign in the equation from \(x^2/a^2 - y^2/b^2 = 1\) to \(x^2/a^2 - y^2/b^2 = -1\), while also noting that switching \(a\) and \(b\) leads to a different hyperbola.
  • There is a consensus that conjugate hyperbolas share the same asymptotes and that the sign change is the primary difference in their equations.

Areas of Agreement / Disagreement

Participants express differing views on the application of the eccentricity formula and the relationship between \(a\) and \(b\) in hyperbolas. While some participants agree on the nature of conjugate hyperbolas, the discussion remains unresolved regarding the eccentricity calculations and the implications of the equations presented.

Contextual Notes

Participants have not fully resolved the implications of the eccentricity formula in different contexts, and there are uncertainties regarding the specific values of \(a\) and \(b\) in relation to the examples discussed.

sachin123
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1)In a hyperbola,
x^2/a^2-y^2/b^2=1(standard form),
b^2=a^2(e^2-1)
This is in the case where b is greater than a.
But if a is greater than b?
Will that hold good correctly?

2)If you consider the conjugate hyperbola(of the standard form),what will be equation relating a,b and e?
Will it be b^2=a^2(e^2-1) or a^2=b^2(e^2-1)?
Some explanation would be helpful.This isn't my homework.I was getting most of my problems wrong,so I wanted to clarify some general things.
Thank You.
 
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hi sachin123! :smile:

(try using the X2 icon just above the Reply box :wink:)
sachin123 said:
b^2=a^2(e^2-1)
This is in the case where b is greater than a.

No, this works for any a and b.

e2 - 1 can be any positive number.

(se the PF Library on eccentricity for some more details :wink:)

And the conjugate hyperbola simply interchanges a and b in every respect.
 
Hey tiny_tim:smile:
Thanks a lot for your help.But I saw a problem in my book,it goes:
x2/5-y2/3=1 is a hyperbola.Is it's eccentricity equal to root(8/5)?
Answer given is NO.
How is so?If I use b2=a2(e2-1),I get root(8/5).
Is it wrong then?
 
sachin123 said:
x2/5-y2/3=1 is a hyperbola.Is it's eccentricity equal to root(8/5)?
Answer given is NO.
How is so?If I use b2=a2(e2-1),I get root(8/5).
Is it wrong then?

You seem to be using e2 = (b2 - 1)/a2 :redface:

It should be 1 + b2/a2 … see the PF Library on eccentricity :wink:
 
Hi tiny_tim
I used e2=b2/a2+1
In the problem stated,
b2=3,a2=5.Correct?
SO when I put them in the above equation,e2 becomes 8/5 correct?
But my book says that isn't the answer(its a yes or no question).
Where am I missing?:confused:

Also,to find the conjugate hyperbola of a given hyperbola,
x2/a2-y2/b2=1, we just have to write,
x2/a2-y2/b2=-1.Am I correct?

If we write,x2/b2-y2/a2=1,it would be totally different hyperbola wouldn't it and it wouldn't be the conjugate ...correct?

Thanks a lot.
 
hi sachin123! :smile:
sachin123 said:
Hi tiny_tim
I used e2=b2/a2+1
In the problem stated,
b2=3,a2=5.Correct?
SO when I put them in the above equation,e2 becomes 8/5 correct?
But my book says that isn't the answer(its a yes or no question).
Where am I missing?:confused:

Yes, you're right … I was completely confused about what a and b are. :redface:

On second thoughts, the book looks wrong to me.
Also,to find the conjugate hyperbola of a given hyperbola,
x2/a2-y2/b2=1, we just have to write,
x2/a2-y2/b2=-1.Am I correct?

If we write,x2/b2-y2/a2=1,it would be totally different hyperbola wouldn't it and it wouldn't be the conjugate ...correct?

Yes, right again …

conjugate hyperbolas have the same asymptotes, so it's only the sign that changes in the equation.

a and b are exchanged in the eccentricity, but not in the equation of the hyperbola itself. :smile:
 
Thank You tim :smile:
 

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