Hyperbola conjugate,a>b questions

[sachin123]

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.

 

[tiny-tim]

hi sachin123!

(try using the X² icon just above the Reply box )

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 )

And the conjugate hyperbola simply interchanges a and b in every respect.

 

[sachin123]

Hey tiny_tim
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?

 

[tiny-tim]

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

It should be 1 + b²/a² … see the PF Library on eccentricity

 

[sachin123]

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?

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.

 

[tiny-tim]

hi sachin123!

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?

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

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.

 

[sachin123]

Thank You tim