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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Hello.

I would appreciate some help in better understanding the key relations between a rectangular hyperbola in standard form,

and a rational function of the form

such that both are symmetric in the same fasion.

What I know: The function f(x) is a hyperbolic that has been rotated 45 degrees and shifted up and over in some manner.

What I want to know: How do the terms a, p, q, and r relate and in what way does one convert from either form to the other and vice versa?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,929

hi Reuel,

I've never met that before. Are you sure that rational function is a hyperbola ?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Certainly. For instance, I believe its center is at

and the vertices, foci, etc. can be found.

*Last edited by Reuel (2013-09-22 07:19:20)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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Well, well well. You've taught me something new.

http://jwilson.coe.uga.edu/EMAT6680Fa11 … tions.html

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

According to your signature I have taught you nothing but have only helped you find it in yourself.

Thanks for the link. I was actually just exploring the topics discussed in the article myself a while ago, experimenting with what effect changing the different parameters has on the graph of a rational function.

So if anyone can explain how to go from one form to the other, from the standard form in terms of a (and maybe a b also) to a rational function, I would appreciate it.

*Last edited by Reuel (2013-09-22 07:27:24)*

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,030

They do not give the same graph. How do you expect to get one from the other???

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
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You should be able to work via the asymptotes. Find those for the rf and then construct a hyperbola with those.

There's a bit about that here

http://en.wikipedia.org/wiki/Hyperbola

Bob

edit:

Stefy wrote:

They do not give the same graph. How do you expect to get one from the other???

I think the idea is to transform one into the other.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

anonimnystefy: They are the same kind of graph only, as I have said, different by 45 degrees in a rotational sense. Suppose you want to get a rational function that is some hyperbola without knowing p, q, or r or anything about it such as its center, foci, vertices, etc. and all you have is a hyperbola in standard form in terms of a and perhaps b. How do you get the hyperbolic "standard" into the form of the rational function? That is the problem I am seeking help with.

Thanks to all who have and will reply. I am grateful for this forum and for your help.

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

bob bundy - I have already successfully converted from a given rational function to a standard hyperbolic by finding a and b (which turn out to be equal because the hyperbola is rectangular) but as I have said in my reply to another user I want to also know how to move in the other direction which is more difficult.

Perhaps a better way of stating my question would be to say that I would like to know how to rotate and shift a hyperbola into a rational function. I was hoping someone knew of a simple trick so I wouldn't have to get into coordinate transformations, though I am willing to learn anything I have to to solve a math problem.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,030

Reuel wrote:

anonimnystefy: They are the same kind of graph only, as I have said, different by 45 degrees in a rotational sense. Suppose you want to get a rational function that is some hyperbola without knowing p, q, or r or anything about it such as its center, foci, vertices, etc. and all you have is a hyperbola in standard form in terms of a and perhaps b. How do you get the hyperbolic "standard" into the form of the rational function? That is the problem I am seeking help with.

Thanks to all who have and will reply. I am grateful for this forum and for your help.

If by a rational function you mean a quotient of two polynomials, then what you are asking is impossible.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
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In general, you are correct. But it seems that a rf, in that format only, is a hyperbola, at least according to the page I linked earlier.

Reuel: You have sparked an interesting debate. I didn't even know about this so I have no quick tricks. If you have the transformation in one direction, you should be able to reverse it. ... I think ... maybe ...

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

A "proof" is here given. I put proof in quotations as this is rather inductive and I still do not know how to go from one to the other. It could be very difficult given the thing is full of transcendentals but maybe there is an easier way than the following I have yet to see. As it is, graphing can show the following is apparently accurate.

That a rational expression of the form

is a rectangular hyperbola of the standard form

...

The value of "a" has been analytically found to be that of

and the center (h,k) is given by

both of which are values crucial to the problem. Please note that "a" in the hyperbola is not the same "a" in f. I went ahead and used the same symbol as we use the hyperbolic a very little. The difference between them is obvious based on the context of my work.

We can solve the equation for y,

but as it is unclear what to do with that we can switch to a parametric form of the hyperbola

The transformation matrix for the rotation of the parametric system by some angle is that of

such that its application for a 7π/4 rotation upon our vector curve <x,y> yields the vector-valued functions

Next, the curve, while the correct shape, is in the wrong location on the plane and simply needs shifted over and up by the original center of the hyperbola which yields the solution

I apologize for any errors made in my work.

Example: If

then f is a rectangular hyperbola. In standard form that hyperbola has the equation

.Omitting all the steps already given in the general form above, the solution is

If f(x) is plotted so as to go from x = 0 to x = 5/3 the parametric equivalent is approximately t=-0.3398369094 to t=0.3398369094 or, if you prefer, the exact value is

.The reason for choosing these bounds is completely by preference; it is where the domain is equal to the range such that f(5/3) = 5/3.

*Last edited by Reuel (2013-09-24 03:06:43)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,929

hi Reuel,

The pair of graphs below show my trial and improvement attempt to find a hyperbola to fit y = 10/(5+3x).

If I'd got much closer the red line would blot the blue out completely so this moment seemed a good one to halt the search.

I'm reasoning like this:

equations of the form y = ax/(b + cx) will always have asymptotes that are parallel to the axes. Therefore a better standard hyperbolic form would be

(x+P)(y+Q) = R

as these will have the 'right' asymptotes.

I'll see if I can find the transform later today. Got to do some gardening now, while the weather holds.

Bob

The equation grapher is at http://www.mathsisfun.com/data/grapher-equation.html

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi Reuel,

Here we are:

I leave the generalisation as an exercise for you.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Oh, thanks. The generalization would be

.It seems I had another question about this subject but I forget now what it was. Oh well. If I think of it I'll speak up.

Thanks again for your input and feedback.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,929

OK, that looks good to me. And I seem (as you pointed out earlier) to have lived up to my signature. That's good.

No rush with your next question; I still have a lot of gardening to do.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Ah, I remember. If the rational expression is a hyperbola that means it is a conic section; if a conic section then it is defined as the intersection of a plane with a double cone.

What is the equation of that plane that intersects the cone - for f(x) not the standard form of a hyperbola? I have read about it on wikipedia but I am not sure how to go about finding the equation for the plane that represents f(x) itself which, as I have shown, is rotated and shifted.

Thanks. And good luck with your gardening. You must not have too much more to do given the time of year unless you live in a warm part of the globe. Where I live we are guaranteed frost-free-ness until early October or so. Or so I have heard. I don't have a garden so I don't pay that much attention.

*Last edited by Reuel (2013-09-26 01:38:16)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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I have a maths book that has the standard equations. It's in the loft so I'll have to go mountaineering. Gardening over for today, I've been laying a turfs for part of a new lawn.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,929

hi Reuel,

Apologies that so many days have passed since my last post. I've found the book and it doesn't have the equation of the cone I'm afraid. So maybe I'll have a go at working it out from scratch.

If so, I'd better be clear which hyperbola we're talking about.

I know now that

is a hyperbola with its lines of symmetry at 45 degrees to the axes, going through (0,0) and with a horizontal, and a vertical, asymptote.

The plane you're after; is it as follows ?:

(i) Find the equation of a cone

(ii) Find the equation of a plane.

(iii) Where this plane intersects the cone, that's the above hyperbola.

If so, I'll see if I can do (i) and (ii). It may be necessary to move the hyp. across so the cone, itself, has an axis as its vertical axis.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Hey.

I am also sorry for not being around. I have been really sick.

Yes, that would be the equation I am looking for. The cone and plane that intersect to form that particular hyperbola.

Thanks so much.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,929

hi Reuel,

Sorry to hear you've been sick. Hope you're better now.

I'll see what I can do about the equations. Never done this before so I can't promise anything.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 178

Thanks. I'm getting there.

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