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

]]>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.

]]>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

]]>Bob

]]>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.

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

Bob

]]>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.

]]>Here we are:

I leave the generalisation as an exercise for you.

Bob

]]>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

]]>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.

]]>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

]]>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.

]]>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.

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

]]>