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Thanks. I'm getting there.
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.
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.
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.
OK, that looks good to me. And I seem (as you pointed out earlier) to have lived up to my signature. That's good.
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.
I leave the generalisation as an exercise for you.
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.
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.
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.
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.
If by a rational function you mean a quotient of two polynomials, then what you are asking is impossible.
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.
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.