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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

{7/3}
2013-06-23 11:06:51

If i can prove that property,i won't need eccentricity for the equation.
Here's a version of the statement for ellipse,the tangents to the end of minor axis lie on the intersections of circles containing ends of major axis and perpendinculars on foci.

{7/3}
2013-06-22 11:04:31

Oh,but how did that site derive the first property?

bob bundy
2013-06-22 02:28:12

hi

Good site.  I like the animations.  smile

Some people have trouble with infinity being a 'number' or, in this case, a position.  So the tangent at infinity definition is neat but difficult to make rigorous.

Having said that I think it would be possible to argue that my curve



has y = x as its tangent at infinity.

Please don't ask me to do so.  smile  I've just spent most of the day checking and filing paperwork and now my head is like this  dizzy

Bob

{7/3}
2013-06-22 01:25:32

Check this- http://www3.ul.ie/~rynnet/swconics/HP%27s.htm first property how was it derived???

{7/3}
2013-06-22 01:18:38

I like the definition that asymptote is tangent to the curve at infinity.
Those are some cool functions,but does the tangent definition apply for them?if not which definition does?

bob bundy
2013-06-21 23:03:57

I've got a better one, in that the curve gets closer and closer to y = x as x tends to infinity.

Bob

bob bundy
2013-06-21 23:00:34

here's one to ponder:

y = x + cos(x)

Is y = x an asymptote ?

Bob

bob bundy
2013-06-21 22:49:29

Arhh.  Now that is a question.

In my opinion, there are no absolutes when it comes to mathematical definitions, although it obviously helps if there is some agreement, or at least, that two definitions are equivalent.  Here's a selection:

http://www.wolframalpha.com/input/?i=as … MathWorld-

https://en.wikipedia.org/wiki/Asymptote

http://www.mathsisfun.com/definitions/asymptote.html

Some include the requirement that the curve never reaches the line.  I'm just wondering if I can construct a function that has an asymptote but also crosses that line elsewhere.  I think so, but I haven't found one yet.

Bob

{7/3}
2013-06-21 18:11:10

Bob,one last question ,what is the definition of an asymptote,there are many on the net,but which one is correct

{7/3}
2013-06-19 20:57:56

Thanks

bob bundy
2013-06-19 20:42:04

hi {7/3}

That's ok but it is difficult to know the properties of the asymptotes before you have the equation. (Strictly, you don't know it has any yet.)

So, my advice is to steer away from b anyway.

The hyperbola has the property



I'm assuming you can use that in the same way we used the



for an ellipse.

From that you can show that the constant is 2a.

The focus and directrix properties will follow once you have the equation, but you might as well define some things in preparation.

eg.  Define c to be ae for some e > 1

Start with (1) and substitute in what we have ( a, c etc)

Simplify as before.

Define b^2 = a^2(e^2 - 1) and the right equation should arrive after some work.  (I haven't tried it yet smile )

This definition for b is actually equivalent to the formula you wanted to prove but the asymptote properties will come as a consequence rather than as a starting point.

Bob

{7/3}
2013-06-19 19:05:57

I define a as the 1/2 of the distance of the two vertace and b as 1/2 of the line segment that is perpendicular to major axis,passes through center and its length corresponds to height of asymptotes over/under a vertex

bob bundy
2013-06-19 16:59:58

hi {7/3}

So did you start with Dandelin spheres again? 

My text book says for a hyperbola



That should enable you to introduce c.

Is that your starting point?

Then, how are you defining b ?

I think you want



e > 0

Bob

{7/3}
2013-06-19 03:06:52

Sorry for not being clear,c is distance of foci from center.if i assume a^2+b^2=c^2 i can derive the equation but how do i prove a^2+b^2=c^2?
[i tried searching in the net,all the websites i've been to say that proving it is tough and i should take it for granted]

bob bundy
2013-06-19 00:52:11

hi {7/3}

What are you starting with?  Not sure what you mean by c^2

Bob

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