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**{7/3}****Member**- Registered: 2013-02-11
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How do you prove that the ratio of a conic section's points' distance from focus and distance from directrix is constant using dandelin spheres?

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi {7/3},

I'd not heard of them so I did a search.

These first two links just do the ellipse and so I was beginning to think that was all, but the third link has a tantalising suggestion that, once you've done the ellipse, the others are easy. Best of luck!

The third looks the easiest to follow.

http://mathworld.wolfram.com/DandelinSpheres.html

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

http://www.mathacademy.com/pr/prime/art … /index.asp

Bob

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

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**{7/3}****Member**- Registered: 2013-02-11
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Sorry,bob but those links prove sum of distance property,not the property i want

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**bob bundy****Moderator**- Registered: 2010-06-20
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Sure but your property then follows.

Bob

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

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

How?

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi {7/3}

Sorry it's been a while. My book's definition starts with the eccentricity property and deduces, firstly the equation, then the sum of distances. It comes out fairly easily this way round.

This morning I started with the sum of distances and tried to work the other way. You'd think it would just be a matter of reversing the steps. At least, that's what I thought when I casually said "your property then follows". Huh! Should have kept quiet, shouldn't I ?

I have a page, densely covered with algebra. The formula nearly dropped out but for one erroneous term. Since I know it will work I've just got to find the error. Then, armed with the formula I will be able to reverse the book proof and get to the eccentricity property. Then convert it to Latex.

I will get there, and I'll be pleased with myself for the method, but it is going to take a few hours yet.

Bob

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

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**{7/3}****Member**- Registered: 2013-02-11
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Found a link it doesn't use algebra-http://xahlee.info/SpecialPlaneCurves_d … ction.html [the link doesn't prove theta is constant for any point ]

*Last edited by {7/3} (2013-06-09 00:33:33)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi {7/3}

Thanks for the link; that's a great page. Don't know if you still want my proof, but I'll do it anyway for my own satisfaction.

Diagram below.

A planar curve has the property that points on the curve, P, follow this rule:

The following properties can be deduced immediately:

If F'F is extended in both directions there will be a point, A, on this line that lies on the curve.

Construct a second line, the perpendicular bisector of F'F.

Let these two lines be the x axis and y axis respectively.

Let F and F' have coordinates (s,0) and (-s,0) respectively.

If P (x,y) is a point on the curve then it is easy using symmetry to see that (x,-y) (-x,y) and (-x,-y) are also on the curve.

Let A be the point (a,0) and let B (on the curve) be the point (o,b)

A is on the curve so F'A + AF = constant = (a + s) + (a - s) = 2a. So the constant is 2a.

let e < 1 be such that b = a(1 - e) then

Let MD be the line

where D is on the x axis and M is the point with the same y coordinate as P.

I think that is the preliminaries dealt with. Now I'm going to work out the equation of this curve.

From the basic property:

Hold onto your hats. This is where it gets hairy!!

Cancelling spree follows

but

so

divide by a^2b^2

I think I need a break now ..... back later to finish off the eccentricity bit

Bob

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**{7/3}****Member**- Registered: 2013-02-11
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That is some proof ! I think i'm going to lose my hair thinking about it. But it is an interesting proof at least now i know how the equation comes

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**{7/3}****Member**- Registered: 2013-02-11
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Hey bob,I tried to derive the equation of hyperbola like this,I've got everything except a^2+b^2=c^2. How do i prove it?

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi {7/3}

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

Bob

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**{7/3}****Member**- Registered: 2013-02-11
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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]

*Last edited by {7/3} (2013-06-18 05:10:53)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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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

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**{7/3}****Member**- Registered: 2013-02-11
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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

*Last edited by {7/3} (2013-06-18 21:09:59)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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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 )

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

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**{7/3}****Member**- Registered: 2013-02-11
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Thanks

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**{7/3}****Member**- Registered: 2013-02-11
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Bob,one last question ,what is the definition of an asymptote,there are many on the net,but which one is correct

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**bob bundy****Moderator**- Registered: 2010-06-20
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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

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**bob bundy****Moderator**- Registered: 2010-06-20
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here's one to ponder:

y = x + cos(x)

Is y = x an asymptote ?

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
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I've got a better one, in that the curve gets closer and closer to y = x as x tends to infinity.

Bob

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**{7/3}****Member**- Registered: 2013-02-11
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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?

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

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

*Last edited by {7/3} (2013-06-21 03:40:31)*

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

Good site. I like the animations.

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. I've just spent most of the day checking and filing paperwork and now my head is like this

Bob

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**{7/3}****Member**- Registered: 2013-02-11
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Oh,but how did that site derive the first property?

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**{7/3}****Member**- Registered: 2013-02-11
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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.

*Last edited by {7/3} (2013-06-22 13:09:51)*

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