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#1 2009-06-17 08:15:37

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Normals

Hey, first post for a while.

I did this question a while ago, but I've lost my solution and I can't seem to reproduce it neutral

__________________________________________________________________________

Show that the equation

has exactly one real solution if
. [this I can do: differentiating shows the curve is strictly increasing for non-negative p]

A parabola C is given parametrically by

where a is a positive constant.

Find an equation which must be satisfied by t at points on C at which the normal passes through the point (h,k). Hence show that, if

, exactly one normal to C will pass through (h,k) [this I can also do: the equation of the normal to C is
, and we can just sub in (h,k) to find the equation]

Find, in Cartesian form, the equation of the locus of the points from which exactly two normals can be drawn to C.

___________________________________________________________________________


I can't do the last bit this time round. I can see that the equation that must be satisfied by h and k needs to have a repeated root, but that's as far as I can get.

Thanks

Last edited by Daniel123 (2009-06-17 08:30:32)

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#2 2009-06-18 03:58:26

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Normals

No one?

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#3 2009-06-18 04:43:02

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Normals

Hi Daniel123;

x^3 + 3x + i =0 has no real roots. You are going to have to define q as real also.
The discriminant of x^3 +px + q =0 is always greater than 0 so this is another way to prove that it only has one real root.

Last edited by bobbym (2009-06-18 05:14:23)


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#4 2009-06-18 07:37:03

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Normals

Thanks. Yes, you're right. I think the question assumes the real numbers though. I wasn't aware cubics had discriminants, either.

You haven't really answered my question though; I'm having trouble finding the Cartesian equation of the locus. When I did this question originally (a few months ago), I remember getting the result that the locus was another parabola (which I thought was rather nice). I can't seem to do it again this time round though sad

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#5 2009-06-18 18:20:02

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Normals

Hi Daniel123;

I know I haven't, I am getting stuck there too.

I got confused trying to adapt what is on this page to your problem.

http://www.goiit.com/posts/list/geometry-show-that-the-locus-of-a-point-such-that-two-of-the-83783.htm

Last edited by bobbym (2009-06-18 19:27:45)


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#6 2009-06-18 20:15:33

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Normals

Okay, thanks for your effort smile. Thoughts so far below:

For the normal to C at some point (at^2,2at) to pass through a point (h,k), h and k must satisfy

. For exactly two normals to be drawn to the curve and pass through (h,k), there must be exactly two different values of t that satisfy this equation. For a cubic to have exactly two distinct roots, it must have a repeated root.

If we differentiate and set the derivative equal to 0,

. This equation gives the value of t corresponding to the two turning points on the curve, one of which is our double root. Solving for t gives
. In other words, when t is either of these values, the cubic we're interested in has exactly two real roots.

Last edited by Daniel123 (2009-06-18 20:22:57)

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#7 2009-06-18 20:41:28

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Normals

Ah. Substituting this back into the cubic and squaring gives

This isn't a parabola though. Maybe I got it wrong first time round.

Last edited by Daniel123 (2009-06-18 20:47:57)

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#8 2009-06-19 01:03:33

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Normals

Hi Daniel123;

I was looking at that exact equation on this page:
http://en.wikisource.org/wiki/1911_Encyclop%C3%A6dia_Britannica/Parabola
(about half way down the page) but I couldn't make it work. Good job!


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2009-06-19 03:55:49

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Normals

smile

Thanks for your help.

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#10 2009-06-19 07:53:28

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Normals

Hi Daniel123;

You did it all, but you did force me to learn quite a bit about parabolas and normals, and for that I can never forgive you.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#11 2009-06-20 05:51:27

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Normals

I still value the time and effort you spent trying to help smile

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#12 2009-06-20 20:25:40

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Normals

Hi Daniel123;

I guess there is only one reply to that, your welcome.

Last edited by bobbym (2009-06-20 21:47:49)


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Offline

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