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

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

__________________________________________________________________________

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

No one?

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

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

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

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

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

Okay, thanks for your effort . 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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**Daniel123****Member**- Registered: 2007-05-23
- Posts: 663

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

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

Thanks for your help.

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

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

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

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

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.**

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