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Confocal conics problem

  1. Dec 25, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Through a given point in the plane of an ellipse prove that exactly two conics (one eliipse and the other hyperbola) confocal with the given ellipse can be drawn.

    2. Relevant equations

    3. The attempt at a solution
    Let the equation of given ellipse be
    [itex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1 [/itex]

    The equation of confocals with this ellipse is
    [itex] \dfrac{x^2}{a^2+\lambda}+\dfrac{y^2}{b^2+\lambda} = 1 [/itex]

    Let the given point be (α,β).
    But I have no idea how to proceed now? Should I substitute the given point in the equation to the confocals?
     
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  3. Dec 25, 2012 #2

    Simon Bridge

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    Recall: you are trying to find the equations of all the conic sections confocal with the ellipse given which also goes through point ##(\alpha,\beta)## ... you are expecting to find two of them.

    You already have the equation for all conic sections confocal with the given ellipse... a particular confocal conic section will have a particular value of ##\lambda##.
     
  4. Dec 26, 2012 #3

    utkarshakash

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    Does that mean I need to substitute the point in the equation of confocal and find λ from it?
     
  5. Dec 26, 2012 #4

    haruspex

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    Yes. You can see this will give you a quadratic in lambda, so two solutions, as desired.
     
  6. Dec 27, 2012 #5

    utkarshakash

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    [itex] \lambda^2 + \lambda(a^2+b^2-\alpha^2-\beta^2)+(a^2b^2-\alpha^2b^2-a^2\beta^2)=0[/itex]

    Don't expect me to solve this equation.... :cry:
     
  7. Dec 27, 2012 #6

    haruspex

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    Solving for lambda is easy - it's just a quadratic. OK, you get a fairly messy expression, but it is a pair of solutions, as required. It remains to show that one is an ellipse and the other a hyperbola.
     
  8. Dec 27, 2012 #7

    Dick

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    I don't think solving the equation is the best way to get the information you need. Write it in a little bit different form first. Suppose a>b and put c^2=a^2-b^2.

    [tex]\dfrac{x^2}{\lambda}+\dfrac{y^2}{\lambda-c^2} = 1[/tex]

    Your original conic is then the case lambda=a^2. Now you can see that if lambda>c^2, you have an ellipse and if 0<lambda<c^2 you have a hyperbola. Now if you turn that equation into a quadratic, your job is to show the equation has two positive roots, one of them larger than c^2 and one less than c^2. I think you've done a few problems before like this. There are tricks that don't involve staring at the quadratic formula.
     
    Last edited: Dec 27, 2012
  9. Dec 28, 2012 #8

    utkarshakash

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    Yes your expression seems very neat and your trick too is appreciable. Let me try out this.
     
  10. Dec 28, 2012 #9

    utkarshakash

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    OK I did what you said and got the following results

    First I converted it into a quadratic

    [itex]\lambda^2 - \lambda (x^2+y^2+c^2)+c^2x^2 = 0 [/itex]

    Now I investigate the nature of roots. Since product of roots is +ve and sum of roots is also +ve this means both roots must be +ve. Now c^2 must lie between the roots. This means f(c^2)<0. Plugging in c^2 in the original equation and simplifying I get f(c^2) = -c^2y^2 which is <0. So the required conditions are met. Tell me if this is correct or not.
     
  11. Dec 28, 2012 #10

    Dick

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    Sure. That's right. You might notice things go a little wrong if x=0 or y=0. The hyperbola becomes degenerate. Also the first argument is assuming that the two roots are real. They may be complex conjugates and still have positive product and sum. But you can use the f(c^2)<0 argument to fill that in.
     
    Last edited: Dec 28, 2012
  12. Dec 28, 2012 #11

    Simon Bridge

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    Of course, as phrased, you don't have to show that there are only those two curves that fit the criteria.

    You can find the foci of an ellipse given the semi-axes a and b.
    Knowing the foci and one point on the curve, you can find the equations for a hyperbola and an ellipse.

    Naturally there can be only the two - there are only two kinds of conic section which have two foci anyway so it becomes a matter of showing that the point provides a unique solution for each type?
     
  13. Dec 28, 2012 #12

    Dick

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    I think that is what we just showed. Sure, you can take the geometric approach. Works fine. But I just knew from looking at some of utkarshakash's past posts there has been a lot of stuff about determining properties of roots. Thought we should continue that way.
     
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