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The equation for any set of lines passing through an ellipse with the same slope

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data

    I was tutoring a student and could not answer one of his questions. Prove that y = mx +- square root( a^2*m^2 + b^2 ) is the equation for two lines passing through an ellipse

    2. Relevant equations

    (x/a)^2 + (y/b)^2 = 1 is the equation of an ellipse

    3. The attempt at a solution

    I started by suggesting he implicitly differentiate the equation for an ellipse and put it in the point slope equation, but it turned out to give an equation for dy/dx = m, but not the desired equation. I worked on different approaches at home for half an hour, but I'm stumped.
     
  2. jcsd
  3. Sep 29, 2007 #2
    Passing through an ellipse? As in they're tangent to it?
     
  4. Sep 29, 2007 #3

    HallsofIvy

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    This is a bit confusing. You title this "the equation for any set of lines passing through an ellipse with the same slope" but then give an equation for exactly two lines.

    Look at [itex]y = mx + \sqrt{ a^2*m^2 + b^2 }[/itex] and [itex]y = mx - \sqrt{ a^2*m^2 + b^2 }[/itex] separately. Where do they cross the ellipse [itex](x/a)^2 + (y/b)^2 = 1[/itex]? I think, like genneth, that it is quite possible they intend these to be the two lines that are tangent to the ellipse on opposite sides.
     
  5. Sep 30, 2007 #4
    Yeah, of course I meant tangent. don't be so nitpicky
     
  6. Sep 30, 2007 #5
    There's the boring and tedious way, which is to show that the lines intercept the ellipse at a single point -- substitute one into the other, show that the ensuing quadratic has a single repeated root. There's probably some other clever geometrical proof.
     
  7. Sep 30, 2007 #6

    HallsofIvy

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    Don't be so nitpicky? You leave out a crucial part of the problem, mis-state another part and you consider that nitpicky? And you are a mathematics tutor?
     
  8. Oct 1, 2007 #7
    So I take for one part: y² = (mx)² + 2mx√(a²m² + b²) + a²m² + b²
    substituting for y² in the equations for an ellipse I get:

    x²(1/a² + m/b²) + x[2m√(a²m² + b²)/b²] + a²m² = 0

    I don't see how to make it clear that this has a repeated root, and I'm unclear on why that means it is a tangent to the ellipse.
     
  9. Oct 3, 2007 #8
    Well, when does the equation [tex]ax^2 + bx + c = 0[/tex] have a repeated root? Chuck it into the quadratic formula and see if you can't spot something obvious.
     
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