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Tangent to Ellipse

  1. Feb 20, 2013 #1
    An ellipse has the equation x^2+5y^2=5
    a line has the equation y=mx+c
    a) show that if the line is a tangent to the ellipse then c^2=5m^2+1
    b) hence find the equation of the tangent parallel to the line x-2y+1=0

    I tried to find the gradient of x^2+5y^2=5 at a point (x1,y1) and then put it into the equation (y-y1)=m(x-x1), but that didnt work out
    any ideas?
     
  2. jcsd
  3. Feb 20, 2013 #2

    Simon Bridge

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    If the line is a tangent to the ellipse, then it intersects at only one point.
     
  4. Feb 20, 2013 #3

    HallsofIvy

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    Simon Bridge's point is that y= mx+ c intersects [itex]x^2+ 5y^2= 5[/itex] then the quadratic equation [itex]x^2+ 5(mx+ c)^2= 5[/itex] has at least one solution. The line is tangent to the ellipse if and only if that quadratic equation has a single solution- its discriminant is 0. That's a method, due to Fermat, that predates Calculus.

    But you certainly can do this by finding the gradient of y. What did you get for the gradient?
     
  5. Feb 20, 2013 #4
    the derivative of x^2+5y^2=5 was dy/dx=-x/5y
     
  6. Feb 21, 2013 #5

    Simon Bridge

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    ... so, if (p,q) is a point on the ellipse, then the gradient of the tangent at that point is
    m=-p/5q and the equations of the tangent line is y=mx+c ... so how do you find c?

    I didn't know the other method was from Fermat :)
     
  7. Feb 21, 2013 #6
    ok, so I did
    y-y1=m(x-x1)
    y-q=-p/5q(x-p)
    y=-px/5q + p^2/5q + q
    y=(-p/5q)x + (p^2/5q + q)
    so then c=p^2/5q + q

    but then when I put that into c^2=5m^2+1, it isn't equal
    what did I do wrong?

    Thanks
     
  8. Feb 21, 2013 #7

    Simon Bridge

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    note: c=p^2/5q + q
    means: c=q(1+5m2) ... encouraging?
    but what you are looking for is c2.

    I, personally, wouldn't have tried it this way.
    If you use Fermat's approach, the relation just drops out.
     
  9. Feb 21, 2013 #8
    sorry, I'm starting to feel stupid now, but I just don't see how (q(1+5m^2))^2=5m^2+1. Also I'm not familiar with fermat's approach, what does that involve?
     
  10. Feb 21, 2013 #9

    Simon Bridge

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    It doesn't. You wouldn't expect it to because that would mean that c=c^2. It is quite close though.
    You just need to figure out how to get from c=q(1+5m^2) to c^2=1+5m^2 ... looks like we are both missing something. Like I said, I wouldn't normally do it this way.
    Hopefully HallsofIvy will pop back in to steer us right ;)
    Quadratic equation.
    See posts #2 and #3.
     
  11. Feb 21, 2013 #10

    HallsofIvy

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    Since you ask (hope I don't get into trouble for this!)

    Fermat's method would be this: the ellipse, [itex]x^2+ 5y^2= 5[/itex] and the line y= mx+ c will intersect when [itex]x^2+ 5(mx+ c)^2= x^2+ 5m^2x^2+ 10mcx+ 5c^2= 5[/itex] or [itex](1+ 5m^2)x^2+ 10mcx+ 5(c^2- 1)= 0[/itex], a quadratic equation for x. They will be tangent when that has a double root- when its discriminant is 0.

    The discriminant is
    [tex]100m^2c^2- 4(1+ 5m^2)5(c^2- 1)= 100m^2c^23- 20(c^2- 1+5m^2c^2- 5m^2)= 0[/tex]
    Can you simplify that?
     
  12. Feb 21, 2013 #11

    Simon Bridge

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    @HallsofIvy: I had hoped you show the next step for the differential method...
     
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