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Reflective property of ellipse.

  1. Jan 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove the reflective property of the ellipse.

    2. Relevant equations
    x^2/a^2 + y^2/b^2 = 1 (equation of ellipse)
    tan theta = m2-m1/(1+m1m2)
    m = (y1-y0)/(x1-x0)

    3. The attempt at a solution
    I apologize if this is to simple for this forum, not sure where else to put it.

    My idea was to draw the tangent to the ellipse, and lines from the Focii F' and F to the tangent. Then prove that the angles formed by the intersection of F' and the tangent, F and the tangent, are equal. It isn't working out as I had hoped.

    Using a tangent to the first quadrant of the ellipse, F' is focus at (-c,0) and F is on the right, (c, 0):

    [tex]m_T = -\frac{b^2x_0}{a^2y_0}[/tex]
    [tex]m_{F} = \frac{y_0}{x_0-c}[/tex]
    [tex]m_{F'} = \frac{y_0}{x_0+c}[/tex]

    I got the slope of the tangent by differentiating the equation of the ellipse.

    [tex]\tan\theta = \frac{\frac{-b^2x_o}{a^2y_0}-\frac{y_0}{x_0-c}}{1-\frac{b^2x_0y_0}{a^2y_o(x_o-c)}}[/tex]

    Which, if my algebra is correct, simplifies to

    [tex]\tan\theta = \frac{-(b^2x_0(x_0-c) + a^2y_0)}{a^2y_0(x_0-c) - b^2x_0y_0}[/tex]

    A similar process leads to the angle between the tangent and the line from the left focus, F', with result:

    [tex]\tan\phi = \frac{-(b^2x_0(x_0+c)+a^2y_0)}{a^2y_0(x_0+c)-b^2x_0y_0}[/tex]

    Which don't seem to be equal :grumpy:

    Can someone point out where I have gone wrong?

    Thank you,
    Last edited: Jan 2, 2007
  2. jcsd
  3. Jan 3, 2007 #2


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    How did you get this equation? (It looks like a tangent of a sum of angles).
    It may be a good idea to check for sign errors (the slope is downwards, but the angles (and thus the tangents) are measured positively).

    An easier way to do the problem is to just use the principle of least time.
  4. Jan 3, 2007 #3
    That's the tangent equation I listed in the 'relevant' equations section. Well, even if I knew what the principle of least time was, I think I am supposed to do it somehow, along this fashion.

    I will check it over yet again...

    Thank you, Galileo.

  5. Jan 3, 2007 #4


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    The general idea you use to solve this question seems correct. I concur with Galileo about checking the signs and the math. In your formula for tan theta = m2-m1/(1+m1m2), make sure you know which slope is m1 and which is m2 so that the signs match.

    Unless you are specifically required to use calculus and slopes to show this, you can easily prove this using geometry and some basic properties of the ellipse.
  6. Jan 3, 2007 #5


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    You don't need to actually find the angles of incidence and reflection and show they are equal.

    If you find unit vectors in the direction from the 2 foci to a point on the ellipse, you can find a vector that bisects that angle. Then show it is perpendicular to the tangent.

    Using the parametric equation for the ellipse (x = a cos t, y = b sin t) may also help.
  7. Jan 4, 2007 #6
    Thank you all so much for your help. I found my problem.

    Solving the equation of the ellipse for y, and substituting this in the equations for the slope, along with the additional substitution of a^2-b^2=c^2, I was able to get




    This second angle (formed by the line from the left focus) is the supplement of the one of interest, so, so the negative of this is the tangent of that angle.

    Thanks again,
  8. Jan 6, 2007 #7


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    http://www.ies.co.jp/math/java/conics/focus_ellipse/focus_ellipse.html [Broken]
    Geometric Proof, Might be worth a look.
    Last edited by a moderator: May 2, 2017
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