Lv 7
Pope asked in Science & MathematicsMathematics · 10 years ago

Reflections in an ellipse?

A ray is projected from a focus of an ellipse and reflected at the point where it intersects the ellipse. The path of the reflection goes through the other focus. This reflective property applies to all ellipses. But what happens after that? Continuing through the focus and onto the ellipse, it is reflected again and returns to the first focus, and so on.

In one trivial case, the first ray is coincident with the major axis, and the path is restricted to that axis. Are there any other cases in which the path would retrace itself? Does the path approach a stable orbit?

1 Answer

  • Indica
    Lv 7
    10 years ago
    Favorite Answer

    Parameterise ellipse as ( acos(t), bsin(t) ) and let the foci be F(ae,0) and F'(−ae,0).

    Let the beam originate from F and first hit the ellipse at P (p). After reflection the beam will pass through F' and meet the ellipse again at Q (q).

    ∴ bsin(q) / { acos(q)+ae } = bsin(p) / { acos(p)+ae }

    → sin(q) / { cos(q)+e } = sin(p) / { cos(p)+e }

    → {e+cos(p)}sin(q) − sin(p)cos(q) − esin(p) = 0

    Now use U=tan(q/2) and above becomes after some algebra

    (1−e)sin(p)U² + 2{e+cos(p)}U − (1+e)sin(p) = 0

    One solution of this is tan(p/2) so the other solution tan(q/2) must be such that

    tan(p/2)tan(q/2) = −(1+e)sin(p)/{(1−e)sin(p)} = −(1+e)/(1−e) … (i)

    For a focal chord QF through F(ae,0) the argument is similar with e replaced by −e

    Hence if QF meets the ellipse again at R (r) then

    tan(q/2)tan(r/2) = −(1−e)/(1+e) … (ii)

    Combining (i) & (ii), tan(r/2) = {(1−e)/(1+e)}².tan(p/2)

    This is after 2 reflections. Applying this repeatedly and noting (1−e)/(1+e)<1 shows that tan(r/2)→0, and hence r→0. Hence the beam will tend towards reflection backwards and forwards along the major axis.

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