Given an ellipse defined by its foci and major axis.
Prove that any line inside the ellipse not passing through the line segment joining the two foci when reflected off the ellipse would be tangent to the same inner ellipse as the initial line, with the same pair of foci.
Accordingly, if the line were to originally pass through the line segment joining the foci, the reflection on the ellipse would be tangent to the same hyperbola as the initial line, with the same pair of foci as the original ellipse.
The other possibility would be if the line started of from one focus, its reflection then passes through the other focus. (This degenerate case is known as the optical property of the ellipse, whose proof is much simpler.)
Let (x_0,y_0) be an arbitrary point P on
x^2 y^2
the outer ellipse  +  = 1.
a^2 b^2
The incident, reflection, and tangent lines
at the point P respectively are,
y = m_i*(x  x0) + y_0,
y = m_r*(x  x0) + y_0, and
y = m_t*(x  x0) + y_0.
WOLOG let x_0,y_0 >= 0.
At x_0 = 0 or y_0 = 0, m_r = m_i.
If the incident line at (0,b) is vertical,
then it cannot be tangent to any ellipse
or hyperbola and it cannot pass through
either foci. If it is not vertical, then
by symmetry it and the reflection line can
be tangent to the same ellipse or hyperbola
and they can pass through the foci.
If the incident line at (a,0) is horizontal,
then it cannot be tangent to any ellipse
or hyperbola, but it and the reflection
line both pass through both foci. If it is
not horizontal, then it cannot be tangent
to any hyperbola or pass through either
foci, but it and the reflection line can
be tangent to the same ellipse.
From now on we will assume x_0,y_0 > 0.
The slope m_t at the point P is easily
calculated to be,
b^2*x_0
m_t = . (1)
a^2*y_0
If the incident line at point P is vertical,
then,
m_t^2  1
m_r =  (2a)
2*m_t
otherwise,
m_i + m_r 2*m_t
 = . (2b)
1  m_i*m_r 1  m_t^2
CASE I: THE INNER ELLIPSE.
x^2 y^2
 +  = 1, (3)
c^2 d^2
where c^2  d^2 = a^2  b^2 since they
both have the same foci.
To find the slopes m_i and m_r we
plugin m*(x  x_0) + y_0 for y in
equation (3) and we get,
(d^2 + c^2*m^2)*x^2 
2*c^2*m*(m*x_0  y_0)*x +
c^2*[(m*x_0  y_0)^2  d^2] = 0.
Solving for x, the radicand of the
quadratic formula is,
4*c^4*m^2*(m*x_0  y_0)^2 
4*(d^2 + c^2*m^2)*c^2*
[(m*x_0  y_0)^2  d^2]
Setting this to zero we get,
(x_0^2  c^2)*m^2 
2*x_0*y_0*m +
(y_0^2  d^2) = 0
If the incident line is vertical
( x_0 = c ) we get
y_0^2  d^2
m_r = 
2*c*y_0
This slope satisfies equation (2a).
If the incident line is not vertical,
then the slopes m_i and m_r are given
by
x_0*y_0 + sqrt(R)
, where
x_0^2  c^2
R = d^2*x_0^2 + c^2*y_0^2  c^2*d^2.
R > 0 if (x_0,y_0) lies outside the
inner ellipse (always).
These two slopes satisfy equation (2b).
CASE II: THE HYPERBOLA.
x^2 y^2
   = 1, (4)
c^2 d^2
where c^2 + d^2 = a^2  b^2 since they
both have the same foci.
To find the slopes m_i and m_r we
plugin m*(x  x_0) + y_0 for y in
equation (4) and we get,
(d^2  c^2*m^2)*x^2 +
2*c^2*m*(m*x_0  y_0)*x 
c^2*[(m*x_0  y_0)^2 + d^2] = 0.
Solving for x, the radicand of the
quadratic formula is,
4*c^4*m^2*(m*x_0  y_0)^2 +
4*(d^2  c^2*m^2)*c^2*
[(m*x_0  y_0)^2 + d^2]
Setting this to zero we get,
(x_0^2  c^2)*m^2 
2*x_0*y_0*m +
(y_0^2 + d^2) = 0
If the incident line is vertical
( x_0 = c ) we get
y_0^2 + d^2
m_r = 
2*c*y_0
This slope satisfies equation (2a).
If the incident line is not vertical,
then the slopes m_i and m_r are given
by
x_0*y_0 + sqrt(R)
, where
x_0^2  c^2
R = c^2*d^2  d^2*x_0^2 + c^2*y_0^2.
R > 0 if (x_0,y_0) lies between the
branches of the hyperbola.
These two slopes satisfy equation (2b).
CASE III: THE LINES PASS THROUGH THE FOCI.
If the incident line is vertical
( x_0 = f ) we get
y_0
m_r = 
2*f
This slope satisfies equation (2a).
If the incident line is not vertical,
then the slopes m_i and m_r are given
by
y_0
, where
x_0 + f
f^2 = a^2  b^2.
These two slopes satisfy equation (2b).
CONCLUSION.
Every incident line to the outer ellipse
and its reflection line either pass
through the foci or are tangent to the
same inner ellipse or hyperbola having the
same foci.
The vertical line through the center of the
outer ellipse seems to be the only
exception to this conclusion.

Posted by Bractals
on 20100505 11:37:38 