 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Simple properties of reflections in ellipse. (Posted on 2010-05-05) 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.)

 See The Solution Submitted by Vee-Liem Veefessional Rating: 4.6667 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Analytic Solution | Comment 1 of 5
`  Let (x_0,y_0) be an arbitrary point P on`
`                   x^2     y^2the outer ellipse ----- + ----- = 1.                    a^2     b^2`
`The incident, reflection, and tangent linesat 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 canbe tangent to the same ellipse or hyperbolaand they can pass through the foci.`
`If the incident line at (a,0) is horizontal,then it cannot be tangent to any ellipseor hyperbola, but it and the reflectionline both pass through both foci. If it isnot horizontal, then it cannot be tangentto any hyperbola or pass through eitherfoci, but it and the reflection line canbe 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 easilycalculated 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 ellipseand its reflection line either passthrough the foci or are tangent to thesame inner ellipse or hyperbola having thesame foci.`
`The vertical line through the center of theouter ellipse seems to be the only exception to this conclusion.`

 Posted by Bractals on 2010-05-05 11:37:38 Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (6)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information