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Here is the Ellipse Image

I know the length of the Major axis and the Minor axis of the ellipse. I know the $x$ and $y$ coordinates of random points. We know the $x,y$ coordinates at the point where the tangent meets the orbit of the ellipse.

At any point, when I shoot light toward the tangent, it reflects and touches the other point of the ellipse.

Is there a way to know the x,y coordinates of that point?

Amour Math
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2 Answers2

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This is really a long comment. You need to do a few things, or at least, I would, to solve your problem. Try to do these things and then let us know where you get stuck. None are particularly challenging, but require some basic calculus.

  1. Transform your ellipse to be centered at the origin with the foci on the $x$-axis. This will give you an equation for your ellipse in the form $\frac{x^2}{a}+\frac{y^2}{b} = 1$.
  2. Find the slope of the tangent line, $m$, by using implicit derivatives, and then the slope of the perpendicular line to this will be $\frac{-1}{m}$. Finding the point on the line closest to your starting point can be done by minimizing the distance between your random point and the perpendicular line.
  3. Find the point that is the reflection of the random point you start with over the line through the point where the light hits the ellipse with slope $\frac{-1}{m}$.
  4. This lets you easily find the equation of the line through this reflected point and your point on the ellipse. Then find where this line intersects your ellipse. (One of the points will be your point where the light initially hits it.)
  5. Transform your point back to your original coordinates.
N. Owad
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Let the ellipse be in the standard orientation $(x/a)^2+(y/b)^2=1$. This defines the coordinate system in use. Let the random point be $(x_r,y_r)$. Tangents to the ellipse are defined by the slope/derivative $2x/a^2+2yy'/b^2=0$, so $yy'/b^2=-x/a^2$, so $y'=-xb^2/(ya^2)$. Let the tangent point be $(x_t,y_t)$, so the tangent line that runs through the tangent point is $(x,y)= (x_t,y_t)+\alpha (1,-x_tb^2/(y_ta^2))$, where $\alpha$ is a parameter along the points of the tangent. The normal to the tangent has a slope which is -1 divided by the slope of the tangent, so it has the direction $(1,y_ta^2/(x_tb^2))$. The vector from $(x_r,y_r)$ to the tangent point is $(x_t-x_r,y_t-y_r)$. It is decomposed to a component parallel and a component perpendicular to the tangent line: $$ \left( \begin{array}{c}x_t-x_r\\ y_t-y_r\end{array}\right) = \beta \left(\begin{array}{c}1 \\ -x_tb^2/(y_t a^2)\end{array}\right) + \gamma \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right) $$ Dot products with the direction vectors give due to the orthogonality $$ \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right) \left( \begin{array}{c}x_t-x_r\\ y_t-y_r\end{array}\right) = \gamma \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right) \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right); $$ $$ x_t-x_r+ \frac{y_ta^2}{x_tb^2}(y_t-y_r) = \gamma (1+(y_ta^2/(x_tb^2)^2); $$ So the factor for the component along the normal of the tangent is known: $$ \gamma=\frac{ x_t-x_r+ \frac{y_ta^2}{x_tb^2}(y_t-y_r)} { 1+[y_ta^2/(x_tb^2)]^2}. $$ The line starting at $(x_t,y_t)$ after the reflection is obtained by flipping the component of the vector before the reflection that is orthogonal. $$ \beta \left(\begin{array}{c}1 \\ -x_tb^2/(y_t a^2)\end{array}\right) - \gamma \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right) = \left( \begin{array}{c}x_t-x_r\\ y_t-y_r\end{array}\right) -2 \gamma \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right). $$ Let $(x_h,y_h)$ with $(x_h/a)^2+(y_h/b)^2=1$ be the point at the next collision with the elliptic billiard wall. so the vector emerging from $(x_t,y_t)$ is $$ \left( \begin{array}{c} x_t \\ y_t \end{array} \right) +\delta [ \left( \begin{array}{c}x_t-x_r\\ y_t-y_r\end{array}\right) -2 \gamma \left(\begin{array}{c}1 \\ y_t a^2/(x_tb^2)\end{array}\right) ] = \left( \begin{array}{c} x_h\\ y_h \end{array} \right) $$ The task is to find the parameter $\delta$ which encodes the distance from $(x_t,y_t)$ to $(x_h,y_h)$ along the (non-normalized) direction. \begin{eqnarray} x_t+\delta[x_r-x_r)-2\gamma] &=& x_h\\ y_t+\delta[y_t-y_r)-2\gamma y_ta^2/(x_tb^2)] &=& y_h\\ \end{eqnarray} Inserting these into $(x_h/a)^2+(y_h/b)^2=1$ gives a quadratic equation for $\delta$, and the solutions which is $\neq 0$ is to be used to find the next collision point.

R. J. Mathar
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