conic sectionsgeometryreflection
Here is my problem. There are two points P and Q outside an ellipse, where the coordinates of the P and Q are known. The shape of the ellipse is also known. A ray comming from point A is reflected by the ellipse and arrivates at Q. The question is how to determine the reflection point on the ellipse. I mean is there any analytical method to calculate the coordinate of the reflection point?
Reflection angle $\theta$ can be found from the sine rule:
$$ {r\over\sin\alpha}={d\over\sin\theta}, $$ where $d$ is the distance from the light source to the centre of the sphere, while $\alpha$ is the angle the incident ray makes with the line joining the source and the centre.
You can obtain the limiting value of $\alpha$ by setting $\sin\theta=1$ in the above formula.
I think one can simplify the computations using vectors. Let's start with a ray emanating from point $A$ on the ellipse and being reflected at another point $B$ on the ellipse. The reflected ray is then $BA'$, where $A'$ is the reflection of point $A$ about the normal $BN$.
If $F$ and $G$ are the foci, finding point $N$ on segment $FG$ is not difficult, because $BN$ is the bisector of $\angle FBG$, which entails $FN:GN=FB:GB$. Hence (remembering that $FB+GB=2a$, the major axis of the ellipse): $$ N=\left(1-{FB\over2a}\right)F+{FB\over2a}G. $$ We can then find $M$, the projection of $A$ on line $BN$, setting $M=B+t(B-N)$ and using that the scalar product $(B-M)\cdot(A-M)$ vanishes. From there we can get $A'=2M-A$. One finds: $$ A'=2B-A+2t(B-N), \quad\text{where:}\quad t=-{(B-A)\cdot(B-N)\over(B-N)\cdot(B-N)}. $$ Once we have $A'$ we can find the other intersection $C$ of line $BA'$ with the ellipse, and repeat the whole process to compute next reflection.
Best Answer
One possible solution can be as follows:
Let $\rm P{(x_1,y_1)} \ Q{(x_2,y_2)}, and\ R{(h,k)}, $ be the points.
Let ellipse ($\rm S$) be a special ellipse with centre (0,0)
$$\begin{align}\rm S &=\rm \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} = 1 \\ \rm Equation \ of \ tangent\ at\ R &:\rm \dfrac{xh}{a^2}+ \dfrac{yk}{b^2}-1 = 0 \ \ \ \ \ \ ... (i)\end{align}$$
Since the line segment $\rm PR$ (slope $\rm m_2$) is equally inclined to the tangent ( slope $\rm m_1$) as the segment $\rm QR$ ( slope $\rm m_3$), we can equalise the angles.
$$\rm\tan\theta=\dfrac{m_1 - m_2}{1+m_1m_2}$$
$$\rm\tan\phi=\dfrac{m_1 - m_3}{1+m_1m_3}$$
$$\rm \implies \tan\theta = \tan\phi \ \ \ \ \ \ ... (ii)$$
Find $\rm m_1, m_2, m_3$
You'll get $2$ equations with $2$ unknowns $\rm h$ and $\rm k$.
Note In my answer, I used eqn of standard ellipse with centre at origin. But this method will work for any ellipse. For finding tangent eqn, just put $T=0$
EDIT
Here is a graph which shows tangent, and a line inclined equally to it (it tangent is the angle bisector of the lines.) : Desmos