The reflective property of ellipses I have following question to proof:
An ellipse is revolved about its major axis to generate an ellipsoid. The inner surface of the ellipsoid is silvered to make a mirror. Show that a ray of light emanating from one focus will be reflected to the other focus. Sound waves also follow such paths, and this property is used in constructing "whispering galleries." (Hint: Place the ellipse in the standard position in the xy-plane and show that the lines from a point P on the ellipse to the two foci make congruent angles with the tangent to the ellipse at P.)
And I don't know how to prove the hint. How to prove the hint? Thanks.
 A: Figure 1: Start with a line $l$ and and two arbitrary points $A$ and $B$ on the same side of the line.

Figure 2: Suppose line $l$ is actually a mirror.  By Fermat's principle we know that light always follows the shortest path.  Therefore of all the possible paths from $A$ to some arbitrary point $X$ along $l$, to $B$, the one that light will follow is that which minimizes distance $AXB$

Figure 3: To find the path minimizing $AXB$, reflect point $A$ about line $l$ (call this reflection $A'$).  Note that by construction $\triangle AXA'$ is an isosceles triangle.  It follows that distance $AXB$ equals distance $A'XB$.  Since this is true, and since the shortest distance between two points is a straight line, distance $AXB$ is minimized when point $X$ takes the position along line $l$ such that $A'XB$ is a straight line.  This position is indicated in figure 3 by point $C$.
Since $\triangle ACA'$ is by construction isosceles, it is not hard to see that (by the properties of isosceles triangles and interior opposite angles) the angles indicated in figure 3 are equal.  That is, the angle of the light ray's incidence equals the angle of its reflection off the mirror.  This what is referred to as the reflective or optical or specular property of light rays.

Figure 4: Finally, to see that the ellipse also has the reflective property, suppose you are told that line $l$ is tangent to an ellipse with focii $A$ and $B$, but you are not given the point where $l$ actually touches the ellipse.  To find this point of tangency the process is the same as for finding the distance-minimizing path of light.  Clearly (by the definition of tangents), for any arbitrary point $X$ along $l$, with $X \neq C$, the inequality $AXB>ACB$ holds.  Thus it is evident that the distance $AXB$ will be minimized at the point of tangency $C$.  We have just seen that this is exactly the optical or reflective property of light.  Therefore the indicated angles are equal and any ray shot from one focus to the ellipse perimeter will end up bouncing to the other focus.

A: Consider the following ellipse with focus $ S_1,S_2.$
Let P($a $ $cos θ $ , $b $ $sin θ $) be any point on the ellipse.

We know that $$PS_1=a(1+ecos θ )$$ $$  PS_2=a(1-ecos θ )$$.
Let the normal through $P$ meet the x axis at $Q$.
Then the  equation of the normal $PQ$ is $$\frac{ax}{cos θ}-\frac{by}{sin θ}=a^2-b^2$$
Hence the coordinate of Q is $(\frac{(a^2-b^2)cos θ}{a},0)$.
Findingthe lenth of $S_1Q$ and $S_2Q$we get $$S_1Q=ae+ae^2cos θ $$ $$S_2Q=ae-ae^2cos θ $$.
Now we get  $$\frac{S_1Q}{S_2Q}=\frac{ae+ae^2cos θ}{ae-ae^2cos θ}=\frac{a(1+ecos θ )}{a(1-ecos θ )}=\frac{PS_1}{PS_2}$$
$$⇒\frac{S_1Q}{S_2Q}=\frac{PS_1}{PS_2}$$
Hence by converse of angle bisector theorem the normal $PQ$ is the angle bisector of $∠P$.
So a ray of light emanating from one focus will be reflected to the other focus.
