TITLE: The locus of the intersection point of two perpendicular tangents to a given ellipse
QUESTION [29 upvotes]: For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.

I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof.

REPLY [14 votes]: Please excuse my English that is probably not quite correct.
Here is a proof that I read somewhere (I am not the finder of it). It may seem long, but it is because of the way I explain it ; actually it is very simple. Is it "synthetic" as you wish ? You will judge yourself.
First of all, imagine a rigid plane that glides on a fixed plane, and consider, at a fixed time t, the vector field of the velocities, with respect to the fixed plane, of all the points of the moving plane. Since the moving plane is rigid, the vector field of the velocities must be of one of the following two types : (i) or it is a constant vector field, and that means that all the points of the moving plane have the same (vectorial) velocity at time t (case of an instant translation) ; (ii) or there is a point C, called the instant centre of rotation, such that the velocity at any point M is orthogonal to the vector CM (case of an instant rotation). Of course, the instant centre of rotation may move to different places during the movement. For a proof of these facts, see the appendix below.
Now draw your ellipse on a fixed plane, and draw a pair of orthogonal axes crossing at point A on some other rigid plane. Then put the second plane on the first one in such a way that the axes are tangent to the ellipse, and finally rotate this second plane on the fixed one while keeping the axes tangent to the ellipse. What we are looking for is the locus of point A during this movement.
Next fix a point P and a point Q on the axes such that, at a given fixed time t, these points coincide with the points of contact of the axes with the ellipse. My first claim is that at that very same time t, the velocities of points P and Q are tangent to the ellipse. Indeed, if they were not, at least one of these points would cross the ellipse at time t, and this would lead to a contradiction with the fact that the axes remain tangent to the ellipse. From this first claim, and since these velocities are orthogonal to each other, we may conclude that there is an instant centre of rotation C located at the intersection of the normals to the ellipse at points P and Q, or in other words, located at the fourth vertex of a rectangle with first three vertices at P, A and Q. It follows that the velocity of point A at that same time t is orthogonal to the diagonal CA of the rectangle (PAQC).
My second claim is that this diagonal CA passes through the centre O of the ellipse. Indeed, this follows from the fact that it passes through the midpoint of P and Q since (PAQC) is a rectangle, and from the well-known fact that when given two tangents (even non orthogonal) to an ellipse, then the intersection point of the two tangents, the midpoint of the two points of contact, and the centre of the ellipse, all lie on a same straight line : it is obvious by symmetry when the ellipse is a circle, and all these facts can be transported to any ellipse since they remain invariant under affine transformation. As a consequence of this second claim, we get that the velocity of point A at time t is orthogonal to the vector OA.
But this result holds for any fixed time t. Therefore, the velocity of point A always remains orthogonal to the vector OA, and it follows that the (square of the) distance OA remains constant during the movement since it has a zero derivative. And this proves that point A moves along a circle centred at O.
The end.
Appendix : the vector field of the velocities of a rigid moving plane.
Let us denote by M' the velocity of point M at some fixed time t. If A and M are two points of the rigid moving plane, by differentiating the square of the constant distance AM we get that M'-A' is orthogonal to the vector AM. Let us mention two consequences of this : (i) if there is a point C such that C'=0, then M' is orthogonal to the vector CM for all points M of the moving plane (case of an instant rotation) ; (ii) if A is such that A'$≠$0, then for all points M lying on the line through A orthogonal to A', the velocities M' are parallel to A' (ie they are linearly dependent).
Then we may conclude as follows : (i) if there are two points A and B such that A' and B' are not parallel (ie are linearly independent), the lines through A orthogonal to A' and through B orthogonal to B' cross each other at some point C where the velocity must be parallel to both A' and B', so that C'$=0$, and then we have an instant rotation as proved above ; (ii) and if (otherwise) the velocities are all parallel to some direction A'≠0, then at any point M such that AM is not orthogonal to A' the velocity has to be M'=A', and it is not difficult to see that this implies that actually M'=A' for all points M of the moving plane, that is, then we have an instant translation.