Problem:
From any fixed point in the axis of a general conic, a line is drawn perpendicular to the tangent at $P$ (on the conic) meeting $SP$ in $R$ where $S$ is the focus. Show that the locus of $R$ is a circle.
I don't really know where to start from, I drew the diagram and now I am stuck. My attempt was to consider four points on conic and show that the four corresponding points $R$ are cyclic but I fail to do so.
On
(I will refer to the excellent drawing and notations in the solution by Aretino, my solution being in the same spirit).
It is a consequence of the following property of an ellipse (see proof below) :
The locus of the (variable) symmetrical point $T$ of one of the foci, say $S'$, with respect to a varying tangent is a circle $C$ centered in the other focus $F$ with radius $2a$ (this is rather well known by the "paper folding construction of an ellipse" : see http://kuzeemath.blogspot.com/2013/11/3-different-methods-to-conic-sections.html). The final circle (red circle in Aretino's drawing) is obtained by taking the homothetic image of circle $C$ using the homothety with center $F$ that sends $S'$ onto $D$ : we find back in this way the radius $2a\dfrac{SD}{2c}$ obtained by Aretino.
Proof : the above property is an immediate consequence of the bi-focal definition of an ellipse :
$$SM+MS'=2a \ \iff SM+MT=2a \ \iff \ ST=2a$$
if we take for $M$ the tangency point.
I'll show the proof for an ellipse, the other cases are analogous.
In an ellipse, the bisector $PH$ of $\angle SPS'$ (where $S'$ is the second focus) is perpendicular to the tangent at $P$. It follows that triangles $SPH$ and $SDR$ are similar ($D$ is the fixed point on axis $SS'$) and $SR:SP=SD:SH$, that is: $$ SR={SP\over SH}SD. $$ On the other hand, from the angle bisector theorem we have $SP:S'P=SH:S'H$, that is: $$ {SP\over SH}={SP+S'P\over SH+S'H}={a\over c}, $$ where $2a$ is the ellipse major axis and $2c=SS'$. Inserting this result into the previous equation we obtain that $SR$ is constant, hence $R$ belongs to a circle of centre $S$ and radius ${a\over c}SD$.