conic sectionsprojective-geometry
Let $S$ be the focus of the parabola and $AB$ be a focal chord. $C$ is a point lying on the parabola. Lines $CA$ and $CB$ intersect the directrix at points $P$ and $Q$. Prove that $\angle PSQ$ is $90^\circ$.
Since the directrix is the polar of $S$, the polar of $P$ must pass through $S$. So, it suffices to prove that the polar of $P$ passes through $Q$,which I was unable to prove, in particular I don't know how to use the fact about point $C$.
(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.
To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.
Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.
Let’s continue with the geometric construction that you started.
Construct the parallelogram $AOBC$. The diagonal $OC$ is parallel to the parabola’s axis. We know that perpendicular tangents meet on the directrix, which is perpendicular to the parabola’s axis, so we now know its directrix $d$. Construct circles centered on $A$ and $B$ and tangent to $d$. Their intersection is the parabola’s focus. You can simply divide $\overline{AB}$ proportionally to find this point: $F = {r_BA+r_AB\over r_B+r_A}$. You can then use the point-point and point-line distance formulas to obtain a Cartesian equation for the parabola.
The construction in the preceding paragraph took advantage of the perpendicularity of the two tangents, but it’s quite easy to obtain a Cartesian equation of a parabola from any pair of points $P_0$ and $P_2$ on it and the tangents at those points: Let $P_1$ be the intersection of the tangents. A Bézier parameterization of the parabola is then $(1-t)^2P_0+2t(1-t)P_1+t^2P_2$. Eliminating $t$ gets you a Cartesian equation for it. In this case, the tangents intersect at the origin, so we have the parameterization $$x = 3(1-t)^2+t^2 = 4t^2-6t+3 \\ y=3(1-t)^2-t^2 = 2t^2-6t+3.$$ Eliminating $t$ produces the equation $$x^2-4xy+4y^2-12x+6y+9=0.$$
Best Answer
The assertion works not just for a parabola but for any conic. From your comment, I assume that a projective explanation is what you want. You need two background lemmas.
Lemma 1: conjugate points on the directrix of a conic subtend an angle of $90 ^{\circ}$ at the focus.
(In fact the focus can be defined projectively as the point through which conjugate lines are perpendicular.)
Lemma 2: For a conic $c$, a line $s$, its pole $S$, a point $C$ on $c$, and a chord $AB$ that passes through $S$, the points $P,Q$ of intersection of the lines $CA$ and $CB$ with $s$ are conjugate.
If $s$ and $S$ are a directrix and its corresponding focus, then Lemma 2 says the points $P$ and $Q$ are conjugate, and Lemma 1 says they subtend a $90^{\circ}$ angle at $S$.
As for the lemmas, background, terminology and proofs can be found in Hatton's Projective Geometry, 1913. For Lemma 1, see pg 175. For Lemma 2, see pg 133. (Although the construction given on pg 133 is for the case of a circle, it can be used for any conic, and the proof works for any conic. Even if the proof used a special property of the circle, a circle can be projectively transformed into any conic and vice versa, which would prove the lemma for any conic. See also Article 97 on pg 179)