[Math] A right angle at the focus of a hyperbola

Question

$P$ is a point on a hyperbola. The tangent at $P$ cuts a directrix at point $Q$.
Prove that $PQ$ subtends a right angle to the focus $F$ corresponding to the directrix.

I have tried to use the general equation of the hyperbola and gradient method to show, but too many unknowns and I can't continue. I tried to show $m_1 m_2 = -1$, but I stuck halfway.

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Note (From @Blue). This property holds for all conics, except circles, which have no directrix. For ellipses and hyperbolas, the property holds for either focus-directrix pair. A proof incorporating this level of generality would be nice to see.

We can restate the property in a way that includes the circle as a limiting case:

$P$ is a point on a conic with focus $F$. The line perpendicular to $\overline{PF}$ at $F$ meets the tangent at $P$ in a point on the directrix corresponding to $F$; if $P$ is a vertex, then the perpendicular, tangent, and directrix are parallel, meeting at a point "at infinity". In the case of a circle, the perpendicular is parallel to the tangent (so that they "meet" in a point on a "directrix at infinity").

Answer 1

This is an euclidean solution to the parabolic case.
To adapt the proof to the hyperbolic case is left to the reader.

Lemma 1 (how to draw a tangent to a parabola). Let $A$ the projection of $P$ on the axis of the parabola and $B$ the symmetric of $A$ with respect to $V$. Then $PB$ is the tangent to the parabola at $P$.

In modern terms, this is just $\frac{d}{dx}x^2 = 2x$.

Lemma 2 (optical property of the parabola). If $C$ is the projection of $P$ on the directrix, the tangent at $P$ bisects the angle $\widehat{FPC}$.

Proof: Let $O$ be the intersection between the axis and the directrix. By Lemma 1 we have $PAF=COB$, so $PFBC$ is a parallelogram. Since $PF=PC$ by the definition of parabola, $PFBC$ is indeed a rhombus.

Corollary. Since $PF=PC$ and $PQ$ bisects $\widehat{FPC}$, $C$ and $F$ are symmetric with respect to $PQ$. It follows that $$ \widehat{PFQ}=\widehat{PCQ} $$ so $\widehat{PFQ}$ is a right angle.

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In the hyperbolic/elliptic case Lemma 2 and the next Corollary have to be replaced with: the tangent at $P$ is the internal/external angle bisector of $\widehat{F_1 P F_2}$, hence $PCQF$ is a cyclic quadrilateral and $\widehat{PFQ}=\widehat{PCQ}$ holds.