A right angle at the focus of a hyperbola 
$P$ 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. 

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"). 

 A: 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.


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.

A: Here is a solution for parabola. You can use the similar way for hyperbola. 
Let $y^2=2px$ be an equation of our parabola, $P(x_1,y_1)$.
Hence, $F\left(\frac{p}{2},0\right)$ and $x=-\frac{p}{2}$ is an equation of the directrix. 
If $x_1=\frac{p}{2}$ then since $yy_1=p(x+x_1)$ is an equation of the tangent,
$Q\left(-\frac{p}{2},0\right)$ and $\measuredangle QFP=90^{\circ}$.
Let $x_1\neq\frac{p}{2}$.
Hence, we can calculate slopes: $$m_{PF}=\frac{y_1}{x_1-\frac{p}{2}}.$$
$x_Q=-\frac{p}{2}$. 
Thus, $yy_Q=p\left(-\frac{p}{2}+x_1\right)$, which gives $Q\left(-\frac{p}{2},\frac{p\left(x_1-\frac{p}{2}\right)}{y_1}\right)$ and
$$m_{FQ}=\frac{\frac{p\left(x_1-\frac{p}{2}\right)}{y_1}-0}{-\frac{p}{2}-\frac{p}{2}}=\frac{\frac{p}{2}-x_1}{y_1}$$
and since
$$m_{PF}\cdot m_{FQ}=\frac{y_1}{x_1-\frac{p}{2}}\cdot\frac{\frac{p}{2}-x_1}{y_1}=-1,$$
we are done!
For hyperbola we obtain:
Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be an equation of our hyperbola, $P(x_1,y_1)$.
Hence, $F\left(\sqrt{a^2+b^2},0\right)$ and $x=\frac{a^2}{\sqrt{a^2+b^2}}$ is an equation of the directrix. 
If $x_1=\sqrt{a^2+b^2}$ then since $\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1$ is an equation of the tangent,
$Q\left(\frac{a^2}{\sqrt{a^2+b^2}},0\right)$ and $\measuredangle QFP=90^{\circ}$.
Let $x_1\neq\sqrt{a^2+b^2}$.
Hence, we can calculate slopes: $$m_{PF}=\frac{y_1}{x_1-\sqrt{a^2+b^2}}.$$
$x_Q=\frac{a^2}{\sqrt{a^2+b^2}}$. 
Thus, $\frac{a^2}{\sqrt{a^2+b^2}}\cdot\frac{x_1}{a^2}-\frac{y_Qy_1}{b^2}=1$, which gives $Q\left(\frac{a^2}{\sqrt{a^2+b^2}},\frac{b^2}{y_1}\left(\frac{x_1}{\sqrt{a^2+b^2}}-1\right)\right)$ and
$$m_{FQ}=\frac{\frac{b^2}{y_1}\left(\frac{x_1}{\sqrt{a^2+b^2}}-1\right)-0}{\frac{a^2}{\sqrt{a^2+b^2}}-\sqrt{a^2+b^2}}=\frac{\frac{b^2}{y_1}\left(x_1-\sqrt{a^2+b^2}\right)}{a^2-(a^2+b^2)}=\frac{x_1-\sqrt{a^2+b^2}}{-y_1}$$
and since
$$m_{PF}\cdot m_{FQ}=-1,$$
we are done!
A: Here's an analytic solution for the general case, using the alternate formulation in my edit to the original question. (We'll ignore the obvious case of circles, which have eccentricity $0$.)

Recall that the focus-directrix definition of a conic with eccentricity $e \neq 0$ is the locus of points such that
$$\text{distance from focus} = \text{eccentricity} \cdot \text{(distance from directrix)} \tag{1}$$
Taking a focus $F$ at the origin, and corresponding directrix at $x=-d$, we can square $(1)$ to get this equation for the locus:
$$x^2 + y^2 = e^2 ( x + d )^2 \tag{2}$$
Let $P=(p,q)$, with $q\neq0$, be a (non-vertex) point on the conic. If you "know" that the equation for the tangent at $P$ is 
$$x p + y q = e^2 ( x + d ) ( p + d ) \tag{3}$$
(see @amd's comment below) then we're practically done. After all, the line perpendicular to $\overline{PF}$ at $F$ is 
$$x p + y q = 0 \tag{4}$$
which matches the left-hand side of $(3)$. Consequently, the intersection of these two lines must have $x=-d$ (to simultaneously zero-out the right-hand side of $(3)$). That is, the point of intersection lies on the directrix, which completes the proof. $\square$

A less-insightful approach might note that the perpendicular $(4)$ meets the directrix at
$$Q = \frac{d}{q} \left( -q, p \right) \tag{5}$$
Then, $\overleftrightarrow{PQ}$ has this equation
$$ x ( q^2 - d p ) - q y ( p+d ) = - d ( p^2 + q^2 ) \tag{6}$$
Since $P$ satisfies $(2)$, we can rewrite $(5)$ by replacing occurrences of $q^2$ with $e^2(p+d)^2-p^2$. Dividing-through by $p+d$ gives
$$x ( e^2(p+d) - p ) - q y = - d e^2 ( p+d ) \quad\to\quad
x p + y q = e^2(x+d)(p+d)
\tag{7}$$
Not recognizing $(7)$ as the equation of the tangent line, one may substitute $y$ from $(7)$ into $(2)$ to get an equation for the $x$-coordinate(s) of the point(s) of intersection; since that equation reduces to
$$(x-p)^2 = 0 \tag{8}$$
we see that the points of intersection coincide; that is, $\overleftrightarrow{PQ}$ is indeed tangent to the conic at $P$. $\square$
A: Consider the hyperbola $\hspace{3cm}\quad\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ 
which has directrix $\hspace{4cm}\quad x=\pm \dfrac {a^2}c$   
and focus  $\hspace{5cm}\qquad F(\pm c,0)$   
where $\hspace{6cm}\quad c=\sqrt{a^2+b^2}$.   
Consider wlog only the right branch of the the hyperbola. 
Let a general point on the hyperbola be $\qquad P(a \sec u, b\tan u)$. 
Differentiating wrt u and diving gives $\qquad \dfrac {dy}{dx}=\dfrac b{a\sin u}$.
Tangent at $P$: $\qquad\qquad\qquad\qquad\qquad\;\; y=\dfrac b{\sin u}\bigg(\dfrac xa-\cos u\bigg)$
Intecept, $Q$, at directrix: $\qquad\qquad\qquad\bigg(\dfrac {a^2}c, \dfrac b{\sin u}\bigg(\dfrac ac-\cos u\bigg)\bigg)$
Slope of $PF$, $m_1$:$\hspace{5cm}\dfrac {b\sin u}{a-c\cos u}$
Slope of $QF$, $m_2$:$\hspace{5cm}\dfrac {\frac{b}{\sin u}(\frac ac-\cos u)}{\frac {a^2}c-c}=\dfrac {c\cos u-a}{b\sin u}$
As $$m_1\cdot m_2=-1$$ hence $PF\perp QF$, i.e. $\angle PFQ$ is a right angle. $\blacksquare$
See desmos implementation here.

