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"). 


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!
