TITLE: A right angle at the focus of a hyperbola
QUESTION [5 upvotes]: $P$ 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").

REPLY [3 votes]: 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 $P$. $\square$