Non-brute-force proof of parabola tangent property I'm working through a classic Calculus book (Morris Kline), and one of the problems is:

Prove that the foot of the perpendicular from the focus to any tangent
  of a parabola lies on the tangent to the vertex.

Basically, it's saying that if the parabola $y=x^2$ has a tangent line T at point $P$, and we draw a line $L$ perpendicular to $T$ that goes through the focus $F$, then $T$ meets $L$ at a point where $y=0$.
I managed to prove this by:


*

*Calculating the equation of $T$ (based on slope of $y'$ and point $P$)

*Calculating the equation of $L$ (based on slope of $\frac{-1}{y'}$ and point $F$)

*Putting $y=0$ in both equations and solving for $x$

*Observe that they both have the same $x$ at $y=0$ and therefore they meet on the $x$ axis.


HOWEVER, this seems like a very brute-force approach.  It's almost like I cheated or I simulated it on the computer with $1000$ points and determined that it is so.  I don't really understand WHY these $2$ lines must meet in this manner.
Is there any sort of geometric or intuitive proof?
 A: Suppose we have a parabola $P$ with focus $F$ ad directrix $l$ (so that $P$ is the set of points equidistant to $F$ and $l$), and let $A$ be a point on the parabola, and $A'$ be the projection of $A$ onto $l$.
Notice that the bisector of the interval $FA'$ intersects $P$ at $A$ (by the definition, as the distance from $A$ to $A'$ is the distance from $A$ to $l$), and it doesn't intersect $P$ at any other point: if it did, we would have, between the two intersection points, another point which is closer to $A'$ (and therefore to $l$) than to $F$. We conclude that the bisector is, in fact, tangent to $P$ at $A$: the only lines which intersect the parabola at exactly one point are the tangents and the vertical lines, and clearly the bisector isn't vertical. How you would prove it depends on how you define the tangent, exactly, but I think the best way would be to show that for any line through $P$ with slope distinct from vertical and tangential, you can find a secant with that slope.
Now, consider the triangle $FAA'$, and denote by $B$ the midpoint of $FA'$. Notice that because $FAA'$ is isosceles, $FB$ is perpendicular to $BA$, which is the tangent to $P$ at $A$, so it's enough to show that $B$ lies on the tangent to vertex of $P$.
To see that last part, notice that if you draw a line $l'$ parallel to $l$ through $B$, you get a line whose distance from $F$ is half of the distance from $F$ to $l'$ (by the intercept theorem, because $B$ is the bisetor of $FA'$ and $A'$ lies on $l$), so $l'$ is just the tangent to $P$ at vertex.

