Lines tangent to a parabola

Let's prove, first of all, that line $IP$ bisects $AB$. Let $C$ and $D$ be the feet of the perpendicular lines drawn from $A$ and $B$ to the directrix, and $S$ be the focus. Tangent $AI$ is the bisector of $\angle CAS$ and $AC=AS$, hence $IC=IS$; by the same reasoning we also get $ID=IS$. If follows that $ICD$ is an isosceles triangle and line $IP$, perpendicular to $CD$, bisects $CD$. By the intercept theorem line $IP$ also bisects $AB$.

Let now the tangent at $P$ meet $AI$ and $BI$ at $E$ and $F$ respectively, and let line $EM$, parallel to the axis, meet $AP$ at $M$. By the above argument, $M$ is the midpoint of $AP$ and, as a consequence, $E$ is the midpoint of $AI$. In an analogous way one proves that $F$ is the midpoint of $BI$, and it follows then from the intercept theorem that $EF$ is parallel to $AB$, as it was to be proved.

This is a projective fact. You can transform your parabola by means of a projective transformation into a circle. Consider a circle and a chord $A'B'$ in it. Consider the diameter conjugate to the direction of $A'B'$, that is, perpendicular to $A'B'$. The tangent to the circle at the point $I'$ of intersection of that diameter with the circle is obviously parallel to $A'B'$. Now there is a unique projective transformation taking the circle into your parabola, namely $A'\to A$, $B'\to B$, $I'$ to $I$ and the opposite point of the diameter of the circle $J'$ into a point at infinity. The pole $P'$ of $A'B'$ becomes the pole $P$ of $AB$ (the pole-polar relation is a projective notion). The circle becomes your parabola and the fact that your tangent at $P$ is parallel to $AB$ corresponds to the obvious parallelism mentioned before.

Actually, you can "see" this fact, because the above mentioned projective transformation is a simple perspective: look at the circle from a point $O$ not on its plane. Consider two rays of light through two opposite ends of the diameter $IJ$ and take as your "drawing" plane one that intersects $OI$ at some finite point but is parallel to $OJ$. The image of the circle on that plane is a parabola. Now rotating the drawing plane conveniently you can make $A$ the image of $A'$ and $B$ that of $B'$ for some $A'B'$ perpendicular to $I'J'$. In that perspective representation of the circle, your tangent at $P$ meets $AB$ on the horizon.

The proofs provided are all correct of course but they invoke unnecessary metric aspects, foreign to the nature of the fact. Analytic proofs lack the beautiful simplicity of the synthetic ones.

Needless to say, the property holds for any conic, all of them are "the conic" projectively speaking.

I consider a parabola $y=ax^2$ with $a \in R$ (other can be obtained by traslation). Let $S$ and $R$ the tangent lines to the parabola at the point $A(x_A,y_A)$ and $B(x_B,y_B)$. The $m$ of the tangent is $m=2ax_0+b$ and so their intersection is: $$x=\frac{x_A+x_B}{2}$$

Now the line throught $A$ and $B$: $m=\frac{ax_B^2-ax_A^2}{x_B-x_A}=a(x_B-x_A)$ and the tangent to the parabola throught $x=\frac{x_A+x_B}{2}$ has: $$m=2\cdot \frac{x_A+x_B}{2}=x_A+x_B$$ So, as you asked in the post, the two lones are parallel.