Prove: The chord joining the points of contact of the tangent lines to a parabola from any point on its directrix passes through its focus. Prove: The chord joining the points of contact of the tangent lines to a parabola from any point on its directrix passes through its focus.
For this chapter/topic, I somehow need to show the points of contact between the tangent lines that intersect at some point $(x_1,-p)$ on the directrix, and then show that the equation of the chord between these two points passes through the focus $(0,p)$.
If I use the parabola $4y=x2$, and take the derivative $y'=\frac{x}{2}$, I get the tangent line equation $(y-y_0)=\frac{x_0}{2}(x-x_0)$. I can't figure out how to use point of intersection  on the directrix $(x_1,-p)$ and the equation of the tangent line to find the two points of tangency on the parabola.
 A: A useful formula can be derived for the equation of the chord joining the two points of contact of tangents drawn from an external point $(x_{0}, y_{0})$ (also known as the chord of contact). Let $(x_{0}, y_{0})$ be a point external to the parabola, and let $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ be the points where the tangents drawn from this point meet the parabola $x^{2} = 4py$.
Differentiating and using the fact that $x_{1}^2= 4py_1$, we find that the equation of the tangent at $(x_{1}, y_{1})$ is
\begin{equation*}
2py = xx_{1} - 2py_{1}.
\end{equation*}
The equation of the tangent at $(x_{2}, y_{2})$ is similar. Since $(x_0, y_0)$ lies on both of these tangent lines, we have
\begin{equation}
2py_0 = x_0x_{1} - 2py_{1} \tag*{(1)}
\end{equation}
and
\begin{equation*}
2py_0 = x_0x_{2} - 2py_{2}. \tag*{(2)}
\end{equation*}
But looking closely at $(1)$ and $(2)$, we see that $(x_{1}, y_{1})$ and $(x_{2}, y_2)$ both lie on the line with equation
\begin{equation}
2py_{0} = xx_{0} - 2py.
\end{equation}
This is the general equation for the chord of contact of tangents from $(x_{0}, y_{0})$.
In the special case where the external point lies on the directrix, $y_0 = -p$, so the equation becomes
\begin{equation*}
xx_0 =2p(y-p).
\end{equation*}
We then see that the focus $(0,p)$ lies on the chord of contact in this case.
A: Although I'd prefer a vector notation of a line $\mathbf{x}=t\mathbf{l}+\mathbf{a}$ or $\mathbf{n}.(\mathbf{x}-\mathbf{a})=0$, the slopes approach works fine for this question, because we have no lines of slope $0$ or $\infty$.
First, OP didn't get the parabola equation right, it should be $4py=x^2$ (see the wiki).
If touching points are $(x_0,y_0)$, the slopes are $4py'=2x$ i.e. $y'(x_0)=\frac{x_0}{2p}$.
Next we consider pencil of lines passing through $(x_1,-p)$ with slope $\frac{x_0}{2p}$: $$\frac{x_0}{2p}(x-x_1)=y+p$$ and $(x_0,y_0)=(x_0,\frac{x_0^2}{4p})$ should fit into it because the tangent line passes through $(x_0,y_0)$:
$$\frac{x_0}{2p}(x_0-x_1)=\frac{x_0^2}{4p}+p$$
$$2x_0(x_0-x_1)=x_0^2+4p^2$$
$$x_0^2-2x_0x_1+x_1^2=x_1^2+4p^2$$
$$(x_0-x_1)^2=x_1^2+4p^2$$
$$x_0-x_1=\pm\sqrt{x_1^2+4p^2}$$
$$x_0=x_1\pm\sqrt{x_1^2+4p^2}$$
These are the $x$-coordinates of the touching points.
The rest of the proof I'd leave to WA or it could be done by hands, rather lengthy.
